Redo Problem 7.198E if the water pump has an isentropic efficiency of 85% (hose, nozzle included).
Redo Problem 7.198E if the water pump has an isentropic efficiency of 85% (hose, nozzle included).
C.V.: pump + hose + water column, total height difference 110 ft. Here V is velocity, not volume.
Continuity Eq.4.3, 4.11: \dot{ m }_{ in }=\dot{ m }_{ ex }=(\rho A V )_{ nozzle }
Energy Eq.4.12: \dot{ m }\left(- w _{ p }\right)+\dot{ m }\left( h + V ^{2} / 2+ g z\right)_{ in }=\dot{ m }\left( h + V ^{2} / 2+ g z\right)_{ ex }
Process:
\begin{array}{l}h _{ in } \cong h _{ ex }, \quad V _{ in } \cong V _{ ex }=0, \quad z _{ ex }- z _{ in }=110 ft , \quad \rho=1 / v \cong 1 / v _{ f } \\- w _{ p }= g \left( z _{ ex }- z _{ in }\right)=32.174 \times(110-0) / 25037=0.141 Btu / lbm\end{array}
Recall the conversion 1 Btu / lbm =25037 ft ^{2} / s ^{2} from Table A.1. The velocity in the exit nozzle is such that it can rise 30 ft. Make that column a C.V. for which Bernoulli Eq.7.17 is:
gz _{ noz }+\frac{1}{2} V _{ noz }^{2}= g z _{ ex }+0
\begin{aligned}V _{\text {noz }} &=\sqrt{2 g\left(z_{e x}-z_{n o z}\right)} \\&=\sqrt{2 \times 32.174 \times 30}=43.94 ft / s\end{aligned}
Assume: v = v _{ F , 70 F }=0.01605 ft ^{3} / lbm
\dot{ m }=\frac{\pi}{ v _{ f }}\left(\frac{ D }{2}\right)^{2} V _{ noz }=(\pi / 4)\left(1^{2} / 144\right) \times 43.94 / 0.01605=14.92 lbm / s\dot{ W }_{\text {pump }}=\dot{ m } w _{ p } / \eta=14.92 \times 0.141 \times(3600 / 2544) / 0.85= 3 . 5 h p
……………………………
Eq.4.3 : \dot{m}=\rho_{ avg } \dot{V}=\dot{V} / v=\int\left( V _{\text {local }} / v\right) d A= V A / v
Eq.4.11 : \dot{m}_{i}=\dot{m}_{e}=\dot{m}
Eq.4.12 : \dot{Q}_{ C.V .}+\dot{m}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g Z_{i}\right)=\dot{m}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g Z_{e}\right)+\dot{W}_{ C . V. }
Eq.7.17 : v P_{i}+\frac{1}{2} V _{i}^{2}+g Z_{i}=v P_{e}+\frac{1}{2} V _{e}^{2}+g Z_{e}