The pump in Fig. 5–25 discharges water at 80 (10³) liters/h. The pressure at A is 150 kPa, whereas the pressure in the pipe at B is 500 kPa. As the water passes through the filter, it causes the internal energy of the water to increase by 50 J/kg due to the effect of friction, while at the same.

Question

The pump in Fig. 5–25 discharges water at 80 (10³) liters/h. The pressure at A is 150 kPa, whereas the pressure in the pipe at B is 500 kPa. As the water passes through the filter, it causes the internal energy of the water to increase by 50 J/kg due to the effect of friction, while at the same time there is a heat conduction loss from the water of 250 J/s. Determine the horsepower that is developed by the pump.

Accepted Answer

Fluid Description. We have steady flow into and out of the pump.
The water is considered incompressible, but viscous friction losses occur. ρ = 1000 kg/m³.

Control Volume. The fixed control volume contains the water within the pump, filter, and pipe extensions. Since there is 1 liter in 1000 cm³ of water, the volumetric and mass flows are

[latex] Q=\left[80\left(10^3 ight) \frac{ ext { liter }}{ h } ight]\left(\frac{1000 cm ^3}{1 ext { liter }} ight)\left(\frac{1 m }{100 cm } ight)^3\left(\frac{1 h }{3600 s } ight)=0.02222 m ^3 / s [/latex]

and

[latex] \dot{m}= ho Q=\left(1000 kg / m ^3 ight)\left(0.02222 m ^3 / s ight)=22.22 kg / s [/latex]

Therefore, the velocities at A (in) and B (out) are

[latex] \begin{aligned} & Q=V_A A_A ; 0.02222 m ^3 / s =V_A\left[\pi(0.075 m )^2 ight] ; V_A=1.258 m / s \ & Q=V_B A_B ; \quad 0.02222 m ^3 / s =V_B\left[\pi(0.025 m )^2 ight] ; V_B=11.32 m / s \end{aligned} [/latex]

Energy Equation. Since there is heat conduction loss, [latex] \dot{Q}_{ ext {in }} [/latex] is negative; that is, the heat flows out. Also, there is no elevation change in the flow from A to B. For this problem, we will apply Eq. 5–12.

[latex] \begin{aligned} & \dot{Q}_{ ext {in }}+\dot{W}_{ ext {pump }}-\dot{W}_{ ext {turbine }}= \ & {\left[\left(\frac{p_{ ext {out }}}{ ho_{ ext {out }}}+\frac{V_{ ext {out }}^2}{2}+g z_{ ext {out }}+u_{ ext {out }} ight)-\left(\frac{p_{ ext {in }}}{ ho_{ ext {in }}}+\frac{V_{ ext {in }}{ }^2}{2}+g z_{ ext {in }}+u_{ ext {in }} ight) ight] \dot{m}} \end{aligned} [/latex] (5-12)

[latex] \begin{aligned} \dot{Q}_{ ext {in }}+\dot{W}_{ ext {pump }}-\dot{W}_{ ext {turbine }}= & {\left[\left(\frac{p_B}{ ho}+\frac{V_B^2}{2}+g z_B+u_B ight) ight.} \ & \left.-\left(\frac{p_A}{ ho}+\frac{V_A^2}{2}+g z_A+u_A ight) ight] \dot{m} \end{aligned} [/latex]

[latex] \begin{aligned} & -250 J / s +\dot{W}_{ ext {pump }}-0 \ & =\left[\left(\frac{500\left(10^3 ight) N / m ^2}{1000 kg / m ^3}+\frac{(11.32 m / s )^2}{2}+g z+50 J / kg ight) ight. \ & \left.-\left(\frac{150\left(10^3 ight) N / m ^2}{1000 kg / m ^3}+\frac{(1.258 m / s )^2}{2}+g z+0 ight) ight](22.22 kg / s ) \ & \dot{W}_{ ext {pump }}=10.54\left(10^3 ight) W =10.5 kW \end{aligned} [/latex]

The positive result indicates that indeed energy is added to the water within the control volume using the pump.

Question 5.11: The pump in Fig. 5–25 discharges water at 80 (10³) liters/h....

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