Answer:
Exit Temperature, T_2 = 107.9 °F
Volume Flow Rate = 47.39 ft^3/s
Explanation:
Given Data
<u>Inlet Conditions
</u>
Pressure, P_1 = 450 psia (psia = lbf/in^2)
Temperature, T_1 = 800 °F
<u>Outlet Conditions
</u>
Pressure, P_2 = 1.2 psia (psia = lbf/in^2)
Saturated Vapor (Hence quality is 01)
Power, W = 12000 hp (1 hp = 2545 Btu/hr)
Heat transfer from the turbine to surroundings, Q = -2000000 Btu/hr
Required
- The exit temperature, in °F
- The volume flow rate of the steam at the inlet, in ft^3/s
Calculations
We will be solving this table using the property tables (English Units)
From table A-4E for steam at inlet condition,
Enthalpy, h_1 = 1415 Btu/lb
Volume, v_1 = 1.608 ft^3/lb
From table A-5E for steam at outlet condition,
Enthalpy, h_2 = 1108 Btu/lb
Exit Temperature, T_2 = 107.9 °F
As kinetic and potential energy are ignored, the energy equation will be:
Q – W = m*(h_2 – h_1)
m is the mass flow rate
m = ((-2000000)-(12000*2545))/(1108-1415)*3600
m = 29.47 lb/s
Mass Flow rate = Volume Flow Rate / v_1
Volume Flow Rate = m*v_1
Volume Flow Rate = 47.39 ft^3/s
