Answer to Question #267500 in Molecular Physics | Thermodynamics for Dhangeunsoo

Question

1. During a steady flow process, the pressure of the working substance drops from 400 to 40 psia, the speed increase s from 300 to 1100 fps, the internal energy of the open system decreases 30 Btu/lb, and specific volume increases from 1.2 to 9 ft^3/lb. no heat is transferred. Sketch the energy diagram. Determine the work per lb. Is it done Or by substance? Determine the work in hp for 11 lb per min.( 1 hp = 42.4 Btu/min).

Answer 1

Steady Flow Energy Equation

$h_1 + \frac{v^2_1}{2} + gz_1 + Q = h_2 + \frac{v^2_2}{2} + gz_2 + W \\ h_1 + \frac{v^2_1}{2} = h_2 + \frac{v^2_2}{2} + W \\ h_1 = p_1V_1 + u_1 \\ h_2 = p_2V_2 + u_2 \\ p_1V_1 + u_1 + \frac{v^2_1}{2} = h_2 + \frac{v^2_2}{2} + W \\ p_1 = 400 \; psia = 400 \times 12^2 \; lb/ft^2 \\ p_2 = 40 \; psia = 40 \times 12^2 \; lb/ft^2 \\ v_1 = 300 \; fps \\ v_2 = 1100 \;fps \\ V_1 = 1.2 \; ft^3/lb \\ V_2 = 9 \; ft^3/lb \\ u_1 -u_2 = 30 \;BTU/lb = 23345.04 \; ft \; lb \\ p_1V_1 +(u_1-u_2) + \frac{v^2_1}{2} = p_2V_2 + \frac{v_2^2}{2} + W \\ 400 \times 144 \times 1.2 + 23345.04 + \frac{300^2}{2} = 40 \times 144 \times 9 + \frac{1100^2}{2} + W \\ 69120 + 23345.04 + 45000 = 51840 + 605000 + W \\ W = -519374.96 \; poundal \; ft/lb \\ W = -667.43 \;BTU/lb$

Hence the work is negative so that the work is done on the system.

$Work \; rate = W \times mass \; flowrate \\ = 667.43 \times 11 \\ = 7341.74 \; BTU/min \\ = \frac{7341.74}{42.4} = 173.15 \;HP$