CHAPTER TWO
FIRST LAW OF THERMODYNAMICS
2.1. Introduction
This law forms the basis for studying the relationships among the various forms of energy and
energy interactions. It states that energy can be neither created nor destroyed during a process: it
can only change form. Every bit of energy should be accounted for during a process. That is
why the first law of thermodynamics is also known as the conservation of energy principle. This
principle can be stated as: Work and heat are mutually convertible energy. Energy cannot be
created or destroyed, but can be converted from one form to another. This is the first law of
thermodynamics. The net change (increase or decrease) in total energy of the system during a
process is equal to the difference between the total energy entering and the total energy leaving
the system. That is,
 Change in the total 
 Total energy

 Total energy







 entering the system -  leaving the system =  energy of the system






Mathematically, Ein  Eout  E System
The above equation is known as energy balance. It is applicable to any kind of system for
undergoing any process. Energy can be transferred to or from a system through heat transfer,
work transfer and mass transfer.
2.2 First Law Applied to Closed System
If Q amount of heat enters into a system (closed system as shown in Fig.2.1) and W amount of
work is done by the system, then (Q - W) amount of energy will be stored in the system and this
stored energy is known as total energy. Energy in storage is neither heat nor work.
Mathematically for a non-flow process, one can write
Q  W  E or Q  E  W
(2.1)
If there are more number of energy transfer as shown in Fig.2.2(b), the energy conservation
equation can then be written as
Q2  Q1  Q3  E  W2  W3  W1  W4
(2.2)
1
Fig 2.1: Closed system: (a) one heat and one work interactions (b) multiple interactions
Fig 2.2: multilpe hea
Now, for a cyclic process, E will be zero, E being a point function. Hence for a cycle
 Q  W or,  Q   dW
cycle
(2.3)
cycle
A steam power plant as a whole may be considered as a closed system as shown in Fig. 2.2.
Applying 1st law of thermodynamics to the steam power cycle, and using Eq.(2.3), one can
write
Qin Q out  Wout  Win
Fig 2.2: A closed system: Schematic of a thermal power plant
2
The symbol E refers to the total energy. Energy can be stored in three forms. They are
(a) Kinetic energy refers to the energy in motion.
(b) Potential energy is associated with external conservative forces such as force of gravity or
internal forces like elastic force.
(c) Molecular energy is stored in the molecule or atomic structure.
So, the total energy of the system is given by
E  Ek  E p  U
In the absence of motion and gravity,
i.e., for Ek  0, E p  0 and hence E  U
 E  E k  E p  U.
In the absenceof motion and gravity, E k  0, E p  0
then E  U
 Q  U  W
In the differntia l form,
dQ  dE  dW
dQ  dU  dW, This equation is known as non - flow energy equation
where dW  dWPdV  dWshaft  dWelastic
If only PdV work is there,
dQ  dE  PdV
dQ  dU  PdV
In the integral form,

Q  U   PdV
Q  E  PdV
Specific heat at constant volume (C v) is defined as the rate of change of specific internal energy
with respect to temperature when the volume is held constant.
T2
 δu 
C v     u v  c v dT
 δT  v
T

1
2.3 Enthalpy
We have
3
dQ  du  pdv
At constant pressure : pdv  d(pv)
 dQ p  du  d ( pv)  d (u  pv)  dh. Where h  u  pv is the specific enthalpy, a property of the system.
Total enthalpy  H  mh
Specific heat at constant pressure (cp) is defined as the rate of change of specific enthalpy with
respect to temperature when pressure is held constant.
T2
 δh 
c p     h p  c p dT
 δT  p
T

