95HE-4
Sr. No. 6
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
HEAT ENGINES
M.E.O. Class IV
(Time allowed - 3hours)
Afternoon Paper
India (2003)
Total Marks 100
NB : (1) All Questions are Compulsory
(2) All Questions carry equal marks
(3) Neatness in handwriting and clarity in expression carries weightage
(4) Illustration of an Answer with clear sketches / diagrams carries weightage.
1.
Consider an ideal gas contained in vessel. If intermolecular interaction
suddenly begins to act , which of the following happens.
a) The pressure increases b) The pressure remains unchanged
c) The pressure decreases d) The gas collapses.
2.
When a system is taken from a state A to state B along the path A-C-B,
180 KJ of heat flows into the system and it does 130 KJ work as shown in
figure.
C
B
A
D
P
V
How much heat will flow into the system along the path A-D-B, if the work
done by it along the path is 40 KJ?
a) 40 KJ
b) 60 KJ
c) 90 KJ
d) 135 KJ
3.
A gas expands from pressure P1 to pressure P2 (P2 = P1 / 10). If the process of
expansion is isothermal the volume at the end of expansion is 0.55m3, If the
process of expansion is adiabatic the volume at the end of compression will
be closer to
a) 0.45m3
b) 0.55 m3 c) 0.65 m3 d) 0.75 m3
4.
A tank contains 25 kg of fresh water at 15. 15.5 Kg of ice at 0C are added
then 2 Kg of steam at 4 bar 0.9 dry are blown into the mixture. Assuming no
heat losses occur and taking the enthalpy of fusion of ice at 0C as 333.5
KJ/Kg,, What is the final temp of the mixture.
a) 8.16C
b) 10.2C
c) 16C
d) 6.12C
5.
A steam pipe is 3.85m long when fitted at a temperature of 18C carrying
steam at a temperature of 260C. Taking the coefficient of linear expansion
of the pipe material as 1.25 x 10-5 / C , the increases in length will be.
a) 10 mm
b) 15 mm
c) 11.65 mm d) 14.2 mm
6.
A mass of 3.5 Kg of gas is cooled at constant volume from 150C to 25C
and then expanded according to the law PV 1.4=C to -40C. If the initial
pressure was 10 bar , determine
a) The final pressure b) Final volume c) The heat extracted during cooling
Take Cv of the gas to be 730 J/Kgk and R=0.287 KJ/Kgk
7.
The steam generation plant supplies 8500 Kg steam per hour at pressure
0.75 MN/m2, the steam is 0.95 dry.
Feed water temp = 41.5C
Coal consumption = 900 Kg/hr
Calorific value of coal = 32450 KJ / Kg
Determine
a) The boiler efficiency b) The equivalent evaporation from and at 100C
c) The saving in fuel consumption. If by installing an economizer it is
estimated that the feed water temp could be raised to 100C assuming that
other conditions remained unchanged and the efficiency of the boiler
increases by 6%
8.
A Refrigerating machine working on a reversed carnot cycle consumes 6KW
for producing a refrigerating effect of 1000 KJ/min for maintaining a region
at -40C, Determine the highest temp of the cycle and the cop of the
Refrigerator machine, when this device is used as a heat pump.
9.
Single acting single stage air compressor has an rpm of 240, has a bore of
200mm, the average piston speed is 180m/min. The air is sucked in at 1 bar
and 15C and compressed to 5.67 bar. Assume R = 0.287 KJ/Kg K.
Determine.
a)The mass flow rate
b)Rate of work done and exit temp for polytopic compression Where n=1.3
10. A steam leaves the nozzle and enters the blade wheel of a single impulse
Turbine at a velocity of 900 m/sec and at an angle of 20 to the plane of
rotation or plane of wheel. The blade velocity is 400m/sec and the exit angle
of the blade is 25. Due to friction the steam losses 15% of its relative
velocity across the blade. Calculate
a)
Inlet blade angle
b)
The magnitude and direction of absolute velocity of steam at exit.
---------------------X-----------------------
95HE-4
Sr. No. 6
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
HEAT ENGINES
M.E.O. Class IV
(Time allowed - 3hours)
Afternoon Paper
India (2003)
Total Marks 100
NB : (1) All Questions are Compulsary
(2) All Questions carry equal marks
(3) Neatness in handwriting and clarity in expression carries weightage
(4) Illustration of an Answer with clear sketches / diagrams carries weightage.
