Question

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

Answer 1

`\dot W_(out) = 3374.289\,(BTU)/(s)`

Step-by-step explanation:

The model for the turbine is given by the First Law of Thermodynamics:

`- \dot W_(out) + \dot m \cdot (h_(in) - h_(out)) = 0`

The turbine power output is:

`\dot W_(out) = \dot m\cdot (h_(in)-h_(out))`

The volumetric flow is:

`\dot V = (\pi)/(4) \cdot \left( (2)/(12)\,ft \right)^(2)\cdot (620\,(ft)/(s) )`

`\dot V \approx 13.526\,(ft^(3))/(s)`

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

`\\u = 1.33490\,(ft^(3))/(lbm)`

The mass flow is:

`\dot m = (\dot V)/(\\u)`

`\dot m = (13.526\,(ft^(3))/(s) )/(1.33490\,(ft^(3))/(lbm) )`

`\dot m = 10.133\,(lbm)/(s)`

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

`h = 1479.74\,(BTU)/(lbm)`

State 2 (Saturated Vapor)

`h = 1146.1\,(BTU)/(lbm)`

The turbine power output is:

`\dot W_(out) = (10.133\,(lbm)/(s) )\cdot (1479.1\,(BTU)/(lbm)-1146.1\,(BTU)/(lbm))`

`\dot W_(out) = 3374.289\,(BTU)/(s)`