1
dQ  dh - vdp, for any process
2.4 Energy of an Isolated System
In this case, there is no energy interaction with the surroundings, i.e.,
dQ  dW  0  dE  0, E  constant,
Energy of an isolated systemis always constant.
2.5 Perpetual Motion Machine of First Kind (PMM 1) :
There can be no machine which would continuously supply mechanical work without some
other form of energy disappearing simultaneously, such a fictitious machine is called PMM 1.
Fig. 2.3: Perpetual Motion Machine
of First Kind (PMM1)
Fig. 2.4: Reverse of PMM1
Problem 2.1: A stationary mass of gas is compressed without friction from an initial state of 0.3
m3 and 0.105 MPa to final state of 0.15 m3 and 0.105 MPa, the pressure remains constant during
the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much
does the internal energy of the gas change?
Solution:
4
For a statitonary system,
Q  E  W,
 Q1-2  E1 - E 2  W12
- - - - - - - - - - - (1)
v2

Now, W1-2  pdv  p(v 2  v1 )  0.105(0.15 - 0.30) MJ  15.75 kJ
v1
Q1-2   37.6 kJ (given)
So, Eq.(1) gives E 2  E1   21.85 kJ.
Therefore, the internal energy of the gas decreases by 21.85 kJ
Problem 2.2: When a system is taken from state a to state b as shown in Fig 2.5 along path acb,
84 kJ of heat flow into the system and the system does 32 kJ of work.
(a) Calculate the amount of heat that flows into the system along the path adb, if the work done
is 10.5 kJ.
(b) When the system is returned from b to a along the curved path, the work done on the system
is 21 kJ. Does the system absorbs or liberate heat?
(c) If Ua = 0 and Ud = 42 kJ, find the heat absorbed in the processes ad & db.
Solution:
5
Given, Q acb  84 kJ, Wacb  32kJ.
Now, Q acb  U b  U a  Wacb
 U b  U a  84  32  52 kJ
(a) Q adb  U b  U a  Wadb
 52  10 .5  62 .5 kJ
(b)Q b-a  U a -U b  Wb-a  -52 - 21  -73 kJ
The systemliberates 73 kJ of heat.
(c) Wadb  Wad  Wdb  Wad  10 .5 kJ [ Wdb  0, dv  0)
 Q ad  U d  0  Wad  42  0  10 .5  52 .5 kJ
and Q adb  62.5  Q ad  Q db
 Q db  62.5 - Q ad  62.5 - 52.5  10 kJ
Problem 3: A piston and cylinder machine contains a fluid system which passes through a
complete cycle of four processes. during cycle the sum of all heat transfer is -170 kJ. The
system completes 100 cycles per minutes. Complete the following table showing the method for
each item and compute neat rate of work output in kW.
Processes
Q (kJ/min)
a-b
0
W (kJ/min)
2170
∆E (kJ/min)
_________
6
b-c
21000
0
_________
c-d
-2100
_______
-36600
d-a
_______
_______
_________
Solution
Process a - b :
Q  E  W  0  E  2170
 E  2170 kJ/min
Process b - c :
Q  E  W  21000  E  0
 E  21000 kJ/min
Process c - d :
Q  E  W  2100  36600  W
 W  34500 kJ/min
Process d - a :
 Q  170 kJ
cycle
The systemcompletes 100 cycles/min
 Q a -b  Q b-c  Q c-d  Q d-a  170  100  17000 kJ/min
 Q d-a  35900 kJ/min

