Question

Air flows through a device in which heat and work is exchanged. There is a single inlet and outlet, and the flow at each boundary is steady and uniform. The inlet flow has the following properties: flowrate 50 kg/s, T 25 °C, and velocity 150 m/s. Heat is added to the device at the rate of 42 MW, and the shaft work is -100 kW (assume the efficiency is 100 %). The exit velocity is 400 m/s Calculate the specific stagnation enthalpy (J/kg or kJ/kg) at the inlet, and use the 1st Law to calculate the specific stagnation enthalpy at the exit. Assume constant cp1.0 kJ/kg -K. Calculate the temperature of the air at the exit. Was the assumption of constant cp a good one?

Answer 1

Answer:

11548KJ/kg

10641KJ/kg

Explanation:

Stagnation enthalpy:

$h_{T} = c_{p}*T + \frac{V^2}{2}$

given:

cp = 1.0 KJ/kg-K

T1 = 25 C +273 = 298 K

V1 = 150 m/s

$h_{1} = (1.0 KJ/kg-K) * (298K) + \frac{150^2}{2} \\\\h_{1} = 11548 KJ / kg$

Answer: 11548 KJ/kg

Using Heat balance for steady-state system:

$Flow(m) *(h_{1} - h_{2} + \frac{V^2_{1} - V^2_{2} }{2} ) = Q_{in} + W_{out}$

Qin = 42 MW

W = -100 KW

V2 = 400 m/s

Using the above equation

$50 *( 11548- h_{2} + \frac{150^2 - 400^2 }{2} ) = 42,000 - 100\\\\h_{2} = 10641KJ/kg$

Answer: 10641 KJ/kg

c) We use cp because the work is done per constant pressure on the system.