Answer to Question #105409 in Molecular Physics | Thermodynamics for Kule

Question

a working substance enters a thermodynamics steady flow system with the following conditions: p1= 20 psia, v1= 11.7 ft^3/lb, u1=101.6 BTU/lb, v1=150 ft/sec. the working substance leaves the system with the following conditions: p2=25 psia, v2=10.3 ft^3/lb, u2=149.0 BTU/lb, v2=500 ft/sec. Changes in elevation through the system are negligible, and 10 BTU/lb transferred heat is added to the fluid as it passes through the system. Determine the work done on or by the fluid, BTU/lb

Answer 1

W is work done by the fluid,

$$Q-W=u_2+P_2V_2+KE_2-u_1-P_1V_1-KE_1$$

$$P_1V_1=\frac{(20)(144)(11.7)}{778}=43.3\frac{Btu}{lb}$$

$$P_2V_2=\frac{(25)(144)(10.3)}{778}=47.7\frac{Btu}{lb}$$

$$KE_1=\frac{(150)^2}{(64.4)778}=0.4\frac{Btu}{lb}$$

$$KE_2=\frac{(500)^2}{(64.4)778}=5.0\frac{Btu}{lb}$$

Thus,

$$-10-W=149+47.7+5-101.6-43.3-0.4$$

$$W=-66.4\frac{Btu}{lb}$$

 The work done on the fluid,

$$W'=66.4\frac{Btu}{lb}$$