SOLUTION
1) The pressure increases
2) 90KJ
Soln:
For the path A-C-B
From 1st Law of TD
Q = ΔU + W
Q = UB - U A + W
180 = UB - UA+ 130
UB - UA=180-130
UB-UA= 50 KJ
For the path A-D-B
Q = ΔU + W
Q = UB - U A + W
Since the initial and final states are same for both the process the change in internal energy will be same.
Q = UB - U A + W
Q = 50 + 40
Q = 90 KJ of heat supplied
3) 0.45m3
soln :
When the gas is expanded isothermally
PV=C
i.e P1V1 = P2V2
P 1 = V2
P 2 V1
(10) = V2
V1
V1 = 0.55 = 0.055m3
10
V1 = 0.055m3
When the gas is expanded adiabatically
PVr = C
i.e P1V1r = P2V2 r
P1 =﴾ V2﴿r
P2 ( V1)
(P1)1/r = V2
(P2 )
V1
(10) 1/1.4 = V2
V1
Since the initial volume for both the process has to be same
(10) 1/1.4 = V2
V2 = 0.284m3
0.055
4) 8.16 o C
Soln:
Amount of heat present in water = Q1 = m cpw ΔT
= 25 x 4.187 x ( 15 – 0)
= 1570.1 KJ
If we take the enthalpy or heat of water at 0 o C as zero
Then the amount of heat present in ice at 0 o C is
Q2 = - ( 15.5 x 333.5 )
= - 5169.3 KJ
Amount of heat present in steam
Q3 = ms ( nf + xnfg ) 4 bar
= 2 (605 + 0.9 x 2134 )
= 5051.2 KJ
ice will gain heat , water and steam loses heat
:. Amount of heat present after mixing will be
Q = Q 1 + Q2 + Q 3
= 1570.1 – 5169.3 + 5051.2
= 1452 KJ
:. We know that Q = m mix cpw x (t-0)
1452 = (2 + 25 + 15.5) 4.187 ( t – 0)
t = 8.16 o C
5) 11.65 mm
Soln :
Force expansion = α x l (T2 – T1 )
= 1.25 x10 –5 x 3.85 ( 260 – 18)
= 11.65 mm
6) Soln:
1
P=C
T
P
2
PV1.4 =C
3
V
V1 = mRT1 = 3.5 x 0.28 x ( 423)
P1
10 x 102
V1 = 0.425 m3
From 1-2
P=C
T
P1 = P2
T1 T2
P2 = P1T2 = 10 x102 x 298
T1
423
P2 = 7.05 x 102 KN/m2
PV1.4 = C
From 2-3
T2 = (V3 / V2) n-1
V3 = V2 x ( T2 / T3 ) 1/ n-1 = 0.425 x ( 298 / 233) 1/1.4 – 1
= 0.786 m3
P 2 V 2n = P 3 V 3 n
P3 =
P2
(V3 /V2 )n
=
7.05 x102
( 0.786 / 0.425) 1.4
P3 = 2. 97 bar
From lst law of TD from 1 – 2 i.e cooling
Q =U2 – U1 + W
W=0
constant volume
:. Heat capacity = U2 – U1
= m.C v (T1 – T2 )
= 3.5 x 0.730 (4.23 – 298)
Heat extracted during cooling ( Q ) = 319 KJ
= 297 KN/m2
1) soln:
ms = msteam
mf
= 8500 = 9.44 Kg / Kg of coal
900
Sp Enthalpy of steam h = ( hγ + x hγg) 0.75
=[ 709.3+ 0.95 ( 2055.5 ) ]
= 2662.025 KJ/Kg
Sp enthalpy of water hw = cpw (T – 0)
= 4.187 ( 41.5 – 0)
= 173.9 KJ/Kg
Boiler efficiency = msteam ( h – hw)
m x c1V
=ms ( h – hw ) = 9.44 (2662.025 – 173.9 )
c1 V
32450
= 72.38%
Equivalent evaporation and at 100oC = ms (h – hw )
(hγ g )
= 9.44 ( 2662.025 – 173.9)
2256.9
= 10.40 Kg/Kg of coal
Energy required to generate steam under new condition
Sp. Enthalpy of BFW at 100oC hw = 4.87 (100 – 0)
= 4187 KJ/Kg
Energy required to generate steam when economizer is incorporated is = ( h – hw )
=2662.025– 418.7
= 2243.3 KJ/Kg
new boiler efficiency = 72.38 + 6
= 78.38%
:. Boiler efficiency = msteam ( h – hw )
mγ C1 V
0.7838
= 8500 x 2242.925
mγ x 32450
mγ
= 749.57 Kg/ hr
saving in full = 900 – 749.57
= 150.43 Kg of coal/hr
2) Soln:
Work done= W = 6KW
Q = 1000 KJ/min
TL = - 40 + 273 = 233o K
PAPER – 2
1) The length of each steel rail is 10m in winter. If the coefficient of linear expansion of steel is
