Answer to Question #235873 in Molecular Physics | Thermodynamics for Unknown346307

Question

The working fluid, in a steady flow process flows at a rate of 220 kg/min.
The fluid rejects 100 kJ/s passing through the system. The conditions of the fluid at inlet and
outlet are given as : C1 = 320 m/s, p1 = 6.0 bar, u1 = 2000 kJ/kg, v1 = 0.36 m3/kg and C2 =
140 m/s, p2 = 1.2 bar, u2 = 1400 kJ/kg, v2 = 1.3 m3/kg. The suffix 1 indicates the condition at inlet
and 2 indicates at outlet of the system.
Determine the power capacity of the system in MW.
The change in potential energy may be neglected.

Answer 1

$m(u_1+p_1v_1+\frac{c_1^2}2+z_1g)-Q=m(u_2+p_2v_2+\frac{c_2^2}2+z_2g)+W,$

$z_1-z_2=0,$

$W=m((u_1-u_2)+(p_1v_2-p_2v_2)+\frac{c_1^2-c_2^2}2)-Q=2.47~MW.$