Question

(30 points) Consider the turbocharger of an internal combustion engine. The exhaust gases enter the turbine at 450C at a rate of 0.02 kg/s and leave at 400C. Air enters the compressor at 708C and 95 kPa at a rate of 0.018 kg/s and leaves at 135 kPa. The mechanical efficiency between the turbine and the compressor is 95 percent (5 percent of turbine work is lost during its transmission to the compressor). Using air properties for the exhaust gases, determine (a) the air temperature at the compressor exit and (b) the isentropic efficiency of the compressor.

Answer 1

(a) T₂ = 126°C , T₂s = 106°C

(b) 64%

Step-by-step explanation:

we will begin with a step by step process;

using the expression

Wт = m(ex)Cp(T₁ - T₂)

where

Wт = turbine power output

m(ex) = mass of the exhaust gas

Cp = specific heat at constant pressure

T₁ = exhaust gas temperature at the turbine inlet

T₂ = exhaust gas temperature at the turbine exit

using from the gas table, the specific heat of exhaust gas at avg temperature of 425°C as 1.075kJ/kg.K

making substitutions we have;

T₁ = 450°C, T₂ = 400°C, m(ex) = 0.02kg/s, Cp = 1.075kJ/kg.K

Substituting values into formula gives;

Wт = m(ex)Cp(T₁ - T₂)

Wт = [(0.02kg/s)(1.075kJ/kg.K)(450+273)K-(400+273)K]

Wт = 1.075 kW

The relationship of mechanical efficiency between turbine and compressor is given as;

Лм = Wc/Wт............................. (1)

where Wт = turbine output

Wc = compressor output

Лм = mechanical efficiency

Wc = Лм × Wт

Wc = 0.95 × 1.075(kW)

Wc = 1.021 kW

compressor power:

Wc = m(air)Cp(T₂ - T₁)

where m(air) = mass of air

T₁ = air temp at the compressor inlet

T₂ = air temp at the compressor exit

Cp = specific heat at constant pressure

T₂ = T₁ + Wc/m(air)Cp .................. (2)

the specific heat of air at the expected average temperature of 100°C is 1.011 kJ/kg.K

substituting T₁ = 70°C, Wc = 1.021 kW, m(air) = 0.018 kg/sm Cp = 1.011 kJ/kg.K

T₂ = 70 °C + (1.021)kW/ (0.018kg/s)(1.011 kJ/kg.K)

T₂ = 126 °C

Considering an isentropic process, the air temperature at the compressor exit;

T₂s = T₁ (P²/P¹)∧(k-1)/(k)

where T₁ = 70°C, P₁= 95kPa, P₂ = 135kPa, k = 1.397

T₂s = (70+273)K (135/95 kPa) ∧ (1.397-1)/(1.397)

T₂s = 379 K = 106 °C

(b). the isentropic efficiency of the compressor

Лc = (T₂s - T₁ / T₂ - T₁)

Лc = (106°C - 70°C) / (126°C - 70°C)

Лc = 0.642 = 64%

the isentropic efficiency of the compressor is 64%

cheers i hope this helps