Now, dE  0
 E ab  Ebc  Ecd  E da  0
 2170  21000  36600  E d a  0
 E d a  17700 kJ/min
 Wd-a  Q d-a - E d a  35900  17700  53670 kJ/min
Q   W
cycle
cycle
 Rate of work input   17000 kJ/min  283 .3 kW
Problem 4: The properties of a certain fluid are related as follows
u = 196+0.718t
pv = 0.287(t+273)
where u is the specific internal energy (kJ/kg), t is in °C, p is pressure (kN/m 2) and v is specific
volume (m3/kg). For this fluid find cv and cp.
7
Problem 2.5: A system undergoes a process A from state 1 to state 2 during which the heat and
work interactions are 290 kJ and 250 kJ, respectively. If the system is taken to state 2 from state
1 through a different process B, the work done by the system is found to be 80kJ. Determine the
heat interaction along path B and the change in internal energy u 2-u1. Suppose the system is
returned from state 2 to state 1 through a third path C, during which 125 kJ work is done on the
system. Determine the heat interactions along this path.
Problem 2.6: A gas contained in a cylinder fitted with a piston expands from state 1 to state 2
and does 55 kJ of work on the surroundings. If the total energies t state 1 and state 2 are 70 kJ
and -20 kJ, determine the heat transfer to or from the system during the process.
Problem 2.7: In a steam power plant the power developed by the turbine is 1200 kW and the
heat supplied to the boiler is 3360 kJ/kg. The heat rejected by the system in the condenser to the
circulating cooling water is 2520 kJ/kg and the pump work required to pump the condensate
back into the boiler is 6 kW. Calculate the steam flow rate in the cycle. draw a schematic
diagram to show different heat and work interactions with proper direction and sign.
Ans. 1.42 kg/s.
2.8: A system undergoes a complete cycle consisting of four processes 1-2, 2-3, 3-4 and 4-1.
The energy transfers in these processes have been listed in the following table. Complete the
table showing the methods for each item. Also determine the net rate of work output in kW.
Process
Q (kJ/min)
W(kJ/min)
E (kJ/min)
1-2
2-3
3-4
4-1
400
200
-200
00
150
75
300
-
Ans. 250, -100, 275, -475, -75, 6.67 kW
2.9: 1.5 kg of liquid having constant specific heat of 2.5 kJ/kg-K is stirred in a well-insulated
chamber causing the temperature to rise by 15 0C. Find E and W for the process.
Ans. E = 56.25 kJ, W = -56.25 kJ.
2.10: A gas of mass 8 kg expands within a flexible container so that the p-v relationship follows
the law pv1.2 = constant. The initial pressure and volume of the gas are 1000 kPa and 1 m3. The
final pressure is 5 kPa. If the specific internal energy of the gas decreases by 40 kJ/kg, find the
heat transfer in magnitude and direction.
8
2.11: A domestic refrigerator is loaded with food and the door closed. During a certain period
the machine consumes 1 kWh of energy and the internal energy of the system drops by 5000 kJ.
Find the net heat transfer for the system. Ans. - 8.6×103 kJ
2.12: A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relation
p  a  bV , where a and b are constants. The initial and final pressures are 0.20 m3 and 1.20 m3.
The specific internal energy of the gas is given by the relation, u =1.5 pv -85 kJ/kg, where p is
in kPa and v is in m3/kg. Calculate the net heat transfer and the maximum internal energy of the
gas attained during expansion. Ans. 660 kJ, 503.3 kJ
2.13: A fluid contained in a cylinder receives 150 kJ of mechanical energy by means of a paddle
wheel, together with 50 kJ in the form of heat. At the same time, a piston in the cylinder moves
in such a way that the pressure remains constant at 200 kN/m2 during the fluid expansion from 2
m3 to 5 m3. What are the changes in internal energy and in enthalpy? Ans. -400 kJ, 200 kJ
2.14: The energy of a closed system can be expressed as a function of temperature as
E  100  50T  0.04T 2 . The heat absorbed is given by Q  5000 20T . The temperature in these
relations are in K (Kelvin), whereas energy and heat are in J (Joule). Calculate the work done
during the process, if the temperature rises from 500K to 1000K.
Ans. -20462 kJ
2.15: A system undergoes a frictionless non-flow process according to the law p  4.5 v  2
,where p is in bar and v is in m3/kg. During the process the volume changes from 0.12 m3/kg to
0.04 m3/kg and the temperature increases by 133 0C. The change in internal energy of the fluid is
given as du  cv dT , where cv  0.71 kJ/kg-K and dT is the temperature change. Calculate heat
transfer and the change in enthalpy.
Ans. 41594 kJ, 784.3 kJ
2.6 First Law Applies to Flow Process
A large number of engineering devices such as turbines, compressors, nozzles operate for a long
periods of time under the same conditions once the transient period is completed and the steady
state operation is established. These are called steady flow devices. For examples, the
9
components of a steam power plant such as turbines, pumps, condensers, boilers operate nonstop for months before the system is shut down for maintenance. Under such condition, any
thermodynamic property of the system will have a fixed value at a particular location, and will
not alter with time. Thermodynamic properties may vary along space coordinates, but do not
vary with time.
During a steady flow process, no intensive or extensive property within the control volume
changes with time. So, the volume, mass and total energy (E) of the control volume remain
constant. The boundary work is therefore zero for steady flow systems. The total mass or energy
entering the control volume must be equal to the total mass or energy leaving it. In other words,
there is no accumulation of mass or energy within the control volume. Also the power delivered
by the system and the rate of heat transfer to or from the system remain constant during a steady
flow process
Steady Flow energy equation
In figure 2.3. a steady flow system has been shown in which one stream of fluid enters and one
stream leaves the control volume. Different parameters and thermodynamic properties at section
1 (inlet) and section 2 (outlet) have been shown and designated with subscript "1" and "2'
respectively.
A1, A2 : cross section of stream, m2
ω1, ω2 : mass flow rate, kg/s
p1, p2 : absolute pressure, N/m2
v1, v2 : specific volume, m3/kg
u1, u2 : specific internal energy, J/kg
V1, V2 : velocity, m/s
Z1, Z2 : elevation from an arbitrary datum, m
dQ
: Net rate of heat trans fer, J/s
d
dW
: Net rate of work, J/s
d
 : Time, sec
Subscript 1 and 2 refers to inlet and outlet sections.
10
In order to introduce the fluid across the boundary an expenditure of energy is required.
Similarly, in order to push the fluid across the boundary at exit, expenditure of energy is
required. Energy required to push an element of length 'l' across boundary
= (p1A1)×l = p1 × (Volume of fluid element)
Energy required for unit mass flow rate of fluid =
p1v1
Similarly, energy required at exit to push unit mass
flow rate of fluid across the boundary = p2v2
1. Mass balance (Continuity equation)
ω1  ω 2 
A1V1 A 2 V2

