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Thermodynamic Basic
UNESCO-NIGERIA TECHNICAL & VOCATIONAL
EDUCATION REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
THERMODYNAMICS II
COURSE CODE: MEC213
YEAR II- SEMESTER III
THEORY
Version 1: December 2008
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MECHANICAL ENGINEERING TECHNOLOGY
THERMODYNAMICS II (MEC 213)
Table of contents
WEEK 1:
1.1.0: THERMAL EFFICIENCY
1.1 Definition And Units
1.2: Mechanical Efficiency (η m ) of an engine
1.3: Thermal Efficiency (ηm ) of an engine
1.4: Work Examples
WEEK 2:
2.0: MEASUREMENT OF INDICATED POWER
2.1
I.P of Multi cylinder Engines
2.2
Worked Examples
2.3 Energy balance account for an I.C. engine regarded as
operating in a closed system:
2.4 Worked Examples
WEEK 3:
3.0 REVERSIBILITY AND IRREVESIBILIT OF
THERMODYAMIC CYCLES/PROCESSES:
3.1.
A Reversible cycle
3.2.
An Irreversible cycle
3.3
Second Law of Thermodynamics
3.3.1 Clauses version of 2nd law of thermodynamics:
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WEEK 4:
WEEK 5:
WEEK 6:
Thermodynamic Basic
nd
3.3.2
Lord Kelvin’s version 2
3.4
Definition of heat engines
3.21
Element of a heat engine cycle
3.3
Deduction from the 2
3.5
Element of a heat engine cycle
3.6.
Deduction from the 2
nd
nd
law of thermodynamics:
Law of thermodynamics
Law of thermodynamics
4.0
THE CARNOT CYCLE EFFICIENCY
4.1
Carnot Principle
4.2
Thermal Efficiency of a Carnot cycle Engine
4.3
Worked Examples
4.3
Exercises and Solutions
5.0
THE SUN SOLAR ENERGY
5.1
Solar Energy -- Energy from the Sun
5.2
Photovoltaic Energy
5.3
Some advantages of photovoltaic systems
5.4
Solar Thermal Heat
5.5
Energy from the Sun
6.0
APPLICATION OF SOLAR ENERGY
6.1
Architecture and urban planning
6.1.1
6.2
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Agriculture and horticulture
Solar lighting
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WEEK 7:
Thermodynamic Basic
6.3
Solar thermal
6.3.1
Heating, cooling and ventilation
6.4.
Water treatment
6.5
Cooking
6.6
Electrical generation
6.6
Concentrating solar power
6.7
Experimental solar power
6.8
6.9
Solar vehicles
Energy storage methods
6.10
Development, deployment and economics
7.0
ENTROPY
7.1
Property Diagrams Involving Entropy
7.1.1
The T-S Diagram
7.2
Determination of Dryness Fraction from Area in T-S
Diagram
WEEK 8:
7.3
For a Perfect Gas
7.4
The h-S Diagram
8.0 WORKED EXAMPLES
8.1 Isentropic Efficiency
8.2 Isentropic Efficiency of Turbine
8.4 Worked Examples
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8.5
WEEK 9:
QUIZ ONE (ENTROPY)
9.0 PURE SUBSTANCES
9.1 Properties and State
9.2 Property Diagram for Phase-Change Processes
9.2.1 P-V Diagram for Pure Substance
9.3
WEEK 10:
WEEK 11:
Two property rules for pure substances
10.0 Pressure Law
10.1
Ideal Gas Law
10.2
Worked Examples
11.0 IDEAL AND REAL GASES
11.1 Worked Examples
WEEK 12:
WEEK 13:
11.2
Exothermic and Endothermic Reactions
11.3
QUIZ (THREE)
12.0 FUELS AND COMBUSTION
12.1
Classification of Fuels into Solid, Liquid and Gaseous
12.2
Combustion Equations
12.3
QUIZ( ONE)
13.0 STOICHIOMETRIC AIR-FUEL RATIO
13.1
Worked Examples
13.2
QUIZ (TWO)
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WEEK 14:
Thermodynamic Basic
14.0 CALORIFIC VALUE OF FUELS
14.0.1 Gross Calorific Value at Constant Volume (Qgross,v)
14.0.2 Net Calorific Value at Constant Volume (Qnet, v)
14.0.3 Gross Calorific Value at Constant Pressure (Qgross,p)
14.0.4 Net Calorific Value at Constant Pressure (Qnet,p)
WEEK 15:
14.1
Practical Determination of Calorific Value
14.2
Solid Fuel – using the bomb calorimeter
14.3
Gas Calorimeter
14.4
Worked Examples
15.0 DENSITY OF GAS MIXTURE
15.1 Atmospheric and Ecological Pollution
15.2 Carbon monoxide (CO)
15.3 Carbon dioxide (CO2)
15.4 Sulphuric dioxide (SO2)
15.5 Oxides of Nitrogen (NO)
15.6
Volatile Organic Compounds (VOCs)
15.7 Photochemical Smog
15.8 Particulates
15.9
Ecological Considerations
15.10 Ecological Considerations
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WEEK ONE
1.0 THERMAL EFFICIENCY:
1.1 Definition and Units:
Power: power is defined as the rate of doing work, and a work rate of one joule per second is named the
WATT.
The Brake power (b.p.): of an engine is the power available for work at the output shaft of the engine,
and is measured using some form of brake. It also means useful work output of an engine. The indicated
power (i.p.): of an engine is the power developed in its cylinder, and is measured by a form of pressure
indicator connected to the cylinder head. The indicated power is always greater than the brake power of
an engine, because there will always be a reduction of power between the cylinder and the output shaft
due to friction between the moving parts and the pumping power required to change the cylinder.
i. p.= b.p + friction power [i.p.= b.p+ f.p.]
1.2 Mechanical Efficiency (ηm ) of an engine is the rate.
It measures the efficiency with which an engine coverts the power developed in its cylinders into useful
power available at the output shaft. An average value of mechanical efficiency for a petrol engine
running at normal speed would be 0.80 or 80%.
1.3 Thermal Efficiency
An engine cannot covert all the heat energy of the fuel into work, for some inevitable reasons. Hence
the thermal efficiency of an engine is the measure of how efficiently the fuel is being used in the engine.
Definition: It is defined as the ratio of the work done per second to the Heat supplied from the fuel per
second.
ηt
=
Work done per second
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Heat supplied from the fuel per second
There are two cases to be considered: they are the indicated thermal efficiency and the brake thermal
efficiency. Take the case of the (I.C.E) internal combustion Engine first. In this case the thermal
efficiencies are follows:
Indicated thermal Efficiency ( ηt ) = Energy equivalent of the i.p per sec. (J/s, watts)
Energy supplied by fuel/sec
Indicated thermal efficiency
=
work done per second (J/S) _
Heat supplied per second (J/kg)
But Heat supplied per second = kg of fuel supplied per sec. x Calorific value of fuel (J/kg)
∴ ηit
=
i. p. (watts)
_
kg of fuel/ sec x C. V.
(J/kg)
Alternatively:
Indicated ηt =
Work done/ hour
=
Heat supplied/ hour
But
kg of fuel/ hour
=
i. p x 3600
_
kg of fuel/hr x c.v
specific fuel consumption (kg/kwh)
Indicated power in kw
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3600 x103
∴ηit =
_
Specific fuel consumption x C. V.
Similarly: Brake thermal Efficiency:
Brake ηt =
b. p. (watts)
_
kg of fuel/sec x c. v.
1.4 WORKED EXAMPLES
Example (1.0):
An oil engine developing 37.5kw uses 9.0kg of oil per hour of calorific value 54MJ/kg. 8.5kg are used to
overcome friction in the engine itself.
Determine:
(a) brake power (b) specific fuel consumption (brake basis) (c) Mechanical Efficiency (d) Indicated
thermal Efficiency of the Engine.
Sol:
Given that: i.p = 37.5kw; fuel consumption = 9.0kg/ hr
Friction power (fp) = 8.5kw;
6
Calorific value = 45MJ/kg (45 x 10 J/kg )
(a) b.p = i.p – f.p
= 37.5 – 8.5 = 29.0kw
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(b) Fuel consumption = 9.0 kg/hr;
but brake power = 29.0kw
∴ specific fuel consumption (sfc) in kg/kwh (Brake basis)
i.e. s. f. c. = Fuel consumption = 9.0kg/hr
Brake power
29.0 kw
(c) Mech. Efficiency ( ηm ) = b.p = 29.0
i.p
37.5
(d) Indicated thermal Efficiency (η it ) =
= 0.310kg/kwh
= 0.773
= 77.3%
i. p ( watts )
kg of fuel /sec. x c.v.
3
=
=
37.5 x 10
9.0
x 45 x 106
3600
37.5 x 103 x 3600
9.0 x 45 x 106
= 33.32%
1.5 MEASUREMENT OF BRAKE POWER
Any method for measuring the brake horse power of an engine involves the application of a torque
which resists the motion of the crank shaft. One method is of use the simple rope brake as throw below:
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A rope makes one complete turn round the rim of a fly wheel keyed to the Engine crankshaft. The rope
carries a dead load at one end and is hooked to a spring balance at the other, the direction of the
rotation being against pull of the dead load. The engine is started with the load off and increasing load
may be applied by adding weights to the dead load hanger.
Spring
Let
S
W = dead load on brake (kgf)
S = spring balance reading (kgf)
wheel
d
D
D = Diameter of brake wheel (m)
d = diameter of rope (m)
W
Dead weight
Rope Brake
By using a brake or dynamometer, it is possible to determine the useful work Output of an engine.
Electrical dynamometer setup showing engine, torque measurement arrangement and tachometer
A dynamometer consists of an absorption (or absorber/driver) unit, and usually includes a means for
measuring torque and rotational speed. An absorption unit consists of some type of rotor in a housing.
The rotor is coupled to the engine or other equipment under test and is free to rotate at whatever
speed is required for the test. Some means is provided to develop a braking torque between
dynamometer's rotor and housing. The means for developing torque can be frictional, hydraulic,
electromagnetic etc. according to the type of absorption/driver unit.
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Work done = force x distance
Braking force, f(N)
r
n rev/s
If a braking force F is applied at the rim of a wheel of radius r, the
product, Fr, which is the work done is called the resisting torque, T.
If the wheel is kept tuning at N rev/s against this braking force; the work done during one revolution of
the brake per second = Resisting force x Distance through which the force is over come per second.
Work done per second = F x 2πr x N
= 2πN x Fr (but Fr = torque T)
Work done per second = 2πNT = ωT
Work done/sec
Hence brake power
= Tω (Watts)
= useful work output of an engine
= Tω (Watts).
Where T = resisting torque (Nm) and
ω = angular velocity (radians/Second)
Now, consider the load to be acting along the centre line of the rope:-
D+d

 2 
Resisting torque due to deed load = 9.81 x W 
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Nm
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where 9.81(N) = g (i.e. acceleration due to gravity)
Torque due to spring pull = 9.81 x S
D+d


 2 
∴ Effective braking torque, T = 9.81 x W
∴Effective braking, T = 9.81 x W
D+d


 2 
D+d


 2 
= 9.81(W- S)
Nm
- 9.81 x S
D+d


 2 
D+d

 2 
– 9.81 x S 
D+d


 2 
Nm
and b. p = Tω (watts)
Note: simple rope brake can only be safely used on
Low speed engine which have their speed kept reasonably constant by a governor. For high speed
engine, a widely used brake is the Heenan and fromde Hydralic Dynamometer.
Example:
In a test on a single-cylinder gas engine using a simple rope brake, the following reading were taken:Dead load 29kgf, spring balance 4kgf: speed 284 rev/min; diameter of brake wheel is 1.05m, diameter
of rope = 20mm. calculate the b. p. being developed by the engine.
Solution:
D+d
 Nm
 2 
Braking torque = 9. 81(W - S) 
= 9. 81 (29 - 4)
 1.05 + 0.02 


2


Nm
= 9. 81 (29 - 4)  1.07  = 9. 81 x 25 x 0.535
 2 
∴T = 131 .2 Nm
∴ b. p = Tω = T x 2πN =
131.2 x 2π x 284
1000 x 60
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=
234..117 x 103
60 x 103
= 3. 902kw
Note: Tachometer- is used for measuring angular velocity of rotating objects usually measured in
rev/min.
Planimeter – is used for measuring area under a graph
WEEK TWO
2.0 MEASUREMENT OF INDICATED POWER:
In order to determine the indicated power of an engine, it is necessary to know the work conditions
which exist in the cylinders. The indicated mean Effective Pres sure (imep) may be obtained direct from
the indicator diagrams and from a knowledge of the (imep) of an engine the i.p can be calculated. An
indicate diagrams in a pressure- volume graph of the condition in the cylinder throughout a complete
cycle.
A typical diagram from a gas engine is shown below
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P
a
Useful work
Ex ansion
Compression
Exhaust
c
b
d
Suction
O
V
Pumping loss (-ve)
Thus, The area under an expansion curve, i.e. when volume is increasing , represents work done by the
gases (i.e. Positive work).
The area under a compression curve i.e. when volume is decreasing represents work done on the gases
(i.e. Negative Work).
By adding algebraically the area for each operation throughout the cylinder, the diagram may be seen to
consist of two enclosed area, the larger of which represent the work done by the engine each cycle. The
smaller enclosed area is called the pumping loop and represents a loss of work from the engine resulting
from the necessity to clean and recharge the cylinder. We shall ignore for the moment, the pumping
loop, because it is usually too small to measure on a normal diagram.
Mean Effective Pressure: the average net pressure which acting on the piston area for one Stroke and
does the same work as that represented by the indicator diagram is known as the Mean Effective
pressure, Pm. The enclosed area of the diagram is irregular in shape but a rectangle of equal area and
having the same base length would have a height equal to the Mean Effective pressure. Hence to
determine the mean Effective pressure from given indicator diagram measure the enclosed area using a
planimeter and then.
Mean Effective pressure, Pm =
Area of diagram (mm 2 )
Base length of diagram (mm )
x
Spring No.
1
Now, the indicated power can be calculated using the mean effective pressure follow:
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Indicated power (i.p) = Work done per second inside the cylinder the cylinder
= (Nm/s =J/s =Watt)
= Work done/cycle x Number of cycles/sec.
= (Nm =J) (Watt)
Now work done per cycle = Area of indicate diagram
= Pm x Vs
= Pm x (A x L)
∴ I. P = Pm x L x A x N (J/s)
= Pm LAN (watts)
Where:
Pm = Mean Effective Pressure N/m2
2
A = Area of piston M ( )
L = Length of Stroke, m
N = Number of working strokes/ sec (Cycles)
Note:
For a four-stroke cycle, no of cycles/sec =½ engine speed. For a two stroke cycle engine, No. of cycle/sec
= the engine speed.
For a gas engine on four-stroke cycle, n =
Where there are misses, n =
Engine speed
Engine speed
2
2
.
- No. of misses/sec.
2.1 I.P of Multi cylinder Engines
For engine having more than one cylinder, the most accurate method is to measure the i.p of each
cylinder separately. The i.p of the engine is the sum of the i.ps of the separate cylinders.
Thus, i.pEngine = i.p one cylinder x Number of cylinders.
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2.2 Worked Examples:
Example 1:
A four-cylinder petrol engine at 1200rev/min gave 25.3KW b.p. when one cylinder was cut out the b.p.
decreased to 17.6 KW. Estimate the ; p of the Engine
Solution:
Given that: The b.p. of 4-cylinder Engine 25.3kw
The b.p. of 3-cylinder the Engine =17.6kw
i.p. = 4(25.3-17.6)
= 30.8kw
Example 2
In a text in a single–cylinder oil engine operating on the four stocker cycle and fitted with a simple rope
brake, the following reading were taken.
Brake which diameter = 0.7m, speed 450rev/min, rope ria. 20mm, dead weight on rape 21kgf, spring
balance reading 3.4kgf, area of indicator diagram = 404mm2 , length of indicator diagram 65mm, spring
No, 140kN/m
2
per mm, bore 100mm, strake 150mm, engine used 0.75kg/h of oil of calorific value
45mJ/kg. calculate the b.p.; i.p.; mechanical efficiency and indicate thermal efficiency of the engine.
Solution :Given Data on above
D+d
 Nm
 2 
(a) b.p. = Tω But T = 9. 81(W - S) 
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= 9.81 (21-3.4)
 700 + 20 


