Question

Five pounds per second of fluid enter a steady flow system with P1=100 psia, ρ1=0.2 lb/cu.ft., v1=100 fps, u1=800 BTU/lb. and leave with P2 = 20 psia, ρ2=0.05 lb/cu.ft, v2=500 fps, and u2=780 BTU/lb. During passage through the open system, each pound rejects 10 BTU of heat. Find the work in horsepower.

Answer 1

P₁=100 psia

ρ₁= 0.2 lb/ft³

v₁=100 fps

u₁= 800 BTU/lb

P₂ = 20 psia

ρ₂=0.05 lb/ft³

v₂=500 fps

u₂=780 BTU/lb

Q = 10 BTU/lb

$u_1 + P_1υ_1 + \frac{v_1^2}{2} + Q = u_2 + P_2υ_2 + \frac{v_2^2}{2} + W$

$W = (u_1 – u_2) + (P_1υ_1 – P_2υ_2) + (\frac{v_1^2}{2} - \frac{v_2^2}{2}) + Q$

$υ_1 = \frac{1}{ρ_1} = \frac{1}{0.2} = 5 \;ft^3/lb$

$υ_2 = \frac{1}{ρ_2} = \frac{1}{0.05} = 20 \;ft^3/lb$

$P_1υ_1 – P_2υ_2 = (100 \times 5) – (20 \times 20) = 100 lbft$

1 lbft = 0.18505 BTO/lb

$P_1υ_1 – P_2υ_2 = 100 \times 0.18505 = 18.505 BTU/lb$

$\frac{v_1^2 -v_2^2}{2} = (\frac{100^2 – 500^2}{2}) \times 0.00003994 = -4.7929\; BTU/lb$

W = 20 + 18.505 – 4.7929 – 10 = 23.7121 BTU/lb