Answer to Question #241560 in Molecular Physics | Thermodynamics for Angelo

Question

A reciprocating compressor draws in 500 ft₃/min of air whose density is 0.079lbm/ft₃ and discharges it with density of 0.304lbm/ft₃. At the suction, Ñ€₁ = 15 psia, at the discharge, Ñ€₂ = 80 psia. The increase in the specific internal energy is 33.8 BTU/lb_m. Determine the work on the air in BTU/min and in hp. Neglect change in kinetic energy.

Answer 1

First, we know that:

$v=500 \frac{ft^3}{min} \\ p_1=15\,psia; p_2=80\,psia \\ \rho_1=0.079\frac{lb}{ft^3}; \rho_2=0.304\frac{lb}{ft^3} \\ \Delta u=33.8\frac{BTU}{lb}; Q=13\frac{BTU}{lb}$

Then we use the energy balance of the system:

$p_1v_1-Q=p_2v_2+W+\Delta u \\ \implies W=p_1v_1-(Q+p_2v_2+\Delta u)$

We proceed to find the work done and if we consider the relation $p_iv_i=p_i/\rho_i$ then we can substitute all the terms and find W:

$W=\cfrac{15\frac{lb_f}{in^2}\cdot \frac{144\,in^2}{ft^2}\cdot \frac{1\,BTU}{778\,lb_fft}}{0.079\frac{lb}{ft^3}}-\Bigg(33.8\frac{BTU}{lb}+13\frac{BTU}{lb}+\cfrac{80\frac{lb_f}{in^2}\cdot \frac{144\,in^2}{ft^2}\cdot \frac{1\,BTU}{778\,lb_fft}}{0.304\frac{lb}{ft^3}} \Bigg)$

After substitution we find

$W=(35.1437-95.5079)\frac{BTU}{lb} \\W=-60.3642\frac{BTU}{lb}\times\frac{0.079\,lb}{ft^3}\times\frac{500\,ft^3}{min} \\ W=-2384.386\frac{BTU}{min}\times\frac{1\,hp}{42.4\,BTU/min} \\ W=-56.236\,hp$

In conclusion, the work on the air in BTU/min and in hp is -2384.386 BTU/min and -56.236 hp, respectively.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.