Answer:
0.64 Kg/s
Explanation:
Given
-Fluid: Air
-Inlet 1:
Pressure P1= 1 bar
Temperature T2 = 300 K
-Exit 2:
Pressure P2 = 5 bar
Temperature T2 = 500 K
-Power input to compressor W = — 150 KW
-The rate of heat transfer from the air to cooling water circulating in a water jacket enclosing the compressor Q = —20 KW
Required:
-Mass flow rate m [kg/s]
Assumption :
-Constant average values
-Steady flow
-Kinetic energy effects are ignored
-Potential energy effects are ignored
-Ideal gas.
Solution:
Energy equation could be defined by.
Q- W=m[(h2-h1)+(V_2^2-V_1^2/2)+g(Z_2-Z_1)]
As kinetic and potential energy are ignored.
Q- W=m(h2-h1)
Specific enthalpy for air at P1 = 1 bar and T2= 300 K from table
h1 = 300.1 KJ/Kg
Specific enthalpy for air at P2 = 5 bar and T2 = 500 K from table
h2 = 503.2 KJ/Kg
The mass flow rate could be calculated as following.
m= Q- W/(h2-h1)
= 0.64 Kg/s
