Change of power in flow of kinetic energy

Question

Air at 20 m/s, 260 K, 75 kPa with 5 kg/s flows into a jet engine and it flows out at 500 m/s, 800 K, 75 kPa. What is the change (power) in flow of kinetic energy?

Answer 1

m.ΔKE = m.1/2(V2e–V2i)

= 5 kg/s×12 (5002 – 202) (m/s)21

1000 (kW/W) = 624 kW

You can consider air and in these condition the first term of the equation will be cancelled out as there will be no role of enthalpy.

Above is an e.g. Of similar question

The only consideration is of Potential energy and kinetic energy and in the question no mass flow rate is given

PV = RT
V = RT/P =8.314*(100+273)/100 =31.0122

Mass Flow Rate, m = V/v = 31.0122/90=.34456

The continuity Equation
(Ve² –V_i2)/2 +g(Ze - Zi)=(Q +W)/m

Also Q = 0

Putting all the values and soling we get

W=1391.17 kw