Question

Steam enters a turbine with an enthalpy of 1292 BTU/lb_m. The transferred heat is 13 BTU/lb_m. What is the work in BTU/min and in hp for a flow of 2lb_m/sec?

Answer 1

First, we know that:

$flow=2 \frac{lb}{s} \\ h_1=1292\frac{BTU}{lb}; h_2=1098\frac{BTU}{lb} \\ q=13\frac{BTU}{lb}$

Then we use the energy balance:

$h_1-q=h_2+W \\ \implies W=h_1-(h_2+q) \\ W=1292\frac{BTU}{lb}-(1098\frac{BTU}{lb}+13\frac{BTU}{lb}) \\ \text{ } \\ W= 181 \frac{BTU}{lb} \times \frac{2\,lb}{s}\times \frac{60\,s}{1\,min} \\ \text{ } \\ W=21720 \frac{BTU}{min} \times \frac{1\,hp}{42.4\,BTU/min}=512.264\,hp$

In conclusion, we find that the work in BTU/min and in hp is 21720 BTU/min and 512.264 hp, respectively.

Reference:

• Sta. Maria, Hipolito B., Thermodynamics 1, National Bookstore, Inc., Philippines, 2005.