v1
v2
- - - - - - - - - - - (1) :
2. Energy balance
ω1e1  ω1p1v1 
dQ
dW
 ω2e 2  ω2 p 2 v 2 
dτ
dτ
- - - - - - - - - - - - - - - - - -(2)
e1, e2 : energy carried into or out of the control volume with unit mass of the fluid,
and p1v1, p2v2 are the flow work associated with the fluid in sections 1 and 2.
The specific energy e is given by e  e k  e p  u 
V2
 Zg  u
2
11
Equation (2) becomes
1 (
V12
dQ
V2
dW
 Z1g  u1 )  ω1p1v1 
 ω2 ( 2  Z 2 g  u2 )  ω2 p 2 v 2 
2
dτ
2
dτ
V12
dQ
V2
dW
 Z1g ) 
 ω2 (h 2  2  Z 2 g ) 
- - - - - - - - - - - - - (3)
2
dτ
2
dτ
dm
dm
Since 1  2 , let   1  2 
. Dividing equation (3) by
d
d
1 (h1 
V12
dQ
V2
dW
 Z1g 
 h2  2  Z2 g 
- - - - - - - - - - - - - (4)
2
dm
2
dm
This is known as steady flow energy equation.
h1 
When more than one stream of fluid enter or leave, it becomes
1. Mass balance : ω1 + ω2 = ω3 + ω4
2. Energy balance :



