Thermodynamics Miscellaneous
- A turbocharged four-stroke direct injection diesel engine has a displacement volume of 0.0259 m3 (25.9 liters). The engine has an output of 950 kW at 2200 rpm. The mean effective pressure (in MPa) is closest to
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B.P = n.pmVN.k 60
where p m = mean effective pressure
n = number of cylinders = 1.
V = volume of displacement = 0.02510 cm3
N = r.p.mk = 1 for 4-stroke 2
= 1 for 2-stroke∴ pm = B.P × 60 n.V.N.k = 950 × 60 for 4-stroke 0.0259 × 2200 × (1 / 2)
= 2000 kPa = 219 PaCorrect Option: A
B.P = n.pmVN.k 60
where p m = mean effective pressure
n = number of cylinders = 1.
V = volume of displacement = 0.02510 cm3
N = r.p.mk = 1 for 4-stroke 2
= 1 for 2-stroke∴ pm = B.P × 60 n.V.N.k = 950 × 60 for 4-stroke 0.0259 × 2200 × (1 / 2)
= 2000 kPa = 219 Pa
- During a Morse test on a 4 cylinder engine, the following measurements of brake power were taken at constant speed. All cylinders firing 3037 kW Number 1 cylinder not firing 2102 kW Number 2 cylinder not firing 2102 kW Number 3 cylinder not firing 2100 kW Number 4 cylinder not firing 2098 kW The mechanical efficiency of the engine is
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Given: B.P. = 3037 kW
(I.P.)1 = 3037 – 2102 = 935 kW
(I.P.)2 = 3037 – 2102 = 935 kW
(I.P.)3 = 3037 – 2100 = 937 kW
(I.P.)4 = 3037 – 2098 = 937 kW
Total I.P. of the cylinder
= (I.P)1 + (I.P)2 + (I.P)3 + (I.P)4
= 935 + 935 + 937 + 937 = 3746 kW
∴ Mechnical efficiency of the engine= B.P. = 3037 = 87.07% I.P. 3746
Correct Option: C
Given: B.P. = 3037 kW
(I.P.)1 = 3037 – 2102 = 935 kW
(I.P.)2 = 3037 – 2102 = 935 kW
(I.P.)3 = 3037 – 2100 = 937 kW
(I.P.)4 = 3037 – 2098 = 937 kW
Total I.P. of the cylinder
= (I.P)1 + (I.P)2 + (I.P)3 + (I.P)4
= 935 + 935 + 937 + 937 = 3746 kW
∴ Mechnical efficiency of the engine= B.P. = 3037 = 87.07% I.P. 3746
- An automobile engine operates at a fuel air ratio of 0.05, volumetric efficiency of 90% and indicated thermal efficiency of 30%. Given that the calorific value of the fuel is 45 MJ/kg and the density of air at intake in 1 kg/m3, the indicated mean effective pressure for the engine is
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Given, Fuel = 0.05 Air
ηvolumetric = 90%
ηthermal = 30%
Calorific value = 45 MJ/kgmf = 0.05 ma
ρair = 1 kg/m3
Volumetric efficiency,ηv= actual volume = vac swept volume vs
∴ Vac = 0.9 Vs
ma = ρairVac
= 1 × 0.9 × Vs
= 0.9 Vs
mf = 0.05 × 0.9 Vs = .045 Vsηthermal = pmep × LAN mf × 45 × 1046 0.3 = pmep × Vs 0.45Vs × 45 × 106
... since LAN = Vs
pmep =0.3 × 0.045 × 45 × 106 = 6.075 × 105N/m²
= 6.075 barCorrect Option: A
Given, Fuel = 0.05 Air
ηvolumetric = 90%
ηthermal = 30%
Calorific value = 45 MJ/kgmf = 0.05 ma
ρair = 1 kg/m3
Volumetric efficiency,ηv= actual volume = vac swept volume vs
∴ Vac = 0.9 Vs
ma = ρairVac
= 1 × 0.9 × Vs
= 0.9 Vs
mf = 0.05 × 0.9 Vs = .045 Vsηthermal = pmep × LAN mf × 45 × 1046 0.3 = pmep × Vs 0.45Vs × 45 × 106
... since LAN = Vs
pmep =0.3 × 0.045 × 45 × 106 = 6.075 × 105N/m²
= 6.075 bar
- Brake thermal efficiency of the three basic types of reciprocating engines commonly used in road vehicles are given in the increasing order as
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NA
Correct Option: A
NA
- BHP of a diesel engine can be increased by
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The BHP of diesel engine is increasing burning more fuel. And burning more fuel can be achieved by increasing density of intake air and increasing the pressure of intake air.
Correct Option: E
The BHP of diesel engine is increasing burning more fuel. And burning more fuel can be achieved by increasing density of intake air and increasing the pressure of intake air.
