HELP NEEDED PLEASE CLEAR SOLUTIONS A four-cylinder, four-stroke, 1.6 L gasoline

Question

HELP NEEDED PLEASE CLEAR SOLUTIONS

A four-cylinder, four-stroke, 1.6 L gasoline engine operates on the Otto cycle with a compression ratio of 10. The air is at 100 kPa and 42 degree C at the beginning of the compression process, and the maximum pressure in the system is 9 MPa. Determine (a) the temperature at the end of the expansion process, (b) the net work output and the thermal efficiency, (c) the engine speed for a net power output of 55 kW. Repeat the calculations for a compression ratio of 12 and compare the results. At an altitude of 9100 m. the ambient conditions are approximately 33 kPa and -32 degree C. A turbojet aircraft is flying at 350 m/s with a pressure ratio across the compressor of 11, a turbine inlet temperature of 1300 K, and a mass flow rate of air into the compressor of 55 kg/s. The jet fuel used by this aircraft has a heating value of 42,500 kJ/kg. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the velocity of the exhaust gases, (b) the propulsive power developed, and (c) the rate of fuel consumption. Consider a steam power plant that operates on the simple Rankine cycle, with water as the working fluid. The condenser operates at a pressure of 40 kPa and the boiler at a pressure of 5500 kPa. The turbine inlet temperature is 425 degree C. The isentropic efficiency of the turbine is 92 percent, consider pressure and pump losses as negligible, and the water that leaves the condenser is subcooled by 5 degree C. The mass flow rate through the boiler is 24 kg/s. Determine (a) the rate at which heat is added in the boiler, (b) the power required to operate the pumps, (c) the net power produced by the cycle, and (d) the thermal efficiency.

Explanation / Answer

Q2.

P1 = 100 kPa

T1 = 42 deg C = 42 + 273 K = 315 K

P3 = 9 MPa = 9000 kPa

Vs = 1.6 L

For each cylinder swept volume = 1.6 / no. of cyl = 1.6 / 4 = 0.4 L = 0.4*10^-3 m^3

Compression ratio r = 10 = (V1 / V2) = (V4 / V3)

r = (Vs + V2) / V2

10 = (0.4*10^-3 + V2) / V2

V2 = 0.044*10^-3 m^3

So, V1 = 10*V2 = 0.44*10^-3 m^3

Trapped mass m = P1*V1 / (R*T1)

m = 100*(0.44*10^-3) / (0.287*315)

m = 4.867*10^-4 kg

a)

T2 / T1 = (V1 / V2)^(gamma - 1)

T2 / 315 = 10^(1.4-1)

T2 = 791.2 K

P2 / P1 = (V1 / V2)^gamma

P2 / 100 = 10^1.4

P2 = 2511.88 kPa

Further, T3 / T2 = P3 / P2

T3 / 791.2 = 9000 / 2511.88

T3 = 2834.8 K

T4 / T3 = (V3 / V4)^(gamma - 1)

T4 / 2834.8 = (1/10)^(1.4-1)

T4 = 1128.6 K = 1128.6 - 273 deg C = 855.6 deg C

b)

Thermal efficiency = 1 - 1 / (r^(gamma-1))...........for air, gamma = 1.4

Thus, Thermal efficiency = 1- 1 / (10^(1.4-1)) = 0.6018 or 60.18 %

Heat input Q = m*Cv*(T3 - T2)

Q = (4.867*10^-4)*0.718*(2834.8 - 791.2)

Q = 0.7141 kJ

Work done = eff*Q = 0.6018*0.7141 = 0.4298 kJ

c)

Work done per cycle = 0.4298 kJ

If engine is running N revs per minute, means N/2 cycles per minute , means (N/2)/60 = N/120 cycles per sec.

Power = 0.4298*(2*pi*N) / 120

55 = 0.4298*2*3.14*N / 120

N = 2445 RPM

If r = 12 we'll get the following values,

V1 = 0.436*10^-3 L

m = 4.8268*10^-4 kg

T2 = 851.1 K

P2 = 3242.3 kPa

T3 = 2362.5 K

T4 = 874.4 K = 601.4 deg C

Thermal eff = 0.6299 or 62.99 %

Q = 0.5238 kJ

W = 0.3299 kJ

N = 3185 RPM

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