1 h1 


 dQ
V12
V2
 Z1 g   2  h 2  2  Z 2 g  


 d
2
2






 dW
V2
V2
 3  h1  1  Z1 g   4  h 2  2  Z 2 g  



 d
2
2




- - - - - - - - - - - - - (5)
The steady flow energy equation applies to a wide variety of processes like pipe line flows, heat
transfer processes, mechanical power generation in engines and turbines, combustion processes,
flows through nozzles and diffusers.
SOME EXAMPLES OF STEADY FLOW PROCESSES
1. Nozzle and diffuser :.
A nozzle is a device that increases the velocity of a fluid at the expense of pressure. On the other
hand, a diffuser increases the pressure of a fluid at the cost of velocity. These devices are
commonly used in jet engines, rockets, spacecraft and also in steam turbine. The cross-sectional
area of nozzle decreases in the flow direction for subsonic flow and increases for supersonic
flow. Opposite is the case with diffuser. The heat transfer from the fluid flowing through the
nozzle or diffuser to the surroundings is usually very small, i.e., Q = 0 due to the high velocity
of fluid. These do not involve any work and the changes in potential energy is also neglected.
Now, the steady flow energy equation is
12
V12
dQ
V2
dW
 Z1 g 
 h2  2  Z 2 g 
2
dm
2
dm
dQ
dW
Here, Heat transfer,
 0, Work done,
 0,
dm
dm
Potential energy change is neglected, Z1  Z2
h1 
The SFEE is then reduced to
V12
V2
 h2  2
2
2
V1 (velocity of approach) is small and sometimes neglected and
h1 
one can write more simplified equation as
V2  2h1  h2  m/s
where enthalpies are in J/kg
2. Throttling device :
Throttling valves are flow-restricting devices that cause significant pressure drop in the fluid.
Ordinary adjustable valves (partially opened valve), an orifice, capillary tubes, a porous plug
are some of the common examples of these types
of valves. They produce a pressure drop without
involving any work. The pressure drop in the fluid
is often accompanied by a large drop in
temperature. These
are
commonly used
in
refrigeration and air conditioning applications. The
magnitude of temperature drop (sometimes may be
temperature rise) during a throttling process is
governed by a property called Joule-Thomson
coefficient of the fluid.
As the throttling valves are small devices, the flow through them are assumed to be adiabatic.
There is neither sufficient time or enough area for effective heat transfer to take place. The
changes in potential energy and kinetic energy may be neglected.
3. Turbine and compressors : Turbine is a device which is used in steam, gas and hydroelectic
power plants for the generation of mechanical power. Compressor, pump and fan are different
devices used to increase the pressure of a fluid. Work is supplied to these devices from external
source through rotating shafts. So, these are work consuming devices. A fan increases the
pressure of a gas slightly just to mobilize it. A compressor compresses a gas to a very high
pressure. Pumps are used to increase the pressure of liquid only. Heat transfer is usually
negligible in case of turbines and compressors. Changes in potential energy and kinetic energy
are also negligible compared to enthalpy change.
13
Compressor
h1  h 2 
Turbine
dW
dW

 h1  h 2 , work is done by the fluid at the expense of enthalpy
dm
dm
For compressor s, h1  h 2 
dW
, enthalpy of the fluid increases by the amount of work input.
dm
Heat Exchangers:
Heat exchangers are devices where two
moving fluid streams exchange heat without
Heat Exchanger
mixing. Heat exchangers are widely used in
many industries. Condensers, boilers and
evaporators are some of the most common
examples. A double pipe heat exchanger
consists of two concentric tubes of different
diameters. One fluid flows in the inner tube
and other in the annular space in between the
two tubes. When the two fluids are allowed to mix, it is called a direct contact heat exchanger.
Heat exchangers typically involve no work interactions and negligible kinetic and potential
energy changes.
For heat exchangers , h1 
dQ
 h2
dm
Problem 1. In the turbine of a gas turbine unit the gases flow through the turbine at 17 kg/s and
the power developed by the turbine is 14000 kW. The enthalpies of the gases at the inlet and
14
outlet are 1200kJ/kg and 360 kJ/kg respectively and the velocities of the gases at the inlet and
outlet are 60 m/s and 150 m/s respectively. Calculate the rate at which heat is rejected from the
turbine. find also the area of the inlet pipe given that the specific volume of the gases at inlet is
0.5 m3/kg.
Soln :
h1 
V12
V2
dQ
dW
 Z1 g 
 h2  2  Z2g 
- - - (1)
2
dm
2
dm
Kinetic energy at inlet 
V12 60 2