3
 2 x 10 
T= 62.16Nm & ω = 2πN
450
T(!) = T x 2πN = 62.16 x 2π x
60
b.p. = 2.93KW
Area of diagram x Spring No.
(b)
i. p. = Pm L A N, but Pm =
Length of diagram
404
=
65
x
140 x 103
1
870.2 KN/m 2
=
∴ i.p = Pm LAN
=
870.2 x 103 x 150 x 10-3 x
π
4
(0.1)2 x
450
60 x 2
i.p = 3.844 KW
(c)
Mechanical Efficiency (ηm) =
b. p
i. p
=
2.93
3.84
= 0.763
ηm = 76.3%
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(d)
Thermodynamic Basic
Indicated thermal Efficiency (ηit)
i. p ( watts )
it
η
=
=
=
kg of fuel/sec x c.v.
3844( watts )
0.75
x 45 x 106
3600
3844 x 3600
0.75 x 45 x 10
Indicated η1 = 0.410
6
= 41.0%
2.3 Energy balance account for an I.C. engine regarded as operating in a closed
system:
To determine the energy balance account of an engine operating in a closed system, the first law of
thermodynamics is applied. The law states that the heat flow across the boundary is equal to the work
flow across the boundary.
Heat outflow to Cooling water, Qc
Heat inflow from
fuel supplied Qf
Engine
Work outflow to brake, W
Heat outflow by radiation, and
out and correction, Qr
Heat outflow to Exploit
Qe Energy balance Account
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Applying the 1st law,
Q f − Qr − Qc − Qe = W
or
Q f − W + Qr + Qe + Qe
This statement may be shown in the form of a balance sheet (often called a Heat or Energy Balance
Account), the left hand side showing the heat energy supplied and right hand side showing its
distribution. The balance may be based on 1kg of fuel, or on a basis of time, and the items may be
expressed as heat quantities and as percentage of energy supplied.
Fir I.C. E, the table is usually as follows:
(a)
Energy supplied per second, Qf = mass of fuel used/sec. x c.v.
= kg of fuel used/sec. x c.v
(b)
Energy distributed per second
(i)
Work outflow, W = b.p (watts).
(ii)
Heat flow to cooling water, Qc = m c (tout – tin) watts
Where m = flow of cooling water kg/s
C = specific heat capacity
tout = leaving temperature of cooling water
tin = Entry temperature of cooling water
(iii)
Heat flow to exhaust and surrounding, Qe + Qr
An addition to the balance sheet which is often asked for is the heat to friction. This accounts for the
reduction of power between the cylinder and the output shaft of an engine due to friction between the
moving parts. This heat to friction = (i.p – b.p.) watts.
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This is already included as heat flowing from the engine partly in the cooling water, partly in the exhaust
gases and partly in the radiation to the surroundings.
2.4 WORKED EXAMPLES:
Example 3:
The following results were obtained during the trial of a four-stroke national oil engine of cylinder bore
200mm and stroke 400mm.
Effective brake wheel diameter = 1.6m; speed = 258rev/min
Effective brake load = 47kgf, area of indicator diagram = 320mm2
Spring No. = 110 KN/m2 per mm; length of diagram = 65mm; fuel used/hour = 3.2kg/hr; c.v. of fuel =
0
45MJ/kg; cooling water = 38 C. (Given Cw = 4187J/kgK.)
Calculate: (a) Mechanical Efficiency
(b) Indicated thermal efficiency, and (c) Draw up an energy balance account on a basis of 1
minute.
Solution:
Mechanical Efficiency
ηm =
b. p
i. p
But b.p = Tω where T = 9.81 x 47 x
and ω = 2πN = 2π
1.6
2
258
60
= 368.9N
= 27.02
∴ b.p = Tω = 368.9 x 27.02
= 9.967 KW
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i.p = Pm LAN;
3
3
Pm = 320 x 110 x 10
= 541 x 10 N/m
2
π
2
258
∴ i.p = Pm LAN = 541 x 103 x 0.4 x 4 x (0.2) x 2 x 60
= 14.62 KW
Hence ηm =
b. p
i. p
=
9.967
14.62
= 0.6817
ηm = 68.17%
(b)
Indicated ηt =
i. p ( watts )
kg of fuel/sec. x c.v.
3
=
ηi.t
14.62 x 10 x 3600
3.2 x 45 x 106
= 0.3655
or 36.55%
(c)
Energy supplied or
Joules
%
6
100
Energy distribution/min.
Joules
%
input/min.
Kg of fuel/s x c.v.
3.2
2.4 x 10
x 45 x 106
60
(i) To useful work = b.p x
60
598.0 x 10
3
24.92
365.9 x 10
3
15.25
(ii) To cooling water = mc
(tout – tin) = 2.3 x 4187
x 38
(iii) To
exhaust
and
radiation (by diff.) [2.4
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x
106
–
(598.0
+
3
365.9)10 ].
3
Total
2400 x 103
100
Total
1436 x 10
59.83
2400 x 103
100
2.5 EXERCISES:
Q1. The following observations were recorded during a four – stroke single- cylinder oil engine:
duration of trial =30 min; oil consumption =4.4kg; calorific value of oil = 42MJ/Kg; Area of indicator
2,
2
diagram = 850mm length of diagram = 80mm; spring rating = 56KN/m per mm; effective brake
wheel diameter = 1.5m; speed = 200 rpm, brake load= 135kgf; spring balance reading = 18kgf; length
0
of stroke = 450mm; cooling water flow = 11kgf/min; temperature rise of cooling water =36 C. If the
engine is 57.2% efficient.
Calculate:
a. The cylinder bore of the engine
b. The brake specific fuel consumption
c.
Indicated thermal efficiency
d. Draw up a heat balance sheet in MJ/min for the engine (Cw=4.18KJ/kgk)
Q2. The following results were obtained during a test on a single- cylinder, double – acting steam
engine fitted with a simple rope brake: swept volume =7.85 liters, speed =300rpm, brake load
=136kg; spring balance reading = 90N, indicated mean effective pressure = 2.32 bar; steam
consumption =0.056kg/s, condenser cooling water =113kg/min; temperature rise of condenser
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0
cooling water =11K; Hot well temperature =38 C; calorific value of fuel =2.514MJ/Kg
(Cw=4.18KJ/kgk).
Determine
a
The effective diameter of the brake wheel, if the efficiency of the engine is 80.5%
b The brake thermal efficiency
c Specific steam consumption on indicated basis
d Draw up a complete energy balance account for the test on a basis of MJ/min.
Q3 .An oil engine on a four stroke cycle has a swept volume of 14 liters and a mean effective
pressure of 5.67 bars. Its rated speed is 6.6rps and is tested at this speeds speed against a brake
which has a torque arm of 0.7m. The net brake load is 755.5N and the fuel consumption is
-1
0.0025Kg/s. calorific value of fuel =44MJkg , cooling water circulation =9.0kg/min at an inlet
temperature of 380C and an outlet temperature of 71 0C, and the energy rejected through the
-1
-1
exhaust pipe is 33.6KW. (Cw=4.18KJg K )
Calculate: (a) the engine torque
(b) The Mechanical efficiency
(c) Brake specific fuel consumption
(d) Draw up an overall energy balance in KJ/s and as
2.6 SOLUTIONS:
Q1.
Given that; Duration = 30min; f c = 4.4kg; c v = 42 MJ/kg;
2
2
Aid = 850mm ; Lid = 80mm; spring No. = 56KN/M /mm
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Wheel dia = 1.5m; N = 200rpm; W = 135kgf;
S = 18kgf; stroke = 450mm; cooling water = 11kg/mm=in.
Temp. rise = 360C; η m = 57.2%
(a)
ηm =
b. p
ip
, but b.p = Tω = (W − S )
9.81(185 − 18)
=
(1.5)
2
x 2πx
(D + d )
2
x 2πN
200
60
9.81 x 117 x 0.75 x 2π x 200 = 18029
20
18.030 KW
=
i. p = Pm LAN = Pm x L x
850
=
80
x
56
1
πD 2
x N , but Pm =
4
Aid
Lid
x
SP.No.
1
= 595 KN/m 2
Ip = 595 x 103 x 0.45 x πD2 x 200
4
60 x 2
`
ip x 4 x 60 x 2
3
594 x 10 x 0.45 x π x 200
Since η m =
(b)
b.p
i.p
b.sf .c =
i. p =
,
f.c.
b.p
=
4.4 x 2
18.03
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=
b.p
ηm
=
=
D2 =
18.03
0.572
=
i.p x 480
168 x 23 x 106
31.52 KW
0.488 kg/kwh
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(c)
ηit =
=
=
i. p. (Watts )
Kg of fuel / s x C.V
31.52 x 103 x 30 x 60
4.4 x 42 x 106
30. 72%
(d)
Energy
MJ
%
Energy Distributed / min
MJ
%
1.082
17.56
1.658
26.92
3.420
55.52
6.160
100.00
supplied/min
(i) to useful work, w =b.p x 60
Qf = kg of oil/s x C.V
4.4kg x 42 x 10
= 18.03 x 60
6
(ii) To cooling water, Qc MC w
30min
∆t = 11 x 4.187 x 36.
6.16
100
(iii) To surrounding Qr
Qf – (W + Qc)
6.16 - (1.082 + 1.658)
Total
6.16
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100
Total
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Q2. Given That:
VS = 7.85 Litres; N = 300rpm;
W = 136kg;
S = 90N
Pm = 2.32bar; f.c = 0.056kg/s, cooling water =
113kg/min; Hot well temp. = 380C; Dt = 11K;
C.V. = 2.514 MJ/Kg; Cw = 4.18KJ/kgk; ∫m = 80.5%
(a)
ηm =
b.p
i.p
, but i.p = Pm LAN = Pm x Vs x N
5
2.32 x 10 x
=
7.85
10 3
x
300
60
Since it is double acting; i.p for the engine = 9.114 x 2
i.p =
18.228KW
Hence b.p = η m x i.p
=
0.805 x 18.228
= 14.674KW
But b. p
=
=
g (W − S )
De
2
x 2π N ; let De rep. effective dia
(9.8 x 136 – 90) x De x 2π x 5
14.674
De
=
=
1244 x
14,674
19,543.22
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De
2
x 2π x 5 = 19,543.22 De
=
0.75m
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Effective dia of brake wheel, De = 0.75m
(b)
ηbt
b. p (watts )
Kg of fuel/S x C.V
=
=
(c)
0.1042
fc
s. f .c (i) =
ip
0.056 x 3600
18.228
(d)
=
=
14674
s
0.056 x 2.514 x 10 6
10.42
Steam consumption
=
Indicated power
=
=
kg. of Steam / hr
i. p
11.06 kg / kwh
Energy balance account
Energy supplied/min
MJ
%
Energy Distributed / min
MJ
%
0.88
10.41
5.20
61.54
Qf = kg of fuell/s x C.V
0.056 x 60 x 2.514 x 10 6
(i) to useful work, w = b.p x 60
= 14.67 x 60
30min
8.45
100
(ii) To cooling water, Qc Qc =
MCw ∆t 113 x 4.18 x 11
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(iii) To Hot well Qw MCw ∆t =
0.056 x 60 x 4.18 x 38 = 0.056 x
60 x 4.18
(iv) To surrounding Qr
0.534
6.32
1.836
21.73
8.450
100.00
Qf – (W + Qc + Qw)
8.45 – (0.88 + 5.20 + 0.534)
Total
Q3.
8.45
100
Total
Given that; Vs = 14 litres, Pm = 5.67bar, N = 6.6 rps; r= 0.7m;
W(F) = 755.4N; f.c.
= 0.0025kg//s; c.v = 44MJ/kg
Cooling water = 9kg/min; ti = 380C; to = 710C
Qe = 33.6KW; Cw = 4.18KJ /kgk;
(a)
Engine torgue, T = Fr
=
(b) ∫m = b.p
= 755.4 x 0.7
528.76 Nm
but b.p = T
=
528.8Nm
= T x 2 πN = 528.8 x 2π x 6.6
= 21.927kw
And i.p = PmLAN
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=
22.0kw
=
Pm x Vs x N
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(c)
Thermodynamic Basic
=
5.67 x 105 x
=
26.2 kw.
∫m
= b. p
bsfc
=
10
=
i. p
f .c
b. p
MJ
x
6.6
2
0.84
0.0025
=
22.0
=
x
3600
90
1
22.0
840%
0.409 Kg/Kwh
=
Energy
14
%
Energy Distributed / min
KJ
%
22.00
20.0
20.7
18.82
33.6
30.55
supplied/min
Qf = kg of fuel/s x
C.V
(i) to useful work, w = b.p 22.0
6
0.0025 x 44 x 10
110
100
(ii) To cooling water, Qc MC w
∆t
9 x 4.18 x (71 – 38)
60
(iii) To exhaust Qe
Qe = 33.6kw given
(iv) To surrounding, Qr
Qf – (w + Qc + Qe)
110 – (22 + 20.75 + 33.6)
33.7
30.63
Total
110
100
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Total
110
100.00
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WEEK THREE
3.0
REVERSIBILITY`AND`IRREVESIBILITY`OF
THERMODYAMIC
CYCLES/PROCESSES:
The thermodynamic cycles in general are classified into two, namely:1.
Reversible or ideal cycle and
2.
Irreversible or natural or real cycle
3,1.
A Reversible cycle:
A thermodynamically reversible cycle consists of reversible processes
only and a reversible process is one which is performed in such a way that at the end of the
process, both the system and the surroundings may be restored to their initial states.
3.2.
An Irreversible cycle:
A cycle will be considered thermodynamically irreversible if any of
the processes constituting the cycle is irreversible. Thus, in an irreversible cycle, the initial
conditions are not restored at the end of the cycle.
In actual practice most of the processes are irreversible to some degree. The main causes for the
irreversibility are: (a)
mechanical and fluid friction (b) unrestricted expansion (c) Heat
transfer with a finite temperature difference.
3.3 SECOND LAW OF THERMODYNAMICS:
Like the first law, the second law of thermodynamics is a very important law and it is based
upon observable phenomena. It states that: “Heat will not transfer up the gradient of
temperature of its accord.” This does not mean that heat cannot be made to transfer up the
gradient of temperature. It can, but it has to be facilitated by the aid of external energy.
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External energy, however it is not required for the transfer heat from a higher temperature
gradient to a lower one it being a purely natural phenomena.
This law imposes limit upon the actual quantity of heat which can be extracted from any given
heat energy supply and hence, by the principle of conservation of energy, imposes a limit upon
the amount of work which can expected theoretically. This limit appears when the temperature
of the heat energy supply becomes the same as its surroundings at which condition heat
exchange ceases.
nd
3.3.1 Clauses version of 2 law of thermodynamics:
He stated that: “It is impossible for a self –acting machine, unaided by an external agency to
covey heat from a body at a low temperature to one at a higher temperature.” This implies that
although the machine may have its own heat energy content in the form of high and low
temperature quantities, heat energy transfer from the low temperature to the high temperature
is impossible unless some external energy supply is used to run the machine. Actually this is the
principle of the Refrigeration and heat pump. Note, that if this is the case, the machine ceases to
be self-acting.
nd
3.3.2 Lord Kelvin’s version 2 law of thermodynamics:
He stated that: “We cannot transfer heat into work merely by cooling a body already below the
temperature of the coldest surrounding objects”.
This implies that when a body researches the temperature of the coldest surroundings objects,
no further natural extraction of heat is possible, and hence no further work can be performed.
Another def.:- The thermodynamic engine is a device in which energy is supplied in the form of
heat and some of this energy is transformed into work.
3.4 Definition of heat engines:
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Since the heat is defined as that of transfer of energy which results from a difference in
temperature, then a heat engine must be an engine in which a transfer of heat occurs.
If heat is introduced into a system and as a result of a cycle process. Some work appears from
that system, together with some heat rejection from the system, then this is a heat engine. In
practice, such as an engine, or plant really, is the closed circuit stream terbium point of a power
station. This can be illustrated as follow:
On the other hand, the open circuit, internal combustion engine since as a petrol engine in
strictly not a heat engine for fuel and air are admitted, which must cross the system boundary,
combustion is internal, as the name implies, and combustion products and heat are rejected,
with some work crossing the system boundary. However, thermodynamic engine are mostly
referred to as heat engines.
3.5 Element of a heat engine cycle:
The essential element of a thermodynamic cycle involving a heat engine are:
1. A working substance: A medium for receiving and rejecting heat, and doing work, it is the
substance that undergo the change of state e.g. steam, air, etc.
2. A source of heat (A hot body or heat Reservior
This is where heat may be added to the working substance.
3. A heat sink (cold body): This is the body to which heat may be rejected by the working
substance. In practice a heat sink in a natural receiver such as the atmosphere, a lake, a
lagoon, a river or ocean.
4. An Engine: An engine in that device in which the working substance may do work or have
work done on it.
When the temperature of any heat energy supply has fallen then the heat energy is said to have
been degraded. Complete degradation occurs when thermal equilibrium has been established.
3.6. Deduction from the 2nd Law of thermodynamics:
1. From this second law of thermodynamics, it follows that in order to run all the various engine
and devices in use today, and hence maintain the present state of civilization a source of supply
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of fuel is absolutely necessary. It is by burning fuel that the various working substances have
their temperature put up above their surroundings thus enabling them to release their heat
nd
energy in a natural manner in accordance with 2
law. Developments in the field of nuclear
physics have shown the way of obtaining a further supply of heat energy. This will augment the
dwindling coal, oil and natural gas stocks of world. For instance, it has been seriously forecasted
that world crude oil stock could be seriously low, if not completely exhausted by the end of this
20th century.
2. By virtue of this 2 nd law, it is essential that all fuels should be used as efficiently as possible in
order that fuel stock may be preserved for as long as possible. It must always be remembered
that when once the heat energy from any particular fuel has been degraded, then further heat
energy is only obtainable at the experience of further fuel.
WEEK 4
4.0 THE CARNOT CYCLE EFFICIENCY
4.1 Carnot Principle:
Reversibility, as it applies to the thermodynamic engine, was discussed by a Frenchman, Sadi
Carnot, in a paper entitled. Reflections on the motive power of heat which was published in
1824. In the paper, Carnot conceived of an engine working on thermodynamically reversible
processes, and from this concept deduced what has since been called Carnot’s principle. This
states that: “No engine can be more efficient than a reversible engine working between the
same limits of temperature. The principle of thermodynamic engine is that it receives heat at
some high temperature from a heat source. The engine then converts some of this heat into
work and then rejects the reminder into a sink
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Consider a thermodynamic reversible engine R working between the temperature limits of
source T1 and sink T2. If the engine receives Q units of heat from the source and temperature T1.
It will convert W units of this heat into work and then reject (Q-W) units of heat into sink at
lower temperature T2. as shown in fig (a).
Source T
Source T
Q – dQ
Q
Carnot
W
E
Q
Q – dQ - W
Q–W
Q
Carnot
(Q – W)
Sink T
Sink T
Assume that source other engine E can be found which is more efficient than the reversible
engine R. Since it is more efficient, this engine E will require less heat supplied to perform the
same amount of work, W. If this engine is used to drive a Carnot heat pump then it will need less
heat than the Carnot engine to produce the work needed to drive the heat pump. This would
mean that the combination would be returning more heat to the source at the higher
temperature T1 than is being taken out. But then would directly contravene the second law of
thermodynamics which says that heat cannot flow from a lower to a higher temperature
without the aid of work from an outside source, and so we conclude that no such more efficient
engine exists. Now we have achieved a very important step in the investigation into engine
efficiencies becomes in have identified an engine with the best possible efficiency i.e. one
consisting only of reversible operation. The next step is to find out the efficiency of the Carnot
engine.
4.2 Thermal Efficiency of a carnot cycle Engine:
P
P1
1
Isothermal expansion
T1 = T2
P
2
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Adiabatic com resssion
2
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By calculating the thermal efficiency of this cycle it is possible to establish the maximum possible
efficiency between the temperature limits taken. The fig (c) above shows the Carnot cycle
illustrated on a P-V diagram. The cycle is made up of four reversible processes i.e. put together
that they form a closed cycle. Thus by proceeding round the cycle, it is possible to return to the
original state and hence the admit of repetition. The processes are as follows:
1-2: Isothermal expansion
Pressure falls from P1 to P2
Volumes increase from V1 to V2
Temperature remain constant at T1 =T2
work done = P1V1 ln
V2
Heat reveived = mRT1 ln
V1
= mRT1 ln
V2
V1
V2
V1
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For Isothermal, Q = W.
2-3:
Adiabatic expansion
Pressure falls from P2 to P3
Volume increase from V2 to V3
Temperature falls from T2 to T3
Work done =
P2V 2 − P3V3
γ −1
=
mR(T2 − T3 )
γ −1
For the adiabatic, Q = O.
No heat transfer during this process.
3-4: Isothermal compression
Pressure increase from P3 to p4
Volume reduced from V3 to V4
Temperature remain constant at T3=T4
Work done = P3V3 ln
V4
V3
= − P3V3 ln
= − mRT3 ln
V3
V4
V3
V4
For the Isothermal, Q = W
Heat rejected = − mRT3 ln
V3
V4
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4-1: Adiabatic compression
Pressure increases from P4 to P1
Volume reduced from V4 to V1
Temperature increases from T4 to T1
Work done =
P4V4 − P1V1
=
γ −1
=
(P1V1 − P4V4 )
γ −1
− mRT (T1 − T4 )
γ −1
For the adiabatic, Q = O
No heat transfer during this process. Note that this process returns the gas its original state 1
The work done during this cycle may be determined by summing the areas beneath the various
processing taking the expansions as positive areas and compression as negative areas.
Thus: work done /cycle = area (1-2) + area (2-3) - area (3-4)
-area (4-1)
= area 1234 =area enclose by cycle
OR
Work done/cycle
= mRT1 In
But
V2
V1
+
mR (T2 - T3 )
γ -1
- mRT3 In
V3
V4
-
mR (T1 - T4 )
γ -1
……….
(1)
from the Isothermal process; T1 = T2 and T3 = T4
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mR (T2 - T3 )
∴
γ -1
mR (T1 - T4 )
=
γ -1
Hence, from equation (1);
V2
Work done/cycle = mRT1 ln
V1
- mRT3 ln
V3
V4
………………………..
(2)
For the Adiabatic, 1-4;
γ −1
V 
=  4 
T4  V1 
T1
……………………………………………………….(3)
γ −1
For the Adiabatic 2 − 3; T2 =  V3
 V 
 2
T3
……………………………………(4)
But T1 = T2 and T3 = T4
T
T
∴ 1 = 2 ………………………………………………………(5)
T4 T3
Hence from Equation (3) and (4);
V4
V1
=
V3
V2
or
V2
V1
=
V3
V4
……………………….………………….(6)
Substituting Equation (6) in equation (2);
Work done/cycle = mR (T1 - T3 ) In
V3
V1
……………………………(7)
This is positive work done and this is always the case if the processes of a cycle proceed in a clockwise
direction. External work can thus be obtained from such cycles.
If the processes proceed in an anti
clockwise direction then the work done is negative, in which case equation (7) now becomes;
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Work done/cycle = - mR (T1 - T3 ) ln
V2
V1
……………………………………….(8)
Negative work means that external work must be put in to carry out such cycle
Now.
Thermal η =
Heat received - Heat rejected
Heat received
From the above analysis
mRT1 In
η th =
V2
V1
− mRT3 ln
V3
V1
mRT1 ln V2
V1
mRT (T1 − T3 ) In
=
mRT1 In
∴ Thermal η =
T1 - T2
T
V2
V1
Since
V2
V2
V1
=
V3
V4
from equation (6)
V1
………………………………………………………(9)
1
=
Max. abs. temp - Min. abs. temp
Max Abs. temp
From Equation (9); Thermal η = 1 −
T3
T1
…………………………….(10)
…………………………….
(11)
And from Equation (3), (4) and (5);
V
=  4
V
T4
 1
T1
γ -1




