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

Answer:

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

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:

$\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )$

$\dot V \approx 13.526\,\frac{ft^{3}}{s}$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,\frac{ft^{3}}{lbm}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu}$

$\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }$

$\dot m = 10.133\,\frac{lbm}{s}$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,\frac{BTU}{lbm}$

State 2 (Saturated Vapor)

$h = 1146.1\,\frac{BTU}{lbm}$

The turbine power output is:

$\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})$

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$