Question #267500

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).


1
Expert's answer
2021-11-17T18:30:36-0500

Steady Flow Energy Equation

h1+v122+gz1+Q=h2+v222+gz2+Wh1+v122=h2+v222+Wh1=p1V1+u1h2=p2V2+u2p1V1+u1+v122=h2+v222+Wp1=400  psia=400×122  lb/ft2p2=40  psia=40×122  lb/ft2v1=300  fpsv2=1100  fpsV1=1.2  ft3/lbV2=9  ft3/lbu1u2=30  BTU/lb=23345.04  ft  lbp1V1+(u1u2)+v122=p2V2+v222+W400×144×1.2+23345.04+30022=40×144×9+110022+W69120+23345.04+45000=51840+605000+WW=519374.96  poundal  ft/lbW=667.43  BTU/lbh_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×mass  flowrate=667.43×11=7341.74  BTU/min=7341.7442.4=173.15  HPWork \; rate = W \times mass \; flowrate \\ = 667.43 \times 11 \\ = 7341.74 \; BTU/min \\ = \frac{7341.74}{42.4} = 173.15 \;HP


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