V
=  3
V
 2
γ −1




= r γ -1
When r = adiabatic compression and expansion volume ratio.
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From Equation (11);
Thermal η = 1 -
1
r y -1
…………………………………………………12)
This thermal efficiency gives the maximum possible thermal efficiency obtained between any this given
temperature limit. It is new possible to suggest that if any engine working between the same
temperature limits heat a thermal efficiency lower than this, then thermal efficiency improvement is
theoretically possible. Most engine have a thermal efficiency much lower than the carnot efficiency.
The militants aim should be an attempt to reach an efficiency as near 100% as possible.
4.3 W0RKED EXAMPLES:
The overall volume expansion ratio of a carnot cycle is 15. The temperature limit of the cycle are 260 0C
0
and 21 . Determine;
(a) The volume rates of the isothermal and adiabatic processes.
(b) The thermal efficiency of the cycle.(take  = 1.4)
SOLUTIONS:
0
0
Given that: T1=260 C; T4 = 21 C; γ = 1.4; overall
Vol. expansion ratio = 15
(a) for the Adiabatic ;
V
=
= 4
T4 T3  V1
T1
T2




γ −1
V
= 3
V
 2




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γ −1
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1
1
 T  γ -1  T2  γ -1

=
=  1 
= 
T 
V2
T
V1
 4
 3 
V3
V4
∴
1
 533 1.4-1


 294 
=
1
= 1.812
0.4
1.812 25 = 4.42
=
Volume ratio of adiabatic = 4.42 =r a
 V3V1 
V4 =  V1 V4 
V3
Volume ratio of isothermals =
Isothermal vol. ratio =
(b) Thermal Efficiency
Overall vol. ratio
Adiabatic vol. ratio
=
T1 - T4
T1
=
15
4.42
= 3.39
x 100%
533 - 294
=
=
533
239
533
x 100%
x 100%
= 44.8%
Alternatively
η = 1 − γ1−1 = 1 −
r
= 1−
1
4.421.4 −1
1
4.421.4 −1
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= 1−
1
1.812
= 1 − 0552
= 0.448 = 44.8%
4.3 EXERCISES AND SOLUTIONS
Q1 (a). In a carnot cycle process, whose temperature limits are 300 0C and 500C, a total of 230g of gas
was taken in. If the hyperbolic expansion volume ratio is 2.5
Determine:
i.
The isentropic expansion volume ratio
ii.
The thermal efficiency of the cycle
iii.
The work done per cycle. (Take R=0.28KJ/kgk;  = 1.4).
0
b.
2
0.5kg of air is first expanded isothermally at a temperature of 235 C from 3.5MN/m to 2.1
MN/m2 and further expanded adiabatically to 140KN/m 2 . The air is then cooled at constant pressure and
is finally returned to is initial state by adiabatic compression.
Calculation:i.
The external work done by the air per cycle
ii.
The thermal efficiency of cycle. (For the air, take = 1.4; Cp =1KJ/Kgk; R=287J/Kgk)
iii.
Q2.
A carnot cycle works with isentropic compression ratio of 5, and hyperbolic expansion ratio of 2.
The volume of air at the beginning of the hyperbolic expansion is 0.3m3 . If the maximum temperature
and pressure is limited to 550K and 21 bar respectively.
Determine;
a.
Minimum temperature in the cycle.
b.
Thermal efficiency of the cycle
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c.
Pressure at all the salient points
d.
Work done per cycle (take ratio of specific heats as 1.4)
Q3 (a) A carnot engine working between 3770C and 370C produce
150KJ of work. Find:
(i)
the thermal efficiency of the engine
(ii)
the heat supplied during the process
(b)
A carnot engine operates between two reservoirs at temperature T1 and T3. The work output of
the engine is 0.6 times the heat rejected. The difference in temperatures between source and the sink is
0
200 C. Calculate:
(i)
the thermal efficiency of the engine
(ii)
the source and the sink temperatures
SOLUTIONS:
Q1.(a) Given that; t1 = 3000C; t3 = 500C; M = 230g (0.23kg)
Ri = 2.5 (i.e T 1 = 573k; T3 = 323k); R = 0.28kg/K
Let
(i)
Isewtropic expansion vol ratio = ra
For adiabatic process
ra = 
T1vr 
T 
 4 
V 
=
=  4
T4 T3  V1 
T1
T2
r −1
V 
=  3
V3 
r −1
r −1
= [ra ]
1
γ −1
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1
ra
=
ra
ii.
ηt =
=
iii.
T1 − T3
T1
=
=
1
 573 1.4 −1


 323 
= (1.774)
(1.774) 2.5
=
0.4
4.19
573 − 333 100%
x
573
1
250 x 100% = 43.6%
573
1
WD/cycle
= MR (T1 – T3) In
V2
V1
= 0.23 x 0.28 (573 – 323) ln 2.5
=0.23 x 0.28 x 250 x 0.916
=
OR WD/cycle
=
MRT1 In
14.75KJ
V2
V1
x ηth = 0.23 x 0.28 x 573 x In 2.5 x 0.436
= 0.23 x 0.28 x 573 x 0.916 x 0.436
=
(b)
Given that; M = 0.5kg; T1 = T2 = 508K;
1.4; CP = IKJ/kgk;
(i)
14.74 KJ
2
2
P1 = 3.5 MN/M2; P2 = 2.1MN/M ; P3 = 140KN/M γ =
R = 28TJ/kgk.
External work done/cycle = Q2 - Q4 = (QS – Q
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MRT1In
P1V1
=
V1 = MRT1
V2
- MCp (T3 – T4)
V1
MRT1
=
0.5 x 287 x 508 =
72898
6
P1
3.5 x 10
3.5 x 10
6
V1 = 2.08
For isothermal process, P1V1 =P2V2
∴
∴ Q12
V2
P1V1
=
P2
=
3.5 2.08
x
2.1
1
=
3.466
=
MRT1 In
V2
V1
=
0.5 X 0.287 X 508 x In (1.666)
= 0.5 X 0.287 X 508 X 0.510
=
37.18KJ
γ −1
T2 =  P2 
T3  P3 
γ
 P3 

 P2 
T3 = T2 
γ −1
γ
But T2 = 508K (Since T1 = T2)
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0.4
=
508
 0.14  1.4