 1800 Nm/s
2
2
V2 2
 11 .25 kJ/kg
2
dW 14000

 823 .5kJ/kg
dm
17
dQ

 7.02 kJ/kg   7.02  17  119 .3 kW
dm
mv 17  0.5
Inlet area 

 0.142 m
V
60
Similarly,
Problem 2: In a Steam power plant station, steam flows steadily through a 0.2 m diameter
pipeline from the boiler to the turbine. At the boiler end, the pressure, temperature, enthalpy and
the specific volume of steam are 4 MPa,400°C, 3213.6 kJ/kg and 0.073 m3/kg respectively. The
corresponding values at the turbine end are 3.5MPa, 392°C, 3202.6 kJ/kg and 0.084 m3/kg
respectively There is a heat loss of 8.5 kJ/kg from the pipeline. Calculate the steam flow rate.
Soln :
V12
V2
dQ
dW
 Z1 g 
 h2  2  Z2g 
- - - (1)
2
dm
2
dm
dW
No change in datum,
 0.
dm
AV A V
Since, 1 1  2 2
v1
v2
h1 
 V2 
A1V1 v 2 v 2
0.084


 V1 
 V1  1.15 V1
v1
A 2 v1
0.073
 From eq n (1),
V


2
2

 V12  10 3
dQ
 h1  h 2 
2
dm
 V12 1.15 2  1  3213 .6  3202 .6  (8.5)
 V1  125 .1 m/s


2
  0.2   125 .1
AV
4
  53 .8 kg/s
m  1 1  
v1
0.073

15
Solution:
The steady flow energy equation is given by
h1 
V12
V2
Q
W
 Z1 g 
 h2  2  Z 2 g 
2
m
2
m
(1)
For the flow through pipe no change in elevation is considered, i.e., Z1  Z 2 , no work is done by
the system or on the system, i.e.
h1 
W
 0 , then the above equation becomes
m
V12 Q
V2

 h2  2
2 m
2
Problem 3 : In a steady flow apparatus, 135 kJ of work is done by each kg of fluid. The specific
volume of the fluid, pressure and velocity at the inlet are 37 m3/kg, 600 KPa and 16m/s. the inlet
is 32 m above the floor and the discharge pipe is at the floor level. The discharge conditions
are0.62 m3/kg, 100 KPa and 270 m/s. The total heat loss between the inlet and the discharge is 9
kJ/kg of fluid. In following through the apparatus, does specific internal energy increase or
decrease?
Soln :
Steady flow energy equation gives:
2
V12
dQ V2
dW
 Z1 g  u1  p1v1 

 Z 2 g  u2  p 2 v 2 
2
dm
2
dm
 u1  u 2  p 2 v 2  p1v1  
V2 2  V12
dW dQ
 Z 2  Z1 g 

