With the schematic diagram of a pump, tanks, and piping determine the following: 1) The velocity of the liquid at point 2 (ft/s) 2) Pump work used by the liquid (Wpump U) (hp) 3) Electric motor power in out (kVW) 4) In figure 3, the shaded (pink) in the U tube is mercury; determine y. specific gravity of mercury 13.7 5) Mass flow rate Assume the whole system is steady flow with no friction loss in the pipes and tank. Use only the following constant and conversion g= 32.2 ft/s 1 hp= 33,000 ft-Ibf/min 1 hp = 0.746 kW Specific gracity of mercury 13.7 In computing for the density of mercury use the density of water =62.4 lbm/ft³ Do not convert English units to SI units except horsepower to kW Given: in in ft psia psia in in ft 1 3.1 2.1 0.7 0.51 14.1 45 60 4.1 2.1 25 Seat Number Diameter of pipe in point 1 Diameter of pipe in point 2 Specific gravity of the liquid h (ft) (see figure 2) Pressure at point 1 Pressure at point 2 Pump and motor efficiency Diameter of pipe in point 4 Diameter of pipe in point 5 Z3 (see figure 1) Figure 1 Pressurized tank Z3 pump Different diameters See figure 3 See figure 2 motor At point 2 Figure 2 V1 Stagnation point At points 4 and 5 Figure 3 D5 D4 y

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Given Data:

Diameter of pipe at point 1 $d_1=3.1 \mathrm{in}=\frac{3.1}{12}=0.258 \mathrm{ft}$

Diameter of pipe at point 2 $d_2=3.1 \mathrm{in}=\frac{2.1}{12}=0.175 \mathrm{ft}$

The specific gravity of the liquid $S_L=0.7$

Difference between the height of liquid in both pipe $h=0.51 \mathrm{ft}$

Pressure at  point 1 $P_1=14.1 \mathrm{psi} =14.1\times 144=2030.4 \raisebox{1ex}{$\mathrm{lbm}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{ft}}^2$}\right.$

Pressure at  point 2 $P_2=45 \mathrm{psi}=45\times 144=6480 \raisebox{1ex}{$\mathrm{lbm}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{ft}}^2$}\right.$

Head at point 3 $z_3=25 \mathrm{ft}$

Efficiency of pump and motor $\eta =0.60$