 2.1 
= 508 x (0.04) 0.286
= 234.2k
T1
T4
 P1 

 P4 
= 
γ −1
γ
λ −1
P 
∴T4 = T1  4 
 P1 
0.4
 0.14 

 3.5 
γ
= 508
1.4
0.286
= 508(0.04 )
= 202.3k
WD/Cycle = mRT1 ln
V2
V1
− mcp - mCp (T3 - T4)
=
37.18 – 0.5 x 1 x (234.2 – 202.3)
=
37.18 – 0.5 x 31.9 = 37.19 – 15.95
= 21.23KJ
Q2
3
Given that, ra = 5, ri = 2, V1, = 0.3m ; T1 = 550k;
P1 = 21bar
But ra =
V4
V1
= 21 x 105 N/M2; γ = Cv = 1.4
= 5 and ri =
Cp
V2
V1
=2
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(a)
Thermodynamic Basic
Minimum temp. in the cycle let the min. temp. =
T3 or T4
First consider Isentropic (adiabetic) compn process (4 – 1)
V 
=  4 
T4
 V1 
T1
γ −1
=
T4
51.4 -1 = 1.9036
T1
=
1.9036
(b)
550
289k
16 C
0
289K or 16 C
Thermal efficiency of the cycle ηt =
T1 − T3
=
550 − 289
550
T1
(c)
=
1.9036
0
=
T3 or T4 =
=
=
0.4745
=
47.45%
Pressure at all the salient points, let P2, P3 and P4 represent pressures at points 2,3, and 4
respectively.
For isothermal expansion process (1 – 2);
P1V1 = P2 V2 , ∴ P2 =
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P1V1
V2
= P1 x
V1
V2
= 21x
1
2
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= 10.5 bar
For isotropic expansion process (2 – 3)
γ
P2V2
1.4
V 
1
= P3V3 , ∴ P3 = P2x 2  = 10.5 
5
 V3 
γ
γ
1.4
= 10.5(0.2)
= 1.103 bar
Since
V4
V1
=
V2
V2
= 5, hence
V2
V3
=
1
5
For isentropic compression process (4 – 1)
P4V4
γ
V 
= P1V11 ; ∴ P4 = P1  11 
 V4 
γ
λ
1 .4
 1 1.4
P4 = 21  = 21 x (0.2 )
5
= 2.206 bar
(d)
Work done/cycle, W = Heat Supplied – Heat rejected.
W = Qs – Qr
= Q12 – Q34
For Heat supplied, Q2 = P1V1 Ln V2
V1
6
= 21 x 10 x 0.3 x In 2
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= 436.68KJ
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For Heat Rejected Q34 = P3V3 In
V3
V4
= 1.103 x 105 x 3.0 x In 2
= 299.26 KI
Work done/cycle, W = 436.68 – 299.36
=
Q3.
207.32KJ
0
0
(a) Given that; t 1 = 377 C; t3 = 37 C, W = 150 KJ
(i)
Thermal efficiency, ηt = T1 − T3 but T1 = 650K; T3 = 300K
T1
=
=
(ii)
ηt
Q12
=
650
0.523 = 52.3%
Heat supplied during the process Q12
Since
650 − 310
=
W
ηt
Work done
=
Heat Supplied
W
η
=
−
VV
Q12
150
=
0.523
286.8KJ
t
(b)
Given that; work output; W = 0.6Qr
=
Temp. difference, T1 – T3
2000C (200k)
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=
0.6Q34
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(i)
Thermal efficiency, η t =
ηt
=
WD
WD + Q r
(ii)
Heat Supplied
Qs - Q r
=
Qs
0.6Qr
0.6Qr + Q r
0.6Qr
=
ηt
=
Work done
1.6Qr
0.6
=
1.6
=
0.375
=
37.5%
Source and sink temperatures, let T1, = Source temp.
T3 = Sink temp.
But
ηt
T1 − T3
=
T1
0.375
=
200
T1 =
=
0.375
200
T1
T1 =
200
ηt
200
T1
= 533.3k (260.3°C )
Since T1 −T 3= 200
T = T − 200
3
1
= 260.3 − 200 = 60.3°C
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WEEK FIVE
5.O: THE SOLAR ENERGY:
5.1 Solar Energy -- Energy from the Sun
has produced energy for billions of years. Solar energy is the sun’s rays (solar radiation) that reach the
earth.
Solar energy can be converted into other forms of energy, such as heat and electricity. In the
1830s, the British astronomer John Herschel used a solar thermal collector box (a device that
absorbs sunlight to collect heat) to cook food during an expedition to Africa. Today, people use
the sun's energy for lots of things.
Solar energy can be converted to thermal (or heat) energy and used to:
•
•
Heat water – for use in homes, buildings, or swimming pools.
Heat spaces – inside greenhouses, homes, and other buildings.
Solar energy can be converted to electricity in two ways:
•
•
Photovoltaic (PV devices) or “solar cells” – change sunlight directly into electricity. PV
systems are often used in remote locations that are not connected to the electric grid.
They are also used to power watches, calculators, and lighted road signs.
Solar Power Plants - indirectly generate electricity when the heat from solar thermal
collectors is used to heat a fluid which produces steam that is used to power generator.
Out of the 15 known solar electric generating units operating in the United States at the
end of 2006, 10 of these are in California, and 5 in Arizona. No statistics are being
collected on solar plants that produce less than 1 megawatt of electricity, so there may be
smaller solar plants in a number of other states.
The major disadvantages of solar energy are:
•
•
The amount of sunlight that arrives at the earth's surface is not constant. It depends on
location, time of day, time of year, and weather conditions.
Because the sun doesn't deliver that much energy to any one place at any one time, a
large surface area is required to collect the energy at a useful rate.
5.2 PHOTOVOLTAIC ENERGY
Photovoltaic energy is the conversion of sunlight into electricity. A photovoltaic cell, commonly
called a solar cell or PV, is the technology used to convert solar energy directly into electrical
power. A photovoltaic cell is a nonmechanical device usually made from silicon alloys.
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Sunlight is composed of photons, or particles of solar
energy. These photons contain various amounts of
energy corresponding to the different wavelengths of
the solar spectrum. When photons strike a
photovoltaic cell, they may be reflected, pass right
through,
or be absorbed.
Only
the absorbed
provide energy
to generate
electricity.
Whenphotons
enough
sunlight (energy) is absorbed by the material (a
semiconductor), electrons are dislodged from the
material's atoms. Special treatment of the material
surface during manufacturing makes the front surface
of the cell more receptive to free electrons, so the
electrons naturally migrate to the surface.
When the electrons leave their position, holes are
formed. When many electrons, each carrying a
negative
charge, travel
toward
frontbetween
surface the
of the
cell, the resulting
imbalance
ofthe
charge
cell's front and back surfaces creates a voltage
potential like the negative and positive terminals of a
battery. When the two surfaces are connected through
an external load, electricity flows.
The photovoltaic cell is the basic building block of a
photovoltaic system. Individual cells can vary in size
from about 1 centimeter (1/2 inch) to about 10
centimeter (4 inches) across. However, one cell only produces 1 or 2 watts, which isn't enough
power for most applications. To increase power output, cells are electrically connected into a
packaged weather-tight module. Modules can be further connected to form an array. The term
array refers to the entire generating plant, whether it is made up of one or several thousand
modules. The number of modules connected together in an array depends on the amount of
power output needed.
The performance of a photovoltaic array is dependent upon sunlight. Climate conditions (e.g.,
clouds, fog) have a significant effect on the amount of solar energy received by a photovoltaic
array and, in turn, its performance. Most current technology photovoltaic modules are about 10
percent efficient in converting sunlight. Further research is being conducted to raise this
efficiency to 20 percent.
The photovoltaic cell was discovered in 1954 by Bell Telephone researchers examining the
sensitivity of a properly prepared silicon wafer to sunlight. Beginning in the late 1950s,
photovoltaic cells were used to power U.S. space satellites (learn more about the history of
photovaltaic cells). The success of PV in space generated commercial applications for this
technology. The simplest photovoltaic systems power many of the small calculators and wrist
watches used everyday. More complicated systems provide electricity to pump water, power
communications equipment, and even provide electricity to our homes.
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5.3 Some advantages of photovoltaic systems are:
1. Conversion from sunlight to electricity is direct, so that bulky mechanical generator
systems are unnecessary.
2. PV arrays can be installed quickly and in any size required or allowed.
3. The
environmental
impact is minimal, requiring no water for system cooling and
generating
no by-products.
Photovoltaic cells, like batteries, generate direct current (DC) which is generally used for small
loads (electronic equipment). When DC from photovoltaic cells is used for commercial
applications or sold to electric utilities using the electric grid, it must be converted to alternating
current (AC) using inverters, solid state devices that convert DC power to AC.
Historically, PV has been used at remote sites to provide electricity. In the future PV arrays may
be located at sites that are also connected to the electric grid enhancing the reliability of the
distribution system.
5.4 SOLAR THERMAL HEAT
Solar thermal(heat) energy is often used for heating swimming pools, heating water used in
homes, and space heating of buildings. Solar space heating systems can be classified as passive
or active.
Passive space heating is what happens to your car on a hot summer day. In buildings, the air is
circulated past a solar heat surface(s) and through the building by convection (i.e. less dense
warm air tends to rise while more dense cooler air moves downward) . No mechanical equipment
is needed for passive solar heating.
.
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5.5: Solar energy technology:
The parabolic dish engine system, which concentrates solar power
Solar energy is the light and radiant heat from the Sun that influences Earth's climate and
weather and sustains life. Solar power is sometimes used as a synonym for solar energy or more
specifically to refer to electricity generated from solar radiation. Since ancient times solar energy
has been harnessed for human use through a range of technologies. Solar radiation along with
secondary solar resources such as wind and wave power, hydroelectricity and biomass account
for most of the available flow of renewable energy on Earth.
Solar energy technologies can provide electrical generation by heat engine or photovoltaic
means, space heating and cooling in active and passive solar buildings; potable water via
distillation and disinfection, daylighting, hot water, thermal energy for cooking, and high
temperature process heat for industrial purposes.
Insolation and Solar radiation
About half the incoming solar energy reaches the earth's surface.
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The Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upper
[1]
atmosphere. Approximately 30% is reflected back to space while the rest is absorbed by
clouds, oceans and land masses. The spectrum of solar light at the Earth's surface is mostly
[2]
spread across the visible and near-infrared ranges with a small part in the near-ultraviolet.
The
absorbed
solar
lightthe
heats
the land
and atmosphere.
warm air containing
evaporated
water
from
oceans
rises,surface,
driving oceans
atmospheric
circulation The
or convection.
When this
air reaches a high altitude, where the temperature is low, water vapor condenses into clouds,
which rain onto the earth's surface, completing the water cycle. The latent heat of water
condensation amplifies convection, producing atmospheric phenomena such as cyclones and
[3]
anti-cyclones. Wind is a manifestation of the atmospheric circulation driven by solar energy.
Sunlight absorbed by the oceans and land masses keeps the surface at an average temperature of
[4]
14 °C. The conversion of solar energy into chemical energy via photosynthesis produces food,
[5]
wood and the biomass from which fossil fuels are derived.
Yearly energy resources & annual energy consumption (TWh)
Solar radiation along
with
secondary
solar
resources
such as
wind
and wave power,
hydroelectricity and
[7]
Wind energy (technical potential)
221,000.0
biomass account for
99.97% of the available
[8]
Electricity (2005)
-45.2
renewable energy on
[10][11]
Earth.
The total
[9]
solar
energy
absorbed by
Primary energy use (2005)
-369.7
Earth's atmosphere,
oceans and land masses
is approximately 3,850 zettajoules (ZJ) per year.[12] In 2002, this was more energy in one hour
[13][14]
than the world used in one year.
Photosynthesis captures approximately 3 ZJ per year in
[15]
biomass.
The amount of solar energy reaching the surface of the planet is so vast that in one
year it is about twice as much as will ever be obtained from all of the Earth's non-renewable
[16]
resources of coal, oil, natural gas, and mined uranium combined.
[6]
Solar energy absorbed by atmosphere, oceans and Earth
751,296,000.0
From the table of resources it would appear that solar, wind or biomass would be sufficient to supply all
of our energy needs, however, the increased use of biomass has had a negative effect on global
warming and dramatically increased food prices by diverting forests and crops into biofuel
production.[17] As intermittent resources
WEEK SIX
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6.0 APPLICATION OF SOLAR ENERGY
Average insolation showing land area (small black dots) required to replace the total world energy
2
supply with
solar electricity. Insolation for most people is from 150 to 300 W/m or 3.5 to 7.0
2
kWh/m /day.
Solar energy refers primarily to the use of solar radiation for practical ends. All other renewable
energies other than geothermal derive their energy from energy received from the sun.
Solar technologies are broadly characterized as either passive or active depending on the way
they capture, convert and distribute sunlight. Active solar techniques use photovoltaic panels,
pumps, and fans to convert sunlight into useful outputs. Passive solar techniques include
selecting materials with favorable thermal properties, designing spaces that naturally circulate
air, and referencing the position of a building to the Sun. Active solar technologies increase the
supply of energy and are considered supply side technologies, while passive solar technologies
reduce the need for alternate resources and are generally considered demand side technologies.
6.1 Architecture and urban planning
Passive solar building design and Urban heat island
Darmstadt University of Technology won the 2007 Solar Decathlon in Washington, D.C. with this passive
house designed specifically for the humid and hot subtropical climate
Sunlight has influenced building design since the beginning of architectural history. Advanced
solar architecture and urban planning methods were first employed by the Greeks and Chinese,
who oriented their buildings toward the south to provide light and warmth.
The common features of passive solar architecture are orientation relative to the Sun, compact
proportion (a low surface area to volume ratio), selective shading (overhangs) and thermal mass.
When these features are tailored to the local climate and environment they can produce well-lit
spaces that stay in a comfortable temperature range. Socrates' Megaron House is a classic
example of passive solar design. The most recent approaches to solar design use computer
modeling tying together solar lighting, heating and ventilation systems in an integrated solar
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design package. Active solar equipment such as pumps, fans and switchable windows can
complement passive design and improve system performance.
6.1.1 Agriculture and horticulture
Agriculture, Horticulture, and Greenhouse
Greenhouses like these in the Netherlands' Westland municipality grow vegetables, fruits and flowers.
Agriculture seeks to optimize the capture of solar energy in order to optimize the productivity of
plants. Techniques such as timed planting cycles, tailored row orientation, staggered heights
between rows and the mixing of plant varieties can improve crop yields. While sunlight is
generally considered a plentiful resource, the exceptions highlight the importance of solar energy
to agriculture. During the short growing seasons of the Little Ice Age, French and English
farmers employed fruit walls to maximize the collection of solar energy. These walls acted as
thermal masses and accelerated ripening by keeping plants warm. Early fruit walls were built
perpendicular to the ground and facing south, but over time, sloping walls were developed to
make better use of sunlight. In 1699, Nicolas Fatio de Duillier even suggested using a tracking
mechanism which could pivot to follow the Sun Applications of solar energy in agriculture aside
from growing crops include pumping water, drying crops, brooding chicks and drying chicken
manure More recently the technology has been embraced by vinters, who use the energy
generated by solar panels to power grape presses.
Greenhouses convert solar light to heat, enabling year-round production and the growth (in
enclosed environments) of specialty crops and other plants not naturally suited to the local
climate. Primitive greenhouses were first used during Roman times to produce cucumbers yearround for the Roman emperor Tiberius. The first modern greenhouses were built in Europe in the
16th century to keep exotic plants brought back from explorations abroad. Greenhouses remain
an important part of horticulture today, and plastic transparent materials have also been used to
similar effect in polytunnels and row covers.
6.2 Solar lighting
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Daylighting features such as this oculus at the top of the Pantheon in Rome have been in use since
antiquity.
The history of lighting is dominated by the use of natural light. The Romans recognized a right to
light as early as the 6th century and English law echoed these judgments with the Prescription
Act of 1832. In the 20th century artificial lighting became the main source of interior
illumination but daylighting techniques and hybrid solar lighting solutions are ways to reduce
energy consumption.
Daylighting systems collect and distribute sunlight to provide interior illumination. This passive
technology directly offsets energy use by replacing artificial lighting, and indirectly offsets nonsolar energy use by reducing the need for air-conditioning. Although difficult to quantify, the use
of natural lighting also offers physiological and psychological benefits compared to artificial
lighting. Daylighting design implies careful selection of window types, sizes and orientation;
exterior shading devices may be considered as well. Individual features include sawtooth roofs,
clerestory windows, light shelves, skylights and light tubes. They may be incorporated into
existing structures, but are most effective when integrated into a solar design package that
accounts for factors such as glare, heat flux and time-of-use. When daylighting features are
properly implemented they can reduce lighting-related energy requirements by 25%.
6.3 Solar thermal
Solar thermal energy
Solar thermal technologies can be used for water heating, space heating, space cooling and
process heat generation.
(a)Water heating
Solar hot water and Solar combisystem
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Solar water heaters facing the Sun to maximize gain
Solar hot water systems use sunlight to heat water. In low geographical latitudes (below
40 degrees)
to 70%
of the domestic
water use
with
temperatures
up to are
60 °C
can be
provided
by from
solar 60
heating
systems.
The mosthot
common
types
of solar
water heaters
evacuated
tube collectors (44%) and glazed flat plate collectors (34%) generally used for domestic hot
water; and unglazed plastic collectors (21%) used mainly to heat swimming pools.
6.3.1 Heating, cooling and ventilation
Solar heating, Thermal mass, Solar chimney, and Solar air conditioning
MIT's Solar House #1, built in 1939, used seasonal thermal storage for year-round heating.
In the United States, heating, ventilation and air conditioning (HVAC) systems account for 30%
(4.65 EJ) of the energy used in commercial buildings and nearly 50% (10.1 EJ) of the energy
used in residential buildings. Solar heating, cooling and ventilation technologies can be used to
offset a portion of this energy.
Thermal mass is any material that can be used to store heat—heat from the Sun in the case of
solar energy. Common thermal mass materials include stone, cement and water. Historically they
have been used in arid climates or warm temperate regions to keep buildings cool by absorbing
solar energy during the day and radiating stored heat to the cooler atmosphere at night. However
they can be used in cold temperate areas to maintain warmth as well. The size and placement of
thermal mass depend on several factors such as climate, daylighting and shading conditions.
When properly incorporated, thermal mass maintains space temperatures in a comfortable range
and reduces the need for auxiliary heating and cooling equipment.
A solar chimney (or thermal chimney, in this context) is a passive solar ventilation system
composed of a vertical shaft connecting the interior and exterior of a building. As the chimney
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warms, the air inside is heated causing an updraft that pulls air through the building.
Performance can be improved by using glazing and thermal mass materials in a way that mimics
greenhouses
6.4 Water treatment
Solar still, Solar water disinfection, Solar desalination, and Solar Powered Desalination Unit
Application of SODIS technology in Indonesia to water disinfection
Solar distillation can be used to make saline or brackish water potable. The first recorded
[49]
instance of this was by 16th century Arab alchemists.
A large-scale solar distillation project
was first constructed in 1872 in the Chilean mining town of Las Salinas. The plant, which had
solar collection area of 4,700 m², could produce up to 22,700 L per day and operated for
40 years. Individual still designs include single-slope, double-slope (or greenhouse type),
vertical, conical, inverted absorber, multi-wick, and multiple effect. These stills can operate in
passive, active, or hybrid modes. Double-slope stills are the most economical for decentralized
domestic purposes, while active multiple effect units are more suitable for large-scale
applications.
Small scale solar powered sewerage treatment plant
Solar energy may be used in a water stabilisation pond to treat waste water without chemicals or
electricity. A further environmental advantage is that algae grow in such ponds and consume
carbon dioxide in photosynthesis.
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6.5 Cooking
Solar cooker
The Solar Bowl in Auroville, India, concentrates sunlight on a movable receiver to produce steam for
cooking.
Solar cookers use sunlight for cooking, drying and pasteurization. They can be grouped into
three broad categories: box cookers, panel cookers and reflector cookers. repositioned to track
the Sun. The solar bowl is a concentrating technology employed by the Solar Kitchen in
Auroville, India, where a stationary spherical reflector focuses light along a line perpendicular to
the sphere's interior surface, and a computer control system moves the receiver to intersect this
line. Steam is produced in the receiver at temperatures reaching 150 °C and then used for process
[60]
heat in the kitchen.
Unglazed transpired collectors (UTC) are perforated sun-facing walls used for preheating
ventilation air. UTCs can raise the incoming air temperature up to 22 °C and deliver outlet
temperatures of 45–60 °C. The short payback period of transpired collectors (3 to 12 years)
[67]
makes them a more cost-effective alternative than glazed collection systems.
As of 2003, over
80 systems with a combined collector area of 35,000 m² had been installed worldwide, including
an 860 m² collector in Costa Rica used for drying coffee beans and a 1,300 m² collector in
Coimbatore, India used for drying marigolds.
6.6 Electrical generation
Sunlight can be converted into electricity using photovoltaics (PV), concentrating solar power
(CSP), and various experimental technologies. PV has mainly been used to power small and
medium-sized applications, from the calculator powered by a single solar cell to off-grid homes
powered by a photovoltaic array. For large-scale generation, CSP plants like SEGS have been the
norm but recently multi-megawatt PV plants are becoming common. Completed in 2007, the
14 MW power station in Clark County, Nevada and the 20 MW site in Beneixama, Spain are
characteristic of the trend toward larger photovoltaic power stations in the US and Europe.
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11 MW Serpa solar power plant in Portugal
A solar cell, or photovoltaic cell (PV), is a device that converts light into direct current using the
photoelectric effect. The first solar cell was constructed by Charles Fritts in the 1880s. The
earliest significant application of solar cells was as a back-up power source to the Vanguard I
satellite, which allowed it to continue transmitting for over a year after its chemical battery was
exhausted. The successful operation of solar cells on this mission was duplicated in many other
Soviet and American satellites, and by the late 1960s, PV had become the established source of
power for them. Photovoltaics went on to play an essential part in the success of early
commercial satellites such as Telstar, and they remain vital to the telecommunications
infrastructure today.
The high cost of solar cells limited terrestrial uses throughout the 1960s. This changed in the
early 1970s when prices reached levels that made PV generation competitive in remote areas
without grid access. Early terrestrial uses included powering telecommunication stations, offshore oil rigs, navigational buoys and railroad crossings. These off-grid applications have proven
very successful and accounted for over half of worldwide installed capacity until 2004.
Building-integrated photovoltaics cover the roofs of the increasing number of homes.
The 1973 oil crisis stimulated a rapid rise in the production of PV during the 1970s and early
[77]
1980s.
Economies of scale which resulted from increasing production along with
improvements in system performance brought the price of PV down from 100 USD/watt in 1971
[78]
to 7 USD/watt in 1985.
Steadily falling oil prices during the early 1980s led to a reduction in
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funding for photovoltaic R&D and a discontinuation of the tax credits associated with the Energy
Tax Act of 1978. These factors moderated growth to approximately 15% per year from 1984
through 1996.
6.7 Concentrating solar power
Concentrating solar power
Solar troughs are the most widely deployed and the most cost-effective CSP technology.
Concentrated sunlight has been used to perform useful tasks since the time of ancient China. A
legend claims that Archimedes used polished shields to concentrate sunlight on the invading
Roman fleet and repel them from Syracuse. Auguste Mouchout used a parabolic trough to
produce steam for the first solar steam engine in 1866, and subsequent developments led to the
use of concentrating solar-powered devices for irrigation, refrigeration and locomotion.
The PS10 concentrates sunlight from a field of heliostats on a central tower.
A solar trough consists of a linear parabolic reflector that concentrates light onto a receiver
positioned along the reflector's focal line. The reflector is made to follow the Sun during the
daylight hours by tracking along a single axis. Trough systems provide the best land-use factor of
any solar technology.
A solar power tower uses an array of tracking reflectors (heliostats) to concentrate light on a central
receiver atop a tower. Power towers are less advanced than trough systems but offer higher efficiency
and better energy storage capability The Solar Two in Barstow, California and the Planta Solar 10 in
Sanlucar la Mayor, Spain are representatives of this technology.
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6.8 Solar vehicles
Solar vehicle, Electric boat, and Solar balloon
Australia hosts the World Solar Challenge where solar cars like the Nuna3 race through a 3,021 km
(1,877 mi) course from Darwin to Adelaide.
Development of a solar powered car has been an engineering goal since the 1980s. The World
Solar Challenge is a biannual solar-powered car race, where teams from universities and
enterprises compete over 3,021 kilometres (1,877 mi) across central Australia from Darwin to
Adelaide.
In 1987,
when
was founded,
thespeed
winner's
speed
was 67
kilometres
hour
(42 mph) and
by 2007
theitwinner's
average
had average
improved
to 90.87
kilometres
perper
hour
(56.46 mph). The North American Solar Challenge and the planned South African Solar
Challenge are comparable competitions that reflect an international interest in the engineering
and development of solar powered vehicles.
Some vehicles use solar panels for auxiliary power, such as for air conditioning, to keep the
interior cool, thus reducing fuel consumption.
Helios UAV in solar powered flight
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In 1974, the unmanned Sunrise II plane made the first solar flight. On 29 April 1979, the Solar
Riser made the first flight in a solar powered, fully controlled, man carrying flying machine,
reaching an altitude of 40 feet (12 m). In 1980, the Gossamer Penguin made the first piloted
flights powered solely by photovoltaics. This was quickly followed by the Solar Challenger
which crossed the English Channel in July 1981. In 1990 Eric Raymond in 21 hops flew from
California
to North
Carolina
using
solar power.
Developments
then
turned culminating
back to unmanned
aerial vehicles
(UAV)
with the
(1997)
and subsequent
designs,
in the
Pathfinder
Helios which set the altitude record for a non-rocket-propelled aircraft at 29,524 metres
(96,860 ft) in 2001. The Zephyr, developed by BAE Systems, is the latest in a line of recordbreaking solar aircraft, making a 54-hour flight in 2007, and month-long flights are envisioned
by 2010.
A solar balloon is a black balloon that is filled with ordinary air. As sunlight shines on the
balloon, the air inside is heated and expands causing an upward buoyancy force, much like an
artificially heated hot air balloon. Some solar balloons are large enough for human flight, but
usage is generally limited to the toy market as the surface-area to payload-weight ratio is
[120]
relatively high.
6.9 Energy storage methods
Thermal mass, Thermal energy storage, Phase change material, Grid energy storage, and V2G
Solar Two's thermal storage system generated electricity during cloudy weather and at night.
Storage is an important issue in the development of solar energy because modern energy systems
[123]
usually assume continuous availability of energy.
Solar energy is not available at night, and
the performance of solar power systems is affected by unpredictable weather patterns; therefore,
storage media or back-up power systems must be used.
Thermal mass systems can store solar energy in the form of heat at domestically useful
temperatures for daily or seasonal durations. Thermal storage systems generally use readily
available materials with high specific heat capacities such as water, earth and stone. Welldesigned systems can lower peak demand, shift time-of-use to off-peak hours and reduce overall
heating and cooling requirements. Phase change materials such as paraffin wax and Glauber's
salt are another thermal storage media. These materials are inexpensive, readily available, and
can deliver domestically useful temperatures (approximately 64 °C). The "Dover House" (in
Dover, Massachusetts) was the first to use a Glauber's salt heating system, in 1948.
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Solar energy can be stored at high temperatures using molten salts. Salts are an effective storage
medium because they are low-cost, have a high specific heat capacity and can deliver heat at
temperatures compatible with conventional power systems. The Solar Two used this method of
energy storage, allowing it to store 1.44 TJ in its 68 m³ storage tank with an annual storage
efficiency of about 99%.Off-grid PV systems have traditionally used rechargeable batteries to
store
excess
electricity.
With give
grid-tied
can be sent
todeliver
the transmission
grid. Net
metering
programs
thesesystems,
systems excess
a creditelectricity
for the electricity
they
to the
grid. This credit offsets electricity provided from the grid when the system cannot meet demand,
effectively using the grid as a storage mechanism.
Pumped-storage hydroelectricity stores energy in the form of water pumped when energy is
available from a lower elevation reservoir to a higher elevation one. The energy is recovered
when demand is high by releasing the water to run through a hydroelectric power generator.
6.10 Development, deployment and economics
Deployment of solar power to energy grids
Nellis Solar Power Plant, the largest photovoltaic power plant in North America
Beginning with the surge in coal use which accompanied the Industrial Revolution, energy
consumption has steadily transitioned from wood and biomass to fossil fuels. The early
development of solar technologies starting in the 1860s was driven by an expectation that coal
would soon become scarce. However development of solar technologies stagnated in the early
20th century in the face of the increasing availability, economy, and utility of coal and
petroleum.[130]
The 1973 oil embargo and 1979 energy crisis caused a reorganization of energy policies around
[131][132]
the
world focused
and brought
renewed programs
attention to
developing
solar technologies.
Deployment
strategies
on incentive
such
as the Federal
Photovoltaic Utilization
Program in
the US and the Sunshine Program in Japan. Other efforts included the formation of research
facilities in the US (SERI, now NREL), Japan (NEDO), and Germany (Fraunhofer Institute for
[133]
Solar Energy Systems ISE).
Between 1970 and 1983 photovoltaic installations grew rapidly, but falling oil prices in the early
1980s moderated the growth of PV from 1984 to 1996. Since 1997, PV development has
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accelerated due to supply issues with oil and natural gas, global warming concerns (see Kyoto
Protocol), and the improving economic position of PV relative to other energy
technologies. [citation needed] Photovoltaic production growth has averaged 40% per year since 2000
[42]
and installed capacity reached 10.6 GW at the end of 2007.
Since 2006 it has been economical
for investors to install photovoltaics for free in return for a long term power purchase agreement.
[134]
50%
of 2009.
commercial
systems
were installed
in this manner
in 2007power
and it is
90% and
will by
Nellis
Air Force
Base is receiving
photoelectric
forexpected
about 2.2that
¢/kWh
[135][136]
grid power for 9 ¢/kWh.
[137]
Commercial solar water heaters began appearing in the United States in the 1890s.
These
systems saw increasing use until the 1920s but were gradually replaced by cheaper and more
reliable heating fuels.[138] As with photovoltaics, solar water heating attracted renewed attention
as a result of the oil crises in the 1970s but interest subsided in the 1980s due to falling
petroleum prices. Development in the solar water heating sector progressed steadily throughout
the 1990s and growth rates have averaged 20% per year since 1999. [41] Although generally
underestimated, solar water heating is by far the most widely deployed solar technology with an
[41]
estimated capacity of 154 GW as of 2007.
Solar installations in recent years have also largely begun to expand into residential areas, with
governments offering incentive programs to make "green" en stallation (sized between 1.3 kW
and 5 kW) is estimated at 18 to 23 years, considering such cost factors as parts, installation and
maintenance, as well as the average energy production of a system on an annual basis.
WEEK 7
7.0
Entropy
The first law of thermodynamics deals with the property energy and the
conservation of it. The second law leads to the definition of a new property
called entropy. Entropy is a somewhat abstract property, and it is difficult
to give a physical description of it.
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Entropy is best understood and
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appreciated by studying its uses in commonly encountered engineering
processes and this is what we intend to do. Entropy is a non conserved
property, and there is no such thing as a conservation of entropy principle.
The second law of thermodynamics of the leads to expressions that
involve inequalities, an irreversible (i.e. actual) heat engine, for example, is
less efficient than a reversible one operating between the same two
thermal energy reservoirs. Likewise, an irreversible refrigerator or a heat
pump has a lower coefficient of performance (COP) than a reversible one
operating between the same temperature limits.
Entropy is an extensive property of a system and some times is referred to
as total entropy. Entropy per unit mass, designated ‘S’ is an intensive
property and has the unit  KJ kg.k  . The term entropy is generally used to


