Answer:
option A
Step-by-step 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