12x10-6/OC and temp increases by 15C in summer the gap to be left between the rail is
a)
b)
c)
d)
0.0018m
0.0012m
0.0022m
0.05m
2) In the figure shown, E is a heat engine with efficiency of 0.4 and R is a refrigerator, Given that
Q2+Q4=3Q1. The cop of the refrigerator is
a)
b)
c)
d)
2.5
3.0
4.0
5.0
3) 80gm of water at 30C are poured on a large block of ice at 0C. The mass of the ice that melt is
a)
b)
c)
d)
1600mgm
30gm
150gm
8gm
4) Three moles of an ideal gas are compressed to half the initial volume at a constant temp of 300K.
The work done in the process is
a) 5186J
b) 2500J
c) –2500J
d) –5186J
5) A heat pump operating between high temperature T1 and lower temperature T2 has its cop
expressed as
a)
T1
T1-T2
b) T1-T2
T1+T2
c)
T2
T1-T2
d) T1+T2
T1-T2
6) Six cylinder, two stroke cycle heavy oil operated main engine has cylinder diameter of 580mm
and a piston stroke of 1700mm. When the engine speed is 116rev/min it uses 1204 kg of heavy
fuel oil of calorific valve 42700 Kj/Kg in one hour. The cooling water amounts to 66.62
tonnes/hour entering at 15C leaving at 63C. The torque transmitted at the engine coupling is 583
KN-M and the indicated mean effective pressure is 15 bar,
Determine,
1) The indicated power
2) The brake power
3) Amount of water required per Kg of level
4) Brake Thermal efficiency
5) Brake mean effective pressure
6) Fuel used per Kw hour on Brake power bases.
7) In a Frean-12 refrigerator, the freon leaves the condenser as a saturated, liquid at 20C and the
evaporator temperature is -10C and the freen leaves the evaporator as a vapour of 0.97dry
calculate
a) The dryness fraction at the evaporator inlet
b) The cooling effect per Kg of refrigerator
c) The volume flow of refrigerant entering the compressor if the mass flow is 0.1 Kg/sec.
8) In a single stage impulse turbine steam leaves the nozzle at a velocity of 600m/sec at 18 to the
plane of rotation of the blades, the linear velocity of the blades is 230 m/sec neglecting friction across
the blades and assuming the steam leaves the blade wheel in an axial direction. Calculate
1) The inlet angle of the blades so that the steam enters without shock
2) The outlet black angle
3) The Black or Diagram efficiency
9) A cold storage compartment is 45 m long by 4m wide by 2.5m high. The four walls ceiling and
flour are covered to a thickness of 150mm with insulating material which has a thermal
conductivity of 5.8x10-2 W/mK. Calculate the quantity of heat leaking through the insulation per
hour when the outside and inside the temperature of the material is 15C and -5C respectively.
10) 3 Kg of wet steam at 14 bar and dryness fraction of 0.95 are blown into 100Kg of water at 22C.
Find the resultant temperature of water.
SOLUTION
1) Ans- A
Increase in length = 2xlx(T 2-T1)
= 12x10-6x10x(15)
= 0.0018m
2) Ans- D
W= Q2-Q1
Also W = Q3-Q4
Q2-Q1= Q3-Q4
Q2+Q4= Q1+Q3
3 Q1= Q3+Q1
2 Q1= Q3
efficiency of heat engine E = W =
Q1
0.4 = W
Q1
W = 0.4 Q1
Cop (Refrigeration) = Q3 =
Q3 =
2Q1
W
Q 4 Q1
Cop = 5
= 5
O1 4Q1
3) Ans – B
Heat gained by ice = Heat lost by water from 35C to C
Mice x latent heat = mwater