Solution of Sheet 2 ICE
Solution of Sheet 2 ICE
Solution of Sheet 2 ICE
(2)
1) In a four stroke cycle S.I. engine the cam shaft runs
(a) at the same speed as crank shaft (b) at half the speed of crank shaft
(c) at twice the speed of crank shaft (d) at any speed irrespective of crank shaft speed
2) Explain with suitable sketches the working of a four stroke Otto engine.
Intake stroke: This starts with the piston at TC and ends with the piston at BC, which draws fresh mixture
into the cylinder.
Compression stroke: when both valves are closed and the mixture inside the cylinder is compressed to a
small fraction of its initial volume. Toward the end of the compression stroke, combustion is initiated and
the cylinder pressure rises more rapidly.
Expansion stroke: This starts with the piston at TC and ends at BC as the high-temperature, high-pressure,
gases push the piston down and force the crank to rotate.
Exhaust stroke: where the remaining burned gases exit the cylinder: first, because the cylinder pressure
may be substantially higher than the exhaust pressure: then as they are swept out by the piston as it moves
toward TC.
Stroke: Movement distance of the piston from one extreme position to the other: tdc to bdc or bdc to tdc.
T.D.C.: is Position of the piston when it stops at the furthest point away from the crankshaft.
Top because this position is at the top of most engines (not always),
Dead because the piston stops at this point.
B.D.C.: is Position of the piston when it stops at the point closest to the crankshaft. It is not always at the
bottom of the engine.
Clearance volume: the volume of the compressed charge which is also the volume contained in the
cylinder above the top or crown of the piston when the piston is at TDC.
Swept volume: the displacement volume swept by the piston in one stroke and is equal to the product of
the inner cross-sectional area of the cylinder and the stroke length of the piston.
Compression ratio: the ratio between the total volume to the clearance volume.
2L N
Piston speed: mean piston speed =
60
4) Discuss the difference between ideal and actual valve timing diagrams of petrol engine
5) Define compression ratio. What is its range for S.I. and C.I. engines? What factors limit the
compression ratio in each type of these engines?
The compression ratio is the ratio between the total volumes of the cylinder to the minimum volume of
the cylinder.
6) Show by suitable sketches the following types of cylinder arrangements indicating their
principal applications
a) In-line engines b) “V” engines c) Opposed piston engines d) Radial engines
2-Thermal efficiency of C.I. engine is higher than that of S.I engine due to
a) fuel used b) higher compression ratio
c) constant pressure heat addition d) none of the above
11) The brake thermal efficiency for S.I. engine varies from:
a)40-50% b)45-60% c)25-30% d)15-25%
13) The indicated thermal efficiency of 4-stroke S.I. engine producing 500 kW, mechanical
efficiency 0.8 and fuel consumption 300kg/hr (HHV=40MJ/kg)is
a) 0.12 b) 0.1875 c) 0.2 d) 0.2175
14) The volumetric efficiency of a 4-stroke engine (stroke volume 30 liters) running at 4000 rpm.
producing 50 kW and consuming 3000kg/hr of air at 1 bar and 300 K is
a) 60% b) 71.8% c) 75% d) 80%
15) The b.mep of a passenger car running on diesel fuel in comparison with the b.mep of a petrol
engine is
a) more b) equal c) less d) dependent on compression ratio
16) The volumetric efficiency of an internal combustion engine will depend upon
1-period of valve overlap 2-density of fresh charge
3-pressure of residual gas 4-design of intake manifold
Of these statements:
a) 1,2,3 are correct b)2,3,4 are correct c)all are correct d)1,3,4 are only correct
18) Select the suitable engine for the following purposes indicating the reasons for your selection.
1-engine power 70kW and its speed 200 rpm used for driving pump to raise crude oil from oil wells
2-engine drive passenger car with variable speed its power 30kW at 4000 rpm
3-Trucks engine, variable speed its power 90kW at 2500 rpm
4-marine engine its power 10000kW at 150 rpm
5-stationary engine used for electrical generation its power 8000kW at 7000 rpm
1 &3 &4 Diesel Engine 2 Petrol Engine 5 Gas turbine
19) What is meant by “quantity control” and “quality control” engines? Compare between these
engines.
The quantity control: it means that we can control of the quantity of mixture which enter to the cylinder of
the engine by control of the amount of throttle valve open.
The quality control: it means that we can control of the quality of mixture which enter to the cylinder of
the engine by control of the amount of fuel inject to the combustion chamber.
1) In petrol engine the pressure in the cylinders at 30% and 70% of the compression
strokes are 1.3 and 2.6 bar respectively. If the compression follows the law PV 1.33
= constant. Find the compression ratio?