refer to both total entropy and entropy per unit mass and mathematically
defined as:
ds =
dQ
T
………………………………………………..(7.0)
Consider a reversible adiabatic process for any system on a P-V diagram.
This is represented by line 1-2 on figure (7), shown below.
P
T2 = T3 = T1
2
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s
s
e
le c
o
b
r
i
s p
r
e c
i
v t
e a
b
R a
i
d
a
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Figure (7): Shows a Cycle on P-V Diagram
Let us suppose that it is possible for the system to undergo a reversible
isothermal process at temperature T 1 from 2 to 3 and then be restored to
its original state by a second reversible adiabatic process from 3 to 1.
Now by definition an adiabatic process is one in which no heat flows to or
from the system. Hence the only heat transferred is from 2 to 3 during the
isothermal process.
enclosed area.
The work done by the system is given by the
We therefore have a system undergoing a cycle and
developing a net work output while drawing heat from a reservoir at one
fixed temperature. This is impossible because it violates the second law of
thermodynamics. Therefore the original supposition is wrong, and it is not
possible to have two reversible adiabatic processes passing through the
same state 1.
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Now one of the characteristics of a property of a system is that there is one
unique line which represents a value of the property on a diagram of
properties.
(For example, the line 2-3 on figure 1.1 represents the
isothermal at T1).
Hence there must be a property represented by a
reversible adiabatic process. This property is called entropy, S.
It follows that there is no change of entropy in a reversible adiabatic
process. Each reversible adiabatic process represents a unique value of
entropy. On a P-V diagram a series of reversible adiabatic processes
appear as shown in figure 1.2 (a), each line representing one value of
entropy. This is similar to figure 1.2 (b) in which a series of isothermals is
drawn, each representing one value of temperature.
P
P
S4
T4
S3
T3
S2
T2
S1
T1
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(a)
V
V
(b)
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Figure 7.1: Shows a Series of Constant Entropy and Constant
Temperature Lines on a P-V Diagram
In order to be able to define entropy in terms of the thermodynamics
property of a system a vigorous approach is necessary, the non-flow
energy equation for a reversible process will be considered to achieve that
as shown below:
dQ = du + pdv …………………………………………..……..(7.1)
For perfect gas
dQ = c r dT + RT dv
v
…………………………….……..……..(7.2)
Equation (1.3) can be integrated after dividing through by T,
i.e.
dQ
T
=
c r dT
T
+
Rdv
v
Also for an adiabatic process, dQ = 0
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dQ
T
=
c r dT
T
+
Rdv
v
= 0 ………………………….. (7.3)
This must mean that dividing through by T is the step which implies the
restriction of the second law, and the important fact that the change of
entropy is zero, we can say, therefore, dQ
T
= 0 for a reversible adiabatic
process. For any other reversible process dQ T
m
0.
This result can be shown to apply to all working substances, i.e.
ds =
Where:
dQ
T
……………………..…………………….. (7.4)
s is entropy.
The change in entropy of a system during a process can be determined by
integrating equation (7.4) between the initial and the final states, i.e.
∆S = S2 - S1 =
∫
2
1
dQ
T
………………..…………………….. (7.5)
The units of entropy are given by kilojoules per kilogram divided by K.
That is, the units of specific entropy, S, are KJ/kgK. The total entropy (S)
is given by
S = ms, the units is KJ
K
from equation (7.5)
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DQ = Tds
Or for any reversible process
Q =
7.1
∫
2
1
Tds ………………..…………………….. (7.6)
Property Diagrams Involving Entropy
In the second law analysis, it is very helpful to plot the processes on
diagrams for which one of the coordinates is entropy. The two diagrams
commonly used in the second law analysis are the temperature-entropy
and the enthalpy-entropy diagrams.
7.1.1 The T-S Diagram
For a Vapour
The T-S diagram for steam only will be considered here, is shown in figure
(1.3). The three lines of constant pressure (P1, P2 and P3) are shown in
the figure. The pressure lines in the liquid region are practically coincident
with the saturated liquid line (i.e. portions AB, EF and JK) and the
difference is usually neglected.
The pressure remains constant with
temperature when the latent heat is added, hence the pressure lines are
horizontal in the wet region (i.e. portions BC, FG and KL). The pressure
lines curve upwards in the superheat region as shown (i.e. portions CD,
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GH and LM). Thus the temperature rises as heating continues at constant
pressure.
T
P3
J
K
P2
M
P3
L
H
P1
P2
E
A
D
G
F
P1
C
B
S
Figure (7.2):
T-S Diagram for a Vapour
7.2 Determination of Dryness Fraction from Area in T-S Diagram
In steam tables the entropy of the saturated liquid and the dry saturated
vapour are represented by Sf and Sg respectively. The difference, Sg – Sf
= Sfg is also tabulated for wet steam with dryness fraction, x, we have
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S = (1 - x )S f + xSg ………………..…………………….. (7.7)
Or
S = Sf + Sg - Sf
⇒ S = Sf + xS fg …………………………………………. (7.8)
The dryness fraction is given by
x =
S − Sf
S fg
Where:
………………………..…………………….. (7.9)
x is the dryness fraction
S is the entropy of wet steam
It can be seen from equation (7.9) that the dryness fraction is proportional
to the distance of the state point from liquid line on T-S diagram shown in
figure (7.3).
T
P1
F
P1
1
G
P1
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Figure (7.3): Dryness Fraction from Area on a T-S Diagram
From figure (7.3), the dryness fraction is given by:
X1 =
distance F1
distance FG
=
S1 − S f 1
S f g1
The area under the line FG on figure (7.3) represents the specific enthalpy
of vaporization hfg. The area under line F1 is given by X 1h fg
7.3 For a Perfect Gas
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It is useful to plot lines of constant pressure and constant volume on T–S
diagram for a perfect gas. In figure (7.4) the pressure line P1 and the
volume line V1 have been drawn passing through the state point 1. Note
that a line of gasket pressure slopes less steeply than a line of constant
volume. This can be proved easily by reference to figure (7.4).
V1
T
P1
A
T2
B
1
T1
S
S1
SA
SB
Figure (7.4)
Consider, equation (7.5) between point 1 and A, we have
A
SA − S1 =
∫
1
dQ
T
Also at a constant volume dQ = Cu dT , therefore
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SA − S1 =
∫
A
dQ
T
1
 TA 

T
 1 
= C ln
v
 T2 
 ……………………………………… (7.10)
 T1 
= Cv ln
Similarly, at constant pressure of gas dQ = C p dT
S B − S1 =
∫
B
1
C p dT
T
 TB 

