Question

The compressor of a large gas turbine receives air from the ambient surroundings at 95 kPa and 20 ºC with a low velocity. At the compressor discharge, air exits at 1.25 MPa and 430 ºC with a velocity of 90 m/s. The power input to the compressor is 5000 kW. Determine the mass flow rate of air through the unit

Answer 1

The mass flow rate is m = 12.0 kg/s

Step-by-step explanation:

From the question we are given the following parameters

inlet temperature ,
`T_1 = 20`°C = 293 K

inlet pressure ,
`P_1 = 95KPa`

Outlet pressure
`P_2`
`= 1.25*10^3 kPa`

Outlet temperature ,
`T_2 = 430`°C

`=703K`

Final velocity ,
`V_2 = 90m/s`

Power Input ,
`W_c =5000kW`

Considering the energy equation we have

`h_1 +(V_1^2)/(2) +q = h_2 +(V_2^2)/(2) +w`

q is the net heat transferred

w is the net workdone

Lets assume that q = 0 and
`V_1 = 0` Hence in this question the specific heat capacity is constant

= >
`-w =h_2 - h_1 +(V_2^2)/(2)`

`=(C_P)_0 (T_2 -T_1) + (V_2^2)/(2)`

`= (1.004)(703-293) + (90^2)/(2(1000))`

The division by 1000 is to convert the kinetic energy to KJ

Note the specific heat of air is 1.004 kJ/kg⋅K

`= 415.5 KJ/kg`

The mass flow rate is given as
`m = (Z)/(-w)`

Where Z is The power input to the compressor which is given as 5000 kW

`m = (5000)/(415.5)`

`12kg/s`