The turbocharger of an internal combustion engine consists o

The turbocharger of an internal combustion engine consists of a turbine and a compressor. Hot exhaust gases flow through the turbine to produce work and the work output from the turbine is used as the work input to the compressor. The pressure of the ambient is increased as it flows through the compressor before it enters the cylinder. Thus more air is delivered to the cylinder and more fuel can be burned to yield more power. In a turbocharger (Fig. 1), exhaust gases enter the turbine at 400 degree C and 120 kPa at a rate of 0.02 kg/s and leave at 350 degree C. Air enters the compressor at 50 degree C and 100 kPa and leaves at 130 kPa at a rate of 0.018 kg/s. The increase in pressure is accompanied by an undesirable increase in temperature, which could lead to early ignition in the cylinder (engine knock). It is estimated that this knocking phenomenon can be avoided if the cylinder inlet temperature is below 80 degree C. To avoid this, an after cooler is placed after the compressor to cool warm air by cold ambient air. Disregarding all frictional losses in the turbine and compressor and treating the exhaust gases as air, determine a) the temperature of the air at the compressor outlet b) the minimum volume flow rate of the ambient air required to avoid knock.

Solution

Assumption

1. Steady operating condition

2. KE and PE change negligible

3. Air Properties used for exhaust gases

Properties:

The specific heat of air at the average temperature of exhaust gas i.e. (400+350)/2 = 375 °C is Cpexhaust = 1.061 kJ/kg K (Air Properties)

Analysis:

Power produced by Turbine = mexhaust Cpexhaust (T1- T2)

Where mexhaust = 0.02kg/sec , Cpexhaust = 1.061 kJ/kg K , T1 = 400 °C = 673K, T2 = 350 C = 623 K

Power Turbine = 1.061 kW

This power is provided to compressor.

Compressor:

Input state of Air : T3 = 50 C , P3 = 100kPa

Exit state of Air : T4 = to find , P4 = 130kPa

mair = 0.018 kg/sec

Cp of air = 1.009 kJ/kg K for this temperature range (Air Properties)

(Note: Value of Cp remains constant from T= 50C to T = 100C at 1.009 kJ/kg K)

Turbine Power = Compressor power consumption

1.061 kW = mair Cpair (T4 – T3) = 0.018 x 1.009 x (T4 – (50+273))

Gives T4 = 378.94 K = 105.94 C

Thus the temperature at the exit of the compressor is 105.94 °C (PART1 answer)

PART 2

In the aftercooler the cooling air enters at 30 °C and leaves at 40 °C . We need to keep the outlet temperature of compressed air limited to 80 °C. Therefore assume the outlet temperature of compressed air as 80 °C to find the minimum cooling air required.

Note:

The value of Cp of air will vary only slightly in this case as the temperature is less. So it can be assumed as same and solved. In this solution exact values of Cp are taken corresponding to respective temperatures according to air properties.

Cp air in temperature range (100 °C – 80 °C) Cp1= 1.009 kJ/kg K

Cp air in temperature range ( 30 °C – 40 °C) Cp2 = 1.005 kJ/kg K

Heat lost by compressed air = Heat gained by cooling air

Heat lost by compressed air = mcompressedair xCp1 x ((105.94+273)- (80+273))

Heat lost = 0.018 x 1.009 x 10^3 x ((105.94+273)- (80+273))

Heat lost = 0.4711kJ

0.4711kJ = Heat gained by cooling air = mcoolair x Cp2 x (40 – 30 )

0.4711kJ = mcoolair x 1.005 x 10^3 x (40 – 30 )

On solving it gives mcoolair = 0.0468 kg/sec

Part2 answer mcoolair = 0.0468 kg/sec

(Note: It can also be solved assuming Cp as constant in part2)