T
 1
= C pln
 T2 
 ……………………………………… (7.11)
T
 1
= Cv ln
7.4
The h-S Diagram
Another diagram commonly used in engineering is the enthalpy-entropy
diagram, which is quite valuable in the analysis of steady-flow devices
such as turbines, compressors and nozzles. The coordinates of an h-s
diagram represent two properties of major interest: enthalpy, which is a
primary property in the first-law analysis of the steady-flow devices and
entropy which is the property that accounts for irreversibilities during
adiabatic processes. In analyzing the steady flow of steam through an
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adiabatic turbine, for example, the vertical distance between the inlet and
the exit states (∆h ) is a measure of the work output of the turbine and the
horizontal distance (∆s ) is a measure of the irreversibilities associated with
the process as shown in figure (7.5).
h
1
Dh
2
DS
S
Figure (7.5): The h-s Diagram
WEEK 8
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8.0 WORKED EXAMPLES
Example (1): 1 kg of steam at 7 bar, entropy 6.5 KJ/kgK, is heated reversibly at
constant pressure until the temperature is 250˚C, calculate the heat
supplied, and show on a T-S diagram the area which represents the heat
flow.
Solution: At 7 bar, Sg = 6.709 KJ/kgK, hence the steam is wet, since the
actual entropy, S, is less than S g.
The dryness fraction is given by:
X =
S - Sf
Sfg
=
6.5 - 1.992
4.717
= 0.955
The enthalpy at liquid state is given by
h1 = h f + xh fg = 697 + (0.955 x 2067 )
= 2672 KJ
kg
At state (2) the steam is at 250˚C at 7 bar and is therefore superheated.
From superheat tables, h2 = 2955 KJ kg
The heat supplied is given by:
Q = h 2 - h1 = 2955 - 2672 = 283 KJ
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The T-S diagram showing the process is given below: The shaded area
represent the heat flow.
T
2
1
S
6.8
6.5
6.709
Example (2): A rigid cylinder of volume 0.025m 3 contains steam at 80 bar
and 350˚C. The cylinder is cooled until the pressure is 50 bar, calculate
the state of the steam at the cooling and the amount of heat rejected by
the steam. Sketch the process on T-S diagram indicating the area which
represents the heat flow.
Solution: Steam at 80 bar and 350˚C is superheated, and the specific
volume from tables is 0.02994 m
3
kg
.
Hence the mass of steam in the
cylinder is given by:
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m =
0.025
0.02994
= 0.835kg
The internal energy for superheated steam above 80 bar is found as
follows:
U1 = h1 − P1v1 = 2990 -
80x105 x 0.02994
103
= 2990 - 339.5 = 2750.5 KJ
kg
3
At state (2), P2 = 50 bar and V2 = 0.02994 m kg , therefore the steam is wet
and the dryness fraction is given by
X2 =
V2
Vg2
=
0.02994
0.03944
= 0.758
The internal energy at state (2)
U 2 = (1 - x 2 )U f 2 + x2Ug 2 = (0.242 x 1149 ) + (0.758 x 2597 )
= 278 + 1969 = 2247 KJ
kg
At constant volume
Q = U 2 - U1 = m(u 2 - u1 ) = 0.835(2247 - 2750.5 ) = 420 KJ
The process drawn on T-S diagram as shown below, the shaded area
representing the heat rejected by the steam.
T
80 bar
1
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T-S Diagram:
Example (3): Air at 15˚C and 1.05 bar occupies 0.02m 3. The air is heated
at instant volume until the pressure is 4.2 bar, and then cooled at constant
pressure back to the original temperature. Calculate the net heat flow to or
from the air and the net entropy changer sketch the process on T-S
diagram.
Solution:
m =
PV
RT
=
1.05 x 105 x 0.02
0.287 x 103 x 288
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T1 = 15 + 273 = 288k
For perfect gas at constant volume, P1
T3 =
4.2 x 288
T1
=
P2
T2
= 1152k
1.05
Heat supplied at constant volume
Q1− 2 = mc v (T2 − T1 ) = 0.0254 x 0.718(1152 - 288) = 15.75KJ
Heat rejected at constant pressure
Q2 − 3 = mc p (T3 − T2 ) = 0.0254 x 1.005(288 - 1152 ) = - 22.05KJ
Therefore,
Net heat flow = Q(1- 2 ) + Q2 − 3 = 15.75 − 22.05 = 6.3KJ i.e. heat rejected = 6.3KJ
Referring to T-S diagram drawn on this process:
Net decrease in entropy = S1 - S3 = (S2 - S2 ) − (S 2 − S1 )
At constant pressure, dQ = mc p dT
m(S2 - S3 ) =
∫
1152
288
mcpdT = 0.0254 x 1.005 x ln 1152  = 0.0354 KJ
k
T
 288 
At constant volume, dQ = mcv dT
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m(S2 - S3 ) =
1152
∫
288
mc v dT
T
 1152 
 = 0.0253 KJ k
288


= 0.0254 x 0.718 x ln
Therefore,
m(S2 - S3 ) = 0.0354 - 0.0253 = 0.0101 KJ
k
i.e. Decrease in entropy of air = 0.0101 KJ
K
The T-S diagram drawn for this process is shown below.
T
0.03m
4.2 bar
2
1.05 bar
3
S3
S1
S2
S
Process on T-S Diagram
8.1 Isentropic Efficiency
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The isentropic efficiency or adiabatic efficiency is defined as a parameter
which is used to measure the deviation of actual processes from the
corresponding idealized ones. Adiabatic efficiencies are defined differently
for different devices since each device is set up to perform different tasks.
Below the isentropic efficiency of turbines and compressors were
computed from figure (1.8).
h
P1
3
h3
h2
h2s
2
2s
P2
h4
h43
4
4s
h1
1
s
Figure (1.8): h-s Diagram
8.2 Isentropic Efficiency of Turbine
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The isentropic efficiency of turbine is defined as the ratio of the actual work
output of the turbine to the work output that would be achieved if the
process between the inlet state and the exit pressure were isentropic, i.e.
mathematically is given by:
ηT =
Actual turbine work (Wa )
Isentropic turbine work (Ws )
=
h3 - h 4
h 3 - h 4s
8.3 Isentropic Efficiency of Compressor
The isentropic efficiency of a compressor is defined as the ratio of the work
input required to raise the pressure of a gas to a specific value in an
isentropic manner to the actual work input, i.e. mathematically given by:
ηC =
Isentropic compressor work (Wa )
Actual compressor work (Ws )
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=
h 2s - h1
h 2 - h1
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8.4 WORKED EXAMPLES
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Example (4): Steam enters an adiabatic turbine steadily at 3mp a and 400˚C and
leaves at 50kpa and 100˚C. If the power output of the turbine is 2mw,
determine (a) the isentropic efficiency of the turbine and (b) the mass flow
rate of the steam flowing through the turbine.
Solution:
P1=3mpa
T
1
˚
400˚C
Steam
turbine
2s
2s
P2=50mpa
S
˚
P1 = 3mpa
h1 = 3209 KJ
T1 = 400°C
S1 = 6.9212 KJ
State 1:
State 2a:
kg
kg
P2 a = 50kpa 
h2 a = 2682.5 KJ kg
T2 a = 100°C 
The exit enthalpy of the steam for the isentropic process h2 s is determined
as follows:
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P2 s = 50kp a → Sf = 1.0910 KJ
State 2b:
X 2s =
S2s - Sf
Sfg
=
kg
6.9212 - 1.0910
, Sg = 7.5939 KJ
kg
, S2s = S1
= 0.897
6.5029
h2 s = h f + x 2s h fg = 340.49 + 0.897(2305.4 ) = 2407.4 KJ
kg
The isentropic efficiency of this turbine is determined by:
η =
h1 - h2s
T
b)
h1 - h2g
=
3230.9- 2682.5
= 0.667or66.7
3230.9 − 2407.4
The mass flow rate of steam through this turbine is determined from the
energy balance for steady-flow systems:
Ein = E out
mh1 = Wa out + mh 2a
Waout = in(h1 − h2 a )
 1000 KJ 
= m(3230.9 − 2682.5) KJ
s 
kg
 1mw
2mw
in = 3.65 KJ s
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Example (5): Air is compressed by an adiabatic compressor from 100kpa
and 12˚C to a pressure of 800kpa at a steady rate of 0.2KJ/s.
If the
isentropic efficiency of the compressor is 80%, determine:
a)
the exit temperature of the air and
b)
the required power input to the compress
P2=800kpa
T
2a
2s
Air
compressor
P1=100kpa
32s=s1
S
=
T1 = 285k → h1 = 285.14 KJ
kg
Pr1 = 1.1584
Pr 2
Pr1
=
p2
 800kp a
⇒
p r 2 = 1.1584
p1
 100kpa
pr 2 = 9.2672 ⇒ h2 s = 517.05 KJ
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
 = 9.2672

kg
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ηc =
h2 s − h1
h2 a − h1
⇒ 0.80 =
h2 a = 575.03 KJ
(517.05 − 285.14)
h2 a − 285.14
→ T2 a = 569.5k
kg
b)
Ein = E out
mh1 = Wa in + mh 2a
Wain = m(ha − h1a ) = 0.2(575.03 − 285.14 ) = 58.0 KW
8.5 QUIZ ONE (ENTROPY)
Q1(a): Find the entropy of 1kg of dry saturated steam at a pressure of 5.2bar. The boiling
o
point of water at this pressure is given as 152.6 C and its total heat at this
temperature is 2110kJ/kg. Ans. 6.82kJ/kgK
(b): Calculate the entropy of 1kg of wet steam with dryness fraction of 0.9 at a
pressure of 8.4bar. Ans. 6.186kJ/kgK
Q2:
Determine the entropy per kg of superheated steam at a pressure of 20bar and
a temperature of 250oC. Assume cp for superheated steam as 2.2kJ/kgk.
Ans.6.5kJ/kgK
Q3 1:1kg of steam at 20bar, dryness fraction 0.9, is heated reversibly at constant
o
pressure to a temperature of 300 C. Calculate the heat supplied, and the change
of entropy, and show the process on a T-s diagram, indicating the area which
represents the heat flow. Answer: 415kJ/kg,0.8173kK/kgK
o
Q4: Steam at 0.05bar, 100 C is to be condensed completely by a reversible constant
pressure process. Calculate the heat rejected per kilogram of steam, and the
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change of specific entropy. Sketch the process on a T-s diagram and shade in the
area which represents the heat flow. Answer: 2550kJ/kg, 8.292kJ/kgK
Q5: 0.05kg of steam at 10bar, dryness fraction 0.84, is heated reversibly in a rigid vessel
until the pressure is 20bar. Calculate the change of entropy and the heat supplied.
Show the area which represents the heat supplied on T-s diagram.
Answer: 0.0704kJ/kgK, 36.85kJ
Q6: 1m3 of air is heated reversibly at constant pressure from 15 to 300oC and is then
cooled reversibly at constant volume back to the initial temperature. The initial
pressure is 1.03bar. Calculate the net heat flow and the overall change of entropy,
and sketch the processes on a T-s diagram. Answer: 101.5kJ; 0.246kJ/K
Q7: 1kg of steam undergoes a reversible isothermal process from 20bar and 250 oC to a
pressure of 30bar. Calculate the heat flow, stating whether it is supplied or reject,
and sketch the process on a T-s diagram. Answer -135kJ/kg
Q8a. Describe the ideal process for an (i) adiabatic turbine (ii) adiabatic compressor
8b. Define the isentropic efficiency for each device.
Q9: Steam enters an adiabatic turbine at 8MPa and 500oC with a mass flow rate of 3kg/s and leaves at
30 kPa. The isentropic efficiency of the turbine is 0.90. Neglecting the kinetic energy change of the
steam, determine: (i) the temperature at the turbine exit (ii) the power output of the turbine. Answer:
(i) 69.1oC (ii) 3052kW
o
o
Q10: Steam enters an adiabatic turbine at 6MPa, 600 C, and 80m/s and leaves at 50kPa, 100 C,
and140m/s. If the power output of the turbine is 5MW, determine (i) the mass flow rate of the steam
flowing through the turbine (ii) the isentropic efficiency of the turbine. Answer: (i) 5.16kg/s (ii) 83.7
percent
o
o
Q11: Air is compressed by an adiabatic compressor from 95kPa and 27 C to 600kPa and 277 C. Assuming
variable specific heats and neglecting the changes in kinetic and potential energies, determine, (i) the
isentropic efficiency of the compressor (ii) the exit temperature of air if the process were reversible.
Answers: (i) 81.9 percent (ii) 505.5K
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WEEK 9
9.0 Pure Substances
A substances that has a fixed chemical composition throughout is called a pure substance,
water, nitrogen, helium and carbon dioxide, for example, are all pure substances.
A pure substance does not have to be of a single chemical element or compound, however. A
mixture of various chemical elements or compounds also qualifies as a pure substance as long as
the mixture is homogenous. Air, for example, is a mixture of several gases, but it is often
considered to be a pure substance because it has a uniform chemical composition. However, a
mixture of oil and water is not a pure substance. Since oil is not soluble in water, it will collect
on top of the water, forming two chemically dissimilar regions.
A mixture of two or more phases of a pure substance is still a pure substance as long as the
chemical composition of all phases is the same. A mixture of ice and liquid water, for example,
is a pure substance because both phases have the same chemical composition. A mixture of
liquid air and gaseous air, however, is not a pure substance since the composition of liquid air is
different from the composition of gaseous air and thus the mixture is no longer chemically
homogenous.
9.1 Properties and State
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Any characteristic of the substance which can be observed or measured is called a property of
the substance. Examples of properties are pressure, volume and temperature. This type of
property which is dependent upon the physical and chemical structure of the substance is called
an internal or thermostatic property.
If a property can be varied at will, quite independently of other properties, then the property is
said to be an independent property. The temperature and pressure of a gas for example, can be
varied quite independently of each other and thus, in this case, temperature and pressures are
independent properties.
It will be found, however, when discussing the formation of a vapour, that the temperature at
which a liquid boils depends upon the pressure at which the formation of the vapour is
occurring. Here, if the pressure is fixed then the temperature becomes dependent upon the
pressure. Hence the pressure is an independent property but the temperature is a dependent
property.
A knowledge of the various thermostatic properties of a substance defines the state of the
substance. If a property, or properties, are changed, then the state is changed. Properties are
independent of any process which any particular substance may have passed through from one
state to another, being dependent only upon the end states. In fact, a property can be identifies
if it is observed to be a formation of state only.
9.2 Property Diagram for Phase-Change Processes
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The variations of properties during phase-change processes are best studied and understand
with the help of property diagram. P-v diagram for pure substances will be discuss as follow:
9.2.1 P-V Diagram for Pure Substance
The general shape of the p-v diagram of a pure substance is shown in figure (1.9), with
temperature constant lines on this diagram have a downward trend.
Consider a piston-cylinder device that contains liquid water at a certain pressure and
temperature.
Water at this state exists as a compressed liquid. The water is allowed to
exchange heat with the surroundings so its temperature remains constant. As the pressure
decreases, the volume of the water will increase slightly. When the pressure increases the
saturation-pressure value at the specified temperature, the water will start to boil. During this
vapourisation process, both the temperature and the pressure remain constant, but the specific
volume increases once the last drop of liquid is vapourized, further reduction in pressure results
in a further increases in specific volume.
If the process is repeated for other temperatures, similar paths will be obtained for the phasechange processes. Connecting the saturated liquid and saturated vapour states by a curve, we
obtain the p-v diagram of a pure substance, as shown in figure (9).
Critical
point
P
Saturated
vapour line
Superheated
vapour region
T2=Constant St1
Compressed
liquid region
Saturated liquidvapour region
Saturated
vapour line
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Figure (9): P-V Diagram of a Pure Substance
9.3 Two property rules for pure substances
The state of a pure substance of given mass can be fixed by specifying two properties, provided that :
i.
ii.
The system is in equilibrium
Gravity, motion, electricity, magnetism and capillarity are without significant
effects.
WEEK 10
10.0 Pressure Law
This states that the pressure of a gas mixture is equal to the sum of the pressures each gas
would exert if it existed alone at the mixture temperature and volume. Mathematically express
as
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Pressure of mixture = Partial pressure of steam + Partial pressure of air
10.1 Ideal Gas Law
Ideal gas law must to obey equation of state that is any equation that relates the pressure,
temperature and specific volume of a substance is called an equation of state. Mathematically
expressed as:
PV = RT
Where:
P is the absolute pressure, T is the absolute temperature and V is the specific
volume
The gas constant ‘R’ is determined from
R =
Ru
m
Where:
Ru is the universal gas constant and m is the molar mass of the gas.
The properties of an ideal gas at two different states are related to each other by
P1V1
T1
=
P2V2
T2
10.2 WORKED EXAMPLES
Example 8: A vessel of volume 0.2m3 contains nitrogen at 1.013 bar and 15°C. If 0.2kg of
nitrogen is now pumped into the vessel, calculate the new pressure when the vessel has
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returned to its initial temperature. The molar mass of nitrogen is 28kg/kmol, and it may be
assumed to be a perfect gas.
Solution:
Specific gas constant, R =
Ru
m
=
8314.5
28
= 296.95 Nm
kgk
for initial conditions.
P1V1 = m1RT1
⇒ m1 =
P1V1
RT1
−
1.013 x105 x 0.2
296.95 x 288
= 0.237 kg
The mass of nitrogen added is 0.2kg, hence m2 = 0.2 + m1
i.e
m2 = 0.2 + 0.237 = 0.437kg
For final conditions
P2V 2 = m 2 RT2
But V2 = V1 and T2 = T1 , therefore
P2 =
m 2 RT2
V2
=
0.437 x 296.95 x 285
10 5 x 0.2
= 1.87 bar
Example 9: A quantity of gas has a pressure of 350 KN/m 2 when its volume is 0.03m3 and its
temperature is 35°C. If the value of a gas constant is 0.29 KJ/kgk, determine the mass of gas
present. If the pressure of this gas is now increased to 1.05 MN/m 2 while the volume remains
constant, what will be the new temperature of the gas?
Solution:
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PV = mRT
PV
⇒m=
RT
=
350 x103 x0.3
0.29 x103 x308
= 0.118kg
For the second part of the pressure,
P1V1
T1
⇒
=
P1
T1
P2V2
T2
=
P2
T2
and in this case V1 = V2
or T2 = T1 =
⇒ T2 = 308 x
1.05 x 10
P2
P1
6
0.35 x 106
= 924k
∴ t 2 = 651°C
10.3 QUIZ (TWO)
Q1: 2kg of gas , occupying 0.7m3, had an original temperature of 15 oC. It was then heated at constant
volume until its temperature becomes 135oC. Determine the heat transferred to the gas and its final
pressure. Take cv= 0.72kJ/kg K and R=0.29kJ/kg K . Answer: 172.8kJ and 338.02KN /m 2
2
3
o
Q2: A gas whose pressure, volume and temperature are 275KN/m ,0.09m and 185 C,respectively, has
o
its state changed at constant pressure until its temperature becomes 15 C. Determine the heat
transferred from the gas and the work done on the gas during the process. Take R=0.29KJ/kg K, cp=1.005
KJ/kg K. Answer: -31.78KJ, -9.19KJ
2
3
Q3: A quantity of gas has an initial pressure of 140KN/m and volume 0.14m . It is then compressed to a
2
pressure of 700KN/m while the temperature remains constant. Determine the final volume of the gas.
3
Answer: 0.028m
Q4: A quantity of gas has an initial volume of 0.06m3 and a temperature and a temperature of 15oC. It is
expanded to a volume of 0.12m 3 while the pressure remains constant. Determine the final temperature
of the gas. Answer: 303oC
o
Q5: A mass of gas has an initial pressure of 1bar and a temperature of 20 C. The temperature of the gas
o
is now increased to 550 C while the volume remains constant. Determine the final pressure of the gas.
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Answer: 2081bar
Q6: A mass of air has an initial pressure of 1.3MN/m2, volume 0.014m3 and temperature 135oC. It is
expanded until its final pressure is 275KN/m 2 and its volume becomes 0.056m 3. Determine:
o
(a) The mass of air (b) the final temperature .Take R=0.287KJ/kgK (Answer: (a) 0.1555kg (b) 72 C
Q7: An air receiver has a capacity of 0.85m3 and contains air at a temperature of 15oC and pressure of
2
275KN/m . An additional mass of 1.7kg is pumped into the receiver. It is then left until the temperature
becomes 15oC once again. Determine:
(a) The new pressure of the air in the receiver.
o
(b) The specific enthalpy of the air at 15 C if it is assumed that the specific enthalpy of the air is zero
o
at 0 C. Take cp=1.005KJ/kgK,
cv= 0.715KJ/kgK .
2
Answer : (a) P2=442KN/m (b) h=15.075KJ/kg
WEEK 11
11.0 Ideal and Real Gases
An ideal gas is defined as a gas whose molecules are spaced far apart so that the behaviour of a
molecule is not influenced by the presence of other molecules, a situation encountered at low
densities. The P-V-T behaviour of an ideal gas is expressed by the simple relation Pv=RT, which
is called the ideal-gas equation of state. The P-v-T behaviour of real gases is expressed by more
complex equations of state or by PV=ZRT, where Z is the compressibility factors.
11.1 WORKED EXAMPLES
3
Example 10: A vessel of volume 0.4m contains 0.45kg of carbon-monoxide and 1kg of air, at
15°C. Calculate the partial pressure of each constituent and the total pressure in the vessel.
The gravimetric analysis of air is to be taken as 23.3% oxygen and 76.79% nitrogen. Take the
molar masses of carbon monoxides, oxygen and nitrogen as 28, 32 and 28 kg/kmol.
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Solution:
Mass of oxygen present =
23.3
100
x 0.233kg
The specific gas constant is give by
~
R =
R
~
m
Equation of state is given by
Pv = mRT
~
mRT
⇒ P =
~
mv
For a constituent
~
Pi = m~i R T
mi v
For oxygen
PO 2 =
0.233 x 8.314 x 288
32 x 0.4
= 43.59 KN
m2
= 0.4359bar
For Nitrogen
PN 2 =
0.767 x 8.3145 x 288
28x 0.4
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= 163.99 KN
m2
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= 1.6399bar
For carbon monoxide
PCO = 0.45 x 8.3145 x 288 = 96.21 KN 2
m
28 x 0.4
= 0.9621bar
The total pressure in the vessel is given by:
P = ∈ pi = PO2 + PN2 + PCO
= 0.436 + 1.640 + 0.962 = 3.038 bar
Example 11: The gravimetric analysis of air is 23.14% oxygen, 75.53% nitrogen, 1.28% argon,
0.05% carbon dioxide. Calculate the specific gas constant for air and the molar mass. Take the
molar masses as 31.999 kg/kmol oxygen, 28.013 kg/kmol nitrogen, 39.948 kg/kmol argon and
44.010 kg/kmol carbon dioxide.
Solution:
~
R =
R
~
m
RO 2 =
RN 2 =
RAR =
8.3145
31.999
8.3145
28.013
8.3145
39.948
= 0.2598 KJ
= 0.2968 KJ
= 0.2081 KJ
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kgk
kgk
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8.3145
RCO 2 =
= 0.1889 KJ
44.010
kgk
PV = mRT
mRT = ∑ mi RiT
R = ∑

