Question

Steam enters a turbine operating at steady state at 800°F and 450 lbf/in^2 and leaves as a saturated vapor at 1.2 lbf/in^2. The turbine develops 12,000 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 2 x 106 Btu/h. Neglect kinetic and potential energy changes from inlet to exit. Determine the exit temperature, in °F, and the volumetric flow rate of the steam at the inlet, in ft3/s.

Answer 1

Answer:

Exit Temperature, T_2 = 107.9 °F

Volume Flow Rate = 47.39 ft^3/s

Explanation:

Given Data  

Inlet Conditions

Pressure, P_1 = 450 psia (psia = lbf/in^2)

Temperature, T_1 = 800 °F

Outlet Conditions

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