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A heat exchanger operates at steady state, and both flows through the exchanger are isobaric.

Water flows through one side of the heat exchanger at 20 bar. The inlet temperature of the water is 280 Celsius. The outlet is saturated liquid. The mass flow rate of water is 2.0 kg/s.

Air flows through the other side of the heat exchanger at 1 bar. The inlet temperature of the air is 300 K. The mass flow rate of air is 9.0 kg/s.

Determine the outlet enthalpy of the air, in kJ/kg.

A)759.7

B) 859.7

C)812.2

D)776.2

E)743.6

Respuesta :

yemmy

Answer:

option A

Explanation:

Consider Water:

Mass flow rate (mwater) = 2 Kg/s

Inlet Pressure = 20 Bar

Inlet temperature = 280 C  

Outlet water is saturated liquid at same pressure i.e. 20 Bar  

Inlet water is superheated steam. From superheated steam table, Specific enthalpy of superheated steam at 20 Bar, 280 C (h_in, water) =  

2975.1 kJ/Kg  

From saturated water table, Specific enthalpy of saturated liq. water at 20 Bar ( h_out, water) = 908.77 kJ/Kg  

Total heat lost by water (Qwater)

= Mass flow rate * Difference between Inlet and Outlet Specific Enthalpies

= mwater * ( h in, water - h out, water)

= 2 * (2975.1— 908.77)

= 4132.66 KJ/s  

Consider Air:

Mass flow rate 9 Kg/s

Pressure of air = 1 Bar  

Inlet temperature of air = 300 K

Since air is considered as ideal gas, its enthalpy only depends on temperature and not pressure.

From Air table,

Specific enthalpy of air at 300 K (hin,air) = 300.19 kJ/Kg

All the heat lost by water is gained by air  

Total heat gained by air (Qair)

= Mass flow rate * Difference between Outlet and Inlet Specific Enthalpies

= m__air * ( h_out,air — h_in,air)

= 9 * ( h_out,air — 300.19)  

Thus, 9 * ( hooter — 300.19) = 4132.66  

On solving,

hout,air = 759.374 kJ/kg (Option A)I  

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