mi
R
m 
i
R = (0.2314 x0.2598) + (0.7553 x0.2968) + (0.0138 x0.2081)
+ (0.0005 x0.1889 ) = 0.2871 KJ
kgK
i.e. Specific gas constant for air = 0.2871 KJ/kgK
~
−
m =
=
R
12
8.3145
0.2871
= 28.960
kg
kmol
Example: A mixture of gases at a temperature of 150 °C has a pressure of 4 bar. A sample is
analysed and the volumetric analysis is found to be CO 2 14%, 02 5%, N81%. Determine the
gravimetric analysis and partial pressure of the gases in the mixture. If 2.3kg of mixture are
cooled at constant pressure to 15°C, find the final volume.
Solution:
The gravimetric analysis can be set as follows:
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i
vi
−
m
v
kg
kmol
 vi m kg
 v  i kmol


mi
m
CO2
0.14
44
6.16
0.202
O2
0.05
32
1.60
0.053
N2
0.81
28
22.68
0.745
1.00
30.44
The partial pressure are:
CO2
PC02 = 0.14 x 4 = 0.56 bar
O2
P02 = 0.05 x 4 = 0.30 bar
N2
PN2 = 0.81 x 4 = 3.24 bar
The amount of substance in 2.3kg is
n =
2.3
30.44
= 0.0756 Kmol
Final volume of mixture is:
~
V =
n RT
P
=
0.0756 x 8.3145 x 288
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100 x 4
= 0.453m 3
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11.2 Exothermic and Endothermic Reactions
In many chemical reactions energy is liberated, in others energy is absorbed. A chemical
reaction in which energy is liberated is called an exothermic reaction. Thus, fuel-burning reactions are
exothermic reactions. A chemical reaction in which energy is absorbed is called an endothermic
reaction. For example, in the manufacture of coal gas there is a reaction between carbon and carbon
dioxide which produces carbon monoxide. Energy is required to carryout this reaction, so it is an
endothermic reaction.
11.3 QUIZ (THREE)
Q1:A gas mixture consists of 8 kmol of H2 and 2kmol of N2. Determine the mass of each gas and the
apparent gas constant of the mixture. Answer: 16kg,56kg, 1.155kJ/kgK
Q2: A mixture of carbon monoxide and oxygen is to be prepared in the proportion of 7kg to 4kg in a
3
o
vessel of 0.3m capacity. If the temperature of the mixture is 15 C, determine the pressure to which the
o
vessel is subject. If the temperature is raised to 40 C, what will then be the pressure in the vessel?
Answer: 29.94bar, 32.54bar
Q3:For the mixture of problem Q1, calculate the volumetric analysis, the molar mass and the
characteristic gas constant. Calculate also the total amount of substance in the mixture. Answer: 33.3%
O2; 66.7% CO; 29.3kg/kmol; 0.283kJ/kgK;0.375kmol
Q4: An exhaust gas is analysed and is found to contain, by volume, 78% N 2, 12% CO2, and 10%O2. What
is the corresponding gravimetric analysis? Calculate the molar mass of the mixture, and the density if
the temperature is 550oC and the total pressure is 1bar. Answer: 72% N2, 17.4% CO2, 10.6% O2;
30.33kg/kmol; 0.443kg/m3
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3
Q5: A vessel of 3m capacity contains a mixture of nitrogen and carbon dioxide, the analysis by volume
o
showing equal quantities of each. The temperature is 15 C and the total pressure is 3.5bar. Determine
the mass of each constituent. Answer: 6.14kg N2; 9.65kg CO2
WEEK 12
12.0 Fuels and Combustion
Any material that can be burned to release energy is called a fuel. Most familiar fuels consist
primarily of hydrogen and carbon. They are called hydro carbon fuels and are denoted by the
general formula CnHm.
A chemical reaction during which a fuel oxidized and a large quantity of energy is released is
called combustion. The oxidizer most often used in combustion processes is air, for obvious
reasons – it is free and readily available. Pure oxygen ‘O 2’ is used as an oxidizer only in some
specialized applications, such as cutting and welding, where air cannot be used.
12.1 Classification of Fuels into Solid, Liquid and Gaseous
The main constituent of coal is carbon. Coal also contains varying amounts of oxygen, hydrogen,
nitrogen, sulphur, moisture and ash. It is difficult to give an exact mass analysis for coal since its
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composition varies considerably from one geographical location to the next and even within the
same geographical location.
Most liquid hydro carbon fuels are a mixture of numerous
hydrocarbons and are obtained from crude oil by distillation. The most volatile hydrocarbons
vaporize first, forming what we know as gasoline. The less volatile fuels obtained during
distillation are kerosene, diesel fuel and fuel oil. The composition of a particular fuel depends
on the source of the crude oil as well as on the refinery.
Although liquid hydrocarbon fuels are mixtures of many different hydrocarbons, they are usually
considered to be a single hydrocarbon for convenience. For example, gasoline is treated as
octane, C8H18 and the diesel fuel as dodecane, C 12H26. Another common liquid hydrocarbon fuel
is methyl alcohol, CH3OH, which is also called methanol and is used in some gasoline blends.
The gaseous hydrocarbon fuel natural gas, which is a mixture of methane and smaller amounts
of other gases, is sometimes treated as methane, CH 4.
Carbon monoxide is an important
gaseous fuel which is constituent of other gas mixtures and is also a product of the incomplete
combustion of carbon.
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Rocket Blast Off
A rocket blasts off from its launching pad at Cape Canaveral, Florida. Most of the rocket is filled with
liquid fuel and a liquid oxidizing agent. The fuel and oxidizing agent mix and ignite in the combustion
chamber; the presence of the oxidizing agent ensures that the fuel burns far more efficiently than it
could if it depended on the surrounding air for oxygen.
12.2 Combustion Equations
During a combustion process, the components that exist before the reaction are called reactants
and the components that exist after the reaction are called products, as shown in figure below.
Reactants
Products
Reaction
Chamber
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Consider, for example, the combustion of 1 Kmol of carbon with 1 Kmol of pure oxygen, forming
carbon dioxide,
C + O2
CO2 ……………………… (12.2)
Hence C and O2 are the reactants since they exist before combustion, and CO2 is the product
since it exists after combustion.
As you will recall from your chemistry courses, chemical by using relative atomic masses,
Therefore,
2 H 2 + O 2 → 2H 2O
2 x (2 x 1) + (2 x 16) → 2 x [(2 x 1) + 16]
4kg H 2 + 32kgO 2 → 36kgH 2O
Or
1kgH 2 + 8kgO 2 → 9kgH 2O …………………………..(12.3)
Since oxygen is accompanied by nitrogen it air is supplied for the combustion, then this nitrogen
should be included in the equation. As nitrogen is inert as far as the chemical reaction is
concerned, it will appear on both sides of the equation.
With 1 kmol of oxygen there are 79 21kmol
2 H 2 + O2 + 79
i)
21
N 2 → 2 H 2O + 79
of nitrogen, hence equation (12.3) becomes:
21
N 2 …………………….. (12.4)
Again equation (1) becomes
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C + O2 +
79
21
N 2 → CO2 +
79
21
N 2 ……………………………… (12.5)
Equation (5) is the complete combustion of carbon to carbon dioxide including the nitrogen.
Considering the volumes of reactants and products
O volume C + 1 volume O 2 + 79
21
volume N 2
⇒ 1 volume CO 2 + 79 21 volume N 2
The volume of carbon is written as zero since the volume of a solid is negligible in comparison
with that of a gas.
By mass
12kg C + (2 x 16)kg O 2 +
79
21
(2 x 14)kg N 2
⇒ 12 + (2 x 16)kg CO2 + 79 (2 x 16) kg N 2
21


i.e
ii)
12kg C + 32kg O 2 + 105.3 kg N 2 → 44kg CO 2 + 105.3 kg N 2
The incomplete combustion of carbon. This occurs when there is an insufficient supply of
oxygen to burn the carbon completely to carbon dioxide, i.e. equations are balanced on the
basis of the conservation of mass principle (or the mass balance), which can be stated as
follows: The total mass of each element is conserved during a chemical reaction. That is, the
total mass of each element on the right-hand side of the reaction equation (products) must be
equal to the total mass of that element on the left-hand side (the reactants) even though the
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elements exist in different chemical compounds in the reactants and products. Also, the total
number of atoms of each element is conserved during a chemical reaction since the total
number of atoms of an element is equal to the total mass of the element divided by its atomic
mass.
H 2 + 1 O2 → H 2O ………………………………………… (12.6)
2
For example, both sides of equation (12.6) contain 12kg of carbon and 52kg of oxygen, event
though the carbon and the oxygen exist as elements in the reactants and as a compound in the
product. Also, the total mass of reactants is equal to the total mass of products, each being
44kg.
Consider the combustion equation of hydrogen, equation (12.6) can be written as:
2 H 2 + O 2 → 2H 2O ……………………………………. (12.7)
This tells us that
i)
hydrogen reacts with oxygen to form steam or water
ii)
two molecules of hydrogen react with one molecule of oxygen to give two molecules of steam
or water, i.e. 2 volume H 2 + 1 volume O2 → 2 volumes H2O
the H2O may be a liquid or a vapour depending on whether the product has been cooled
sufficiently to cause condensation. The properties by mass are obtained
2C + O2 → 2CO ………………………………………….. (12.8)
2C + O2 +
79
21
N 2 → 2CO +
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79
21
N 2 …………………………. (12.9)
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If a further supply of oxygen is available then the combustion can continue to completion, i.e.
2CO + O2 +
79
21
N 2 → 2CO2 +
79
21
N 2 ………………………. (12.10)
12.3 QUIZ (ONE)
Q1: What are the approximate chemical composition of gasoline, diesel ,fuel and natural gas?
Q2: Write three statements that express the conservation of mass principle for chemical reaction.
Q3: Will a fuel start burning when it is brought into intimate contact with oxygen?
Q4: What are the causes of incomplete combustion?
Q5 Which is more likely to be found in the products of an incomplete combustion of a
hydrocarbon fuel, CO or OH? Why?
Q6: What does 100 percent theoretical air represent?
Q7 Are complete combustion and theoretical combustion identical? If not, how do they differ?
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WEEK 13
13.0 Stoichiometric Air-Fuel Ratio
A frequently used quantity in the analysis of combustion processes to quantify the amounts of
fuel and air is the air-fuel ratio i.e. A/F. It is usually expressed on a mass basis and is defined as
the ratio of the mass of air to the mass of fuel for a combustion process as shown in figure
below.
Fuel
Combustion Chamber
Products
A/F = 17
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Air
18kg
Mathematical expression of A/F is given by
A
F
=
massofair
massoffuel
……………………………………….. (13.0)
A Stoichiometric mixture of air and fuel is one that contains just suffient oxygen for the
complete combustion of the fuel. A mixture which has an excess of air is termed a weak mixture
and one which has a deficiency of air is termed a rich mixture.
In actual combustion processes, it is common practice to use more air than the Stoichiometric
amount to increase the chances of complete combustion to control the temperature of the
combustion chamber. The amount of air in excess of the Stoichiometric amount is called excess
air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent
excess air.
Percentage excess =
actual A/F ratio - Stoichiome tric A/F ratio
StoichiometricA / Fratio
….. (13.1)
For gaseous fuels the ratios are expressed by volume and for solid and liquid fuels the ratios are
expressed by mass.
The mixture strengths of combustion is defined as follow:
Mixture strenght =
Stoichiometric A/F ratio
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actualA / Fratio
……………….. (13.2)
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13.1 WORKED EXAMPLES
Example (1): One kmol of octane (C8 H18 ) is burned with air that contains 20 kmol of O2. Assuming the
products contain only CO2, H2O, O2 and N2, determine the mole number of each gas in the
products and the air-fuel ratio for this combustion process.
Solution:
The amount of fuel and the amount of oxygen in the air are given. The amount of the products
and the A/F ratio are to be determined. Assumptions: The combustion products contain CO 2,
H2O, O2 and N2 only.
Analysis: The chemical equation for this combustion process can be written as:
C8 H18 + 20(O2 + 3.76 N 2 ) → xCO2 + yH 20 + ZO2 + wN 2
Where: x, y, z and w represent the unknown mole numbers of the gases in the products. These
unknowns are determined by applying the mass balance to each elements i.e., by requiring that
the total mass or mole number of each element in the reactants be equal to that in the
products:
C:
8 = x → x = 8
H:
18 = 24 → y = 9
O:
40 = 2x + y + 2z → z = 7.5
N2 :
(20)(3.76) = w → w = 75.2
Substituting in the chemical equation of this combustion process, it yields:
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C8 H18 + 20(O2 + 3.76 N 2 ) → 8CO2 + 9 H 2O + 7.5O2 + 75.2 N 2
Note:
Total mole of air = 75.2 mole of nitrogen + 20 mole of oxygen = 95.2
A
F
=
=
mair
m fuel
( Nm )air
( Nm )c + ( Nm )H 2
=
(20 x 4.76kmol ) 29kg kmol 