2
dm dm
 1 0.62  6  0.37  
270 2  16 2
2  10
3


   32  9.81  10 3  135  9
 20.136 kJ/kg
 Specific internal energy decreases.
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Problem 4 : In a turbo machine handling an incompressible fluid with a density of 1000 kg/m 3.
The conditions of the fluid at the rotor entry and exit are given below
Inlet
Exit
Pressure
1.15 MPa
0.05 MPa
Velocity
30 m/s
15.5 m/s
Height above datum
10 m
2m
If the volume flow rate of the fluid is 40 m3/s, estimate the net energy transfer from the fluid as
work.
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Problem 2. 5: Air having pressure 100 kPa and specific volume 0.95m3/g enters a compressor
with velocity 7 m/s. After compression the air leaves the compressor with a velocity of 5 m/s,
pressure of 700 kPa and specific volume of 0.19 m3/kg respectively. The internal energy of the
air leaving is 90 kJ/kg greater than that of the air entering. Cooling water in the compressor
jackets absorbs heat from the air at the rate of 58 kW. Calculate the power input to the
compressor and the ratio of the inlet pipe diameter to the outlet pipe diameter.
Ans. -122 kW, 1.89
Problem 2.6: Steam enters a turbine with a velocity of 40 m/s and specific enthalpy of 2500
kJ/kg and leaves with a velocity of 90 m/s and specific enthalpy of 2030 kJ/kg. Heat losses from
the turbine to the surroundings are 240 kJ/min and the steam rate is 5040 kg/hr. Neglecting the
change in potential energy, estimate the power developed by the turbine.
Ans. 650 kW
Problem 2.7: Air at the rate of 75 kg/min enters a steady flow device at 2 bar, 100 0C at an
elevation of 100 m above the datum. The same mass leaves the control volume at 150 m
elevation from datum with a pressure of 10 bar and temperature of 300 0C. The inlet and exit
velocities are 40 m/s and 20 m/s. During the process heat is entering the device at the rate of
54000 kJ/hr and the enthalpy of air increases by 8 kJ/kg. Calculate the power developed by the
device.
Ans. 5.13 kW
Problem 2.8: Steam at the rate of 1.5 kg/s enter a turbine at a pressure of 2 MPa and
temperature of 3500C and leaves the turbine as saturated steam at a pressure o 0.1 MPa. The
velocities at the inlet and outlet are 50 m/s and 200 m/s respectively. The inlet is situated at a
height of 6 m and the height of the outlet point is 3m from a datum. If the heat transfer from the
turbine is 8.5 kW, determine the power output of the turbine.
Ans. 655.7 kW
Problem 2.9: Air flows at the rate of 2.3 kg/s in a 15 cm diameter pipe. It has a pressure of 7
bar and a temperature of 95 0C before it is throttled by a valve to 3.5 bar. Find the velocity of air
just after throttling. Ans.
Problem 2.10: At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg
and the velocity is 60m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The nozzle is
horizontal and there is negligible heat loss from it. Find the velocity at the exit of the nozzle and
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mass flow rate, if the inlet area is 0.1 m2 and the specific volume at the inlet is 0.187 m3/kg. If
the specific volume at the nozzle exit is 0.498, find the exit area of the nozzle.
Ans. 692.5 m/s, 32.08 kg/s, 0.023 m2
Problem 2.11: A room for four persons has two fans, each consuming 0.18 kW power and three
100 W lamps. Ventilation air at the rate of 80 kg/hr enters with enthalpy of 84 kJ/kg and leaves
with an enthalpy of 59 kJ/kg. If each person puts out heat at the rate of 630 kJ, determine the
rate at which heat is to be removed by a room cooler so that a steady state is maintained in the
room. Ans. 1.92 kW
Problem 2.12: A turbo-compressor delivers 2.33 m3/s of air at 0.276 MPa and 430C which is
heated at this pressure to 430 0C and finally expanded in a turbine which delivers 1860 kW.
During the expansion, there is a heat transfer of 0.09 MJ/s to the surroundings. Calculate the
exhaust temperature if changes in kinetic and potential energy are negligible.
Ans. 1570C
Problem 2.13: An air turbine forms part of an aircraft refrigerating plant. Air at a pressure of
295 kPa and a temperature of 580C flows steadily into the turbine with a velocity of 45 m/s. The
air leaves the turbine at a pressure of 115 kPa, a temperature of 2 0C and a velocity of 150 m/s.
The shaft work delivered by the turbine is 54 kJ/kg of air. Neglecting changes in elevation,
determine the magnitude and sign of heat transfer per unit mass of air flowing. Take, cp for air =
1.005 kJ/kg-K and enthalpy h = cpt
Ans. +7.96 kJ/kg
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