Given:
Pa = 1.3 bar Pb = 2.6 bar
The solution:
Here: take V1 is total volume and V2 is the clearance volume be unit =1
V
The compression ratio r = 1 = V1
V2
The stroke volume = V1 - V2 = r - 1
Va = 1 + 0.7 (r − 1) = 0.3 + 0.7r
Vb = 1 + 0.3(r − 1) = 0.7 + 0.3r
Pa Va1.33 = Pb Vb1.33
1 1
Va Pb 1.33 2.6 1.33
= = = 1.684
Vb Pa 1.3
also :
Va 0.3 + 0.7 r
= 1.684 =
Vb 0.7 + 0.3r
1.18 + 0.505 r = 0.3 + 0.7 r
r = 4.51
2) The ideal Otto engine working on the air standard has a temperature and pressure
at the beginning of 25ºC and 1 bar respectively, and a thermal efficiency of 48%.
Determine the pressure and temperature at the end of compression.
Given:
P1 = 100 kPa T1= 298 K η = 0.48
The solution:
Point 1: P1 = 100 kPa & T1= 298 K
k −1
1
where : th = 1 − = 0.48
r
= (1 − th ) k −1 r =
1 1 1 1
= = 5.13
r (1 −th ) 1k −1 (1 − 0.48) 10.4
P2 = P1 (r ) P2 = 100 (5.13) = 986.7 kPa
k 1.4
Point 2:
T2 = T1 (r ) = 298 (5.13)
k −1
= 573.14 K
0.4
3) A petrol engine works on Otto cycle. the pressure, temperature and volume of air
at the beginning of compression are 0.93 bar, 38 oC and 0.028 m3 and the pressure
at the beginning and end of heat addition are 7.87 bar and 22 bar. If there are 250
cycles per minute, find power developed by the engine and the cycle efficiency.
Given:
P1 = 93 kPa T1= 311 K V1 = 0.028 m3 P2 = 7.87 bar
P3 = 22 bar N = 250 cycle per minute
The solution:
Point 1: P1 = 93 kPa & T1= 311 K & V1= 0.028 m3
Point 2: P2 = P1 (r )k 787 = 93 (r )1.4 r = 4.6
T2 = T1 (r )k −1 = 311 (4.6)0.4 = 572.6 K
V1 0.028
V2 = = = 0.0061 m 3
r 4.6
P 2200
Point 3: P3 = Pmax = 2200 kPa & T3 = T2 3 = 572.6 = 1600.7 K
2
P 787
k 1.4
1 1
Point 4: P4 = P3 = 2200 = 259.8 kPa
r 4.6
k −1 0.4
1 1
T4 = T3 = 1600.7 = 869.4 K
r 4.6
P1 (V1 − V2 ) 93 (0.028 − 0.0061)
where : m= = = 0.0228 kg
R T1 0.287 311
Wnet = Qadd − Qrej = m Cv (T3 − T2 ) − m Cv (T4 − T1 )
Wnet = 0.0228 0.718 (1600.7 − 572.6) − 0.0228 0.718 (869.4 − 311) = 7.7 kJ
250
Power = Wnet No. of cycles per sec. = 7.7 = 32.1 kW
60
Wnet 7.7
th = = = 0.46
Qadd 16.83
4) The rotational speed of 4-strock engine working on an ideal Otto cycle is 4000 rpm.
The initial pressure and temperature are 100 kPa and 300 k respectively. If the
maximum volume to the minimum volume ratio is 11,heat add to system 2000 kJ/kg
air and stork volume is 2.2 lit, find the following:
a) The theoretical thermal efficiency
b) The engine's power and m.e.p.
Given:
P1 = 100 kPa T1= 300 K r = 11 Qadd = 2000 kJ/kg
Vs = 2.2 lit
The solution:
Point 1: P1 = 100 kPa & T1= 300 K
Point 2: P2 = P1 (r ) = 100 (11)1.4 = 2870.1 kPa
k
Wnet 1727
th = = = 0.864
Qadd 2000
Wnet 4.41
mep = = = 2004.5 kPa
Vs 0.0022
4000 1
Power = Wnet No. of cycles per sec. = 4.41 = 147 kW
60 2
5) An engine working on a theoretical diesel cycle a suction pressure of 100 kPa, air
temperature of 300k, and a compression ratio of 16.if the heat add to the cycle is
1500 kJ/kg of air . Find the pressure and temperature at the key points in the cycle.
Given:
P1 = 100 kPa T1= 300 K r = 16 Qadd = 1500 kJ/kg
The solution:
Point 1: P1 = 100 kPa & T1= 300 K
P2 = P1 (r ) = 100 (16) = 4850.3 kPa
k 1.4
Point 2:
T2 = T1 (r ) = 300 (16) = 909.4 K
k −1 0.4
Point 3: Qadd = Cp (T3 − T2 ) 1500 = 1.005 (T3 − 909 .4) T3 = 2401 .94 K
P3 = P2 = 4850 .3 kPa
T 2401.94
rcut = 3 = = 2.64
Point 4: T2 909.4
k 1.4
r 2.64
P4 = P3 cut = 4850.3 = 389.3 kPa
r 16
k −1 0.4
r 2.64
T4 = T3 cut = 2401.94 = 1168.3 K
r 16