(8kmol )12 kg kmol  + (9kmol ) 2kg kmol 

= 24.2kg air kg



f uel
Example 2: Octane (C8 H18 ) is burned with dry air. The volumetric analysis of the products on a
dry basis is
CO2 :
O2 :
10.02%
5.62%
CO :
0.88%
N2 :
83.48%
Determine:
i)
the air-fule ratio
ii)
the percentage of Stoichiometric or theoretical air.
Solution: The combustion equation can be written as:
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xc8 H18 + a (O 2 + 3.76N 2 ) → 10.02CO2 + 0.88CO + 5.62O2 + 83.48 N 2 + 6 H 2O
N 2 : 3.76a = 83.48 ⇒ b = 12.24, C : 8 x = 10.02 + 0.88 ⇒ x = 1.36
H : 18 x = 2b ⇒ b = 12.24, O2 a = 10.02 + 0.44 + 5.62 + b
2
→ 22.20 = 22.20
1.36C8 H18 + 22.2(O2 + 3.76 N 2 ) → 10.02CO2 + 5.6202 + 83.48 N 2 + 12.24 H 2 0
The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1.36,
C8 H18 + 16.32(O2 + 3.76 N 2 ) → 7.3 + CO2 + 0.65CO + 4.1302 + 61.38 N 2 + 9 H 2 0
i)
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of
the fuel i.e.
A
=
ii)
F
=
mair
m fuel
19.76kg air
=
(16.32 x4.76kmol ) 29kg kmol 


 8kmolx12kg
 + (6kmol ) 2kg

kmol
kmol




kg fuel
To find the percentage of theoretical air used, the need to know the theoretical amount
of air, which is determined from the theoretical combustion equation of the fuel.
C8 H18 + 9 m (0 2 + 3.76 N 2 ) → 8C0 2 + 9 H 2 0 + 3.76 9mN 2
O2 : 9m = 8 + 4.5 → 9 m = 12.5
Percentage of theoretical air =
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m air , actual
mair , theore
=
N air , act
N air , theore
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=
(16.32)(4.76)Kmol
(12.50)(4.76)Kmol
= 131%
That is, 31% excess air was used during this combustion process.
Example 3: A sample of dry anthracite has the following composition by mass
C 98&, H 3%, O 2.5%, N 1%, S 0.5%, ash 3%
Calculate:
i)
Stoichiometric A/F ratio
ii)
The A/F ratio when 20% excess air is supplied
Solution:
i)
Each constituent is taken separately and the amount of oxygen required for complete
combustion is found from the relevant chemical equation.
Carbon:
C + O 2 → CO 2 12kg C + 32kg 0 2 → 44kg CO 2
i.e.
Oxygen required = 0.9 x
32
12
=
2.4kg
kgcoal
Where the carbon content is 0.9kg per kilogram of coal
Carbon dioxide produced = 0.9 x 44
12
= 3.3kgCO2
Hydrogen:
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H2 +
1
2
O 2 → H10
2kg H 2 + 16kg 02 → 18kg H 2O
1kgH 2 + 8kg02 → 9kgH 2O
i.e.
Oxygen required = 0.03 x 8 =
and steam produced = 0.03 x 9 =
0.24kg
kg coal
0.27kg
kg coal
Sulphur:
S + O 2 → SO 2
32kg S + 32kg 02 → 64kg SO 2
1kgS + 8kg02 → 2kgSO 2
i.e.
Oxygen required =
0.005kg
kg coal
Sulphur dioxide produced = 2 x 0.005 =
0.01kg
kg coal
Oxygen Required
Constituent
Mass Fraction
kg
Product Mass
kg
kg
kg
coal
coal
Carbon (C)
0.900
2.400
3.30 (CO2)
Hydrogen (H)
0.030
0.240
0.27 (H2O)
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Sulphur (S)
0.005
0.005
0.01 (SO2)
Oxygen (O)
0.025
-0.025
-
Nitrogen (N)
0.010
-
0.01 (N2)
Ash
0.030
-
-
Total
2.630
Note: the oxygen in the fuel is a negative quantity
O2 required per kilogram of coal = 2.62kg
Therefore,
Air required per kilogram of coal =
2.62
0.233
= 11.245kg
Where air is assumed to contain 23.3% O2 by mass
i.e. Stoichiometric air-fuel ratio = 11.245
ii)
For an air supply which is 20% in excess,
Percentage excess air =
actual A/F ratio - Stoichiometric A/F ration
Stoichiometric A/F ratio
Actual A/F ratio = Stoichiome tric A/F ratio + percentage excess air x Stoichimet ric A/F
 20

x11.245 
 100

= 11.245 + 
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= 11.245 + 2.249
= 13.494
Or A/F ratio = 1.2 x 11.245
= 13.494
13.2 QUIZ (TWO)
Q1: What is the air- fuel ratio? How is it related to the fuel – air ratio?
Q2: How does the presence of N2 in the air affect the outcome of a combustion process?
Q3: How does the presence of moisture in the air affect the outcome of a combustion
process?
Q4: Propane (C3H8) is burned with 50 percent excess air during a combustion process. Assuming
complete combustion, determine the air- fuel ratio. Answer: 23.5kg air/kg fuel.
o
Q5: Liquid propane (C3H8) enters a combustion chamber at 25 C at a rate of 0.05kg/min where it is
o
mixed and burned with 50 percent excess air that enters the combustion chamber at 7 C. An analysis of
the combustion gases reveals that all the hydrogen in the fuel burn to H 2O but only 90 percent of the
carbon burns to CO2 with the remaining 10 percent forming CO. If the exit temperature of the
combustion gases is 1500K, determine (a) the air fuel ratio of the combustion process (b) the mass flow
rate of air. Answer: 23.53 kg air/kg fuel and 1.18 kg air/min.
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Q6: In a combustion chamber, ethane (C2H6) is burned at a rate of 6kg/h with air that enters the
combustion chamber at a rate of 132kg/h. Determine the percentage of excess air used during this
process. Answer: 37 percent.
WEEK 14
14.0 Calorific Value of Fuels
The heat energy released by the complete combustion of unit quantity of fuel is called the
calorific value of the fuel. The conditions under which the combustion is carried out must be
specified and the calorific value so determined is for the said conditions. Calorific value is used
in the gas and solid fuel industries as a contractual and legal term. An alternative term in the oil
industry is heat of combustion; calorific value is also called specific energy.
The determination of the calorific value of fuels is carried out in specially designed calorimeters.
The type of calorimeter used will depend upon the form which the fuel takes. In the case of
solid and some liquid fuels the calorific value is usually determined in a bomb calorimeter. In
the case of gaseous and some liquid fuels the calorific value is determined in a gas calorimeter.
14.0.1 Gross Calorific Value at Constant Volume (Q gross,v)
Gross calorific value at constant volume is the amount of heat liberated per unit quantity of fuel
when burned in oxygen (O2) in a bomb calorimeter under standard conditions and the products
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of combustion are cooled to the original standard conditions. Correction should be made for
products of combustion such as O2, CO2, SO2, N2, liquid water (H2O), etc.
14.0.2 Net Calorific Value at Constant Volume (Q net, v)
Net calorific value at constant volume is the amount of heat liberated per unit quantity of fuel
when burned in oxygen (O2) in a bomb calorimeter under standard conditions and the products
of combustion are cooled to the original standard conditions. Corrections are made for the
products of combustion but the water (H2O) formed remains as vapour (i.e. it remains
uncondensed).
14.0.3 Gross Calorific Value at Constant Pressure (Qgross,p)
Gross calorific value at constant pressure is used mostly in the case of gaseous fuels and is the
amount of heat liberated per unit volume of fuel when burned in oxygen (O 2) at constant
pressure. The products of combustion are carbon dioxide (CO2), sulphur dioxide (SO2), oxygen
(O2), nitrogen (N2) and liquid water H 2O)
14.0.4 Net Calorific Value at Constant Pressure (Q net,p)
Net calorific value at constant pressure is used mostly in the case of gaseous fuels and is the
amount of heat liberated per unit volume of fuel when burned in oxygen (O 2) at constant
pressure. The products of combustion are CO2, SO2, O2, N2 and H2O, in this case the H2O
remains as vapour (i.e. it remains uncondensed).
The above quantities are related as follows for constant volume process.
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Fig. 14
A Schole’s stainless steel bomb calorimeter is illustrated in Fig.(14), in which a small mass of the fuel is
held in a crucible portion as shown in Fig(14). If the fuel is solid, it is usually crushed,passed through a
sieve, and then pressed into the form of a pellet in a special press. The size of pellet is estimated from
the expected heat release, and is such that the temperature rise to be measured does not exceed 2-3K. .
Apparatus required to prepare specimens for the bomb calorimeter are: a pestle and mortar (coal), a
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small press ( for compressing powdered coal into a small pellet) and a chemical balance. A high-pressure
oxygen supply, usually from a high pressure cylinder, is also required along with a stop-clock.
14.2 Solid Fuel – using the bomb calorimeter
Assume that it is coal under test. The coal is first crushed in a pestle and mortar until it is a fine powder.
The coal should be ground to pass through a sieve. The crucible from the calorimeter is then weighed.
The powdered coal is pressed into the small press to form a pellet then approximately 1g is placed in the
crucible. The crucible plus coal is weighed; the weight of coal is determined by subtracting the known
weight of the crucible.
14.3 GAS CALORIMETER
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Fig. 14.1
Figure (14.1) shows the fundamental principle behind the construction of one kind of gas calorimeter
used for the determination of the calorific value of a gaseous fuel. The basic elements are a central
funnel surrounding a burner at its base. A cooling coil is fitted around the outside of the funnel: in some
calorimeters the coil is replaced by a nest of tubes. Underneath the coil and arranged round the funnel
is a condensate trap. Round the outside is fitted a cover, and the whole apparatus is heat insulated by
lagging. The cooling water inlet passes through the cover at the bottom: the outlet is at the top.
Thermometers are installed to measure the cooling water inlet and outlet temperatures.
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The gas products are arranged to pass out at the bottom through an exhaust. The exhaust is fitted with a
thermometer to measure its temperature. Most gases under test contain H2; during combustion this will
form H2O, which will condense as it passes over the cooling coil.
14.4 WORKED EXAMPLES
Example 4:
In a bomb calorimeter test on petrol the GCV was determined and found to be
46,900KJ
kg
. If the fuel contains 14.4% H by mass, calculate the NCV.
Solution:
2 H 2 + O2 → 2 H 2O
Therefore,
4kg H + 32kg O 2 → 36kg H 2O
1kg H + 8kg O 2 → 9kg H 2O
∴1kg of H gives 9kg H 2O, therefore
0.1441kg H gives 0.144 x 9 = 1.296kg H 2O
Then using equation (12)
Qnet , v = Q gr, v - m c u fg
= 46,900 - (1.296 x 2304.4 )
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= 43910KJ
kg
Example 5: The Stoichiometric A/F ratio for ethyl alcohol, C2 H 6O , was found to be 8.957/1. If
the NCV of ethyl alcohol is 27800 KJ
kg
, calculate the calorific value of the combustion mixture
per cubic meter at 1.013 bar and 15˚C.
Solution: 1 kmol of ethyl alcohol is
per kilogram of fuel =
1
46
The molar mas of air is
8.957
28.96
= 0.3093kmol
46kg
kmol
= 0.02174kmol
28.96kg
kmol
, therefore we have amount of substance of fuel
kg
, therefore amount of substance of air per kg of fuel =
kg
i.e.
Total amount of substance of mixture = 0.02174 + 0.3093
=
0.3310kmol
kg
~
Pv = n R T
V =
0.331 x 8314.5 x 288
1.013 x10
5
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= 7.824m 3 per kg of fuel
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NCV of mixture =
Q net, v
Volume
=
27.8
7.824
= 3.55MJ
m3
of mixture
WEEK 15
15.0 DENSITY OF GAS MIXTURE
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From the volumetric analysis of a gas mixture, the average molar mass is
nM
∑
=
......................................(15)
M
∑n
~
~
~
By definition, one mole of the gas mixture will have a mass of
M av kg.
The volume of one mole of the gas mixture at its pressure and temperature can be calculated from
~
V =
RT ..........................................................................(15.1)
P
From equations (15) and (15.1)
~
Density = ρ =
M av
V
3
kg / m .........................................(15.2)
15.1 Atmospheric and Ecological Pollution
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The combustion of fuels pollutes the atmosphere. And fuels are burned in increasing amounts,
so the problems of pollution, both atmospheric and ecological, need serious consideration.
15.2 Carbon monoxide (CO)
Carbon monoxide is produced by the incomplete combustion of carbon with oxygen.
If
breathed in, it reacts with the haemoglobin in the blood and prevents the uptake of oxygen. It
impairs thought, alertness and reflexes. It is toxic enough to kill in sufficient dosage.
15.3 Carbon dioxide (CO2)
Carbon dioxide is produced by the complete combustion of carbon with oxygen. Large masses
are now emitted into the atmosphere from the various combustion processes. This is especially
true of motor car and lorries. It has become possibly the main contributor to the green house
effect, producing global warming.
15.4 Sulphuric dioxide (SO2)
Sulphur dioxide is produced by the complete combustion of sulphur with oxygen. It is an
atmospheric pollutant mostly from diesel engines and flue gas emission in industrial areas. It is
a toxic gas and can be a respiratory irritant. It can produce atmospheric acid, a precursor of acid
rain.
15.5 Oxides of Nitrogen (NO)
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Various oxides of nitrogen can be produced from the nitrogen in the air reacting with the
oxygen in the air as the result of high temperatures obtained during the combustion of a fuel.
The oxides are toxic and can be respiratory irritants. They can produce atmospheric acids,
precursors of acid rain. They also contribute to photochemical smog.
15.6 Volatile Organic Compounds (VOCs)
Volatile organic compounds (VOCs) come from the evaporation of fuels from petrol stations,
petrol tanks, leaks of hydrocarbons, etc. They contain cancer-forming compounds e.g. benzene.
They also contribute to photochemical smog.
15.7 Photochemical Smog
Photochemical smog occur on sunny, windless days, which perhaps have a temperature
inversion in the lower atmosphere (i.e. when atmospheric temperature increases with height
instead of falling as normal). If there are high concentrations of NOx and VOCs, they may react
with the ultraviolet light from the sun to produce ozone (O3) an irritating, bluish gas with a
pungent smell. Ozone in the very high atmosphere is formed naturally and protects the earth
from some of the sun’s ultraviolet radiation. But near to the ground it is photochemical smog.
15.8 Particulates
Particulates are small particles which mix with atmospheric air. They are produced during
inefficient fuel burning (e.g. black, sooty smoke from some diesel engines, coal and oil burning).
Dust particles may be thrown up by moving vehicles or may come from their brakes.
15.9 Ecological Considerations
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Thermodynamic Basic
The acquisition and use of coal and oil present ecological problems. Coal has to be mined, both
on the surface and in deep mines, mining scars the earth’s surface, produces subsidence and
creates large mounds of earth spoil and coal storage. In the case of oil, both land and sea rigs
and used for drilling and for receiving the acquired oil. Large tanks are used for storage.
Inevitably, some spillage occurs; tanker disasters during sea transport cause large oil discharge
into the sea and may severely damage the local ecology.
15.10 ECOLOGICAL CONSIDERATIONS
The acquisition and use of coal and oil present ecological problems. Coal has to be
mined, both on
the surface and in deep mines. Mining scars the earth’s surface, produces subsidence and creates large
mounds of earth spoil and coal storage. After burning, coal can produce atmospheric particulates and
quantities of ash which need disposal. In the case of oil, both land and sea rigs are used for drilling and
for receiving the acquired oil. Large tanks are used for storage. Inevitably, some spillage occurs; tanker
disasters during sea transport because large oil discharges into the sea and may severely damage the
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