Question 5.13: The irrigation pump in Fig. 5–27a is used to supply water to......
The irrigation pump in Fig. 5–27a is used to supply water to the pond at B at a rate of 0.09 m³/s. If the pipe is 150 mm in diameter, determine the required power of the pump. Assume the frictional head loss per meter length of pipe is 0.1 m/m. Draw the energy and hydraulic grade lines for this system.
Fluid Description. Here we have steady flow. The water is assumed to be incompressible, but viscous friction losses occur. \gamma = 9.81(10^3) N/m^3.
Control Volume. We will select a fixed control volume that contains the water within the reservoir A, along with that in the pipe and pump. For this case, the velocity at A is essentially zero, and the pressure at the inlet and outlet surfaces A and B is zero. Since the volumetric flow is known, the average velocity at the outlet is
Q = V_BA_B; 0.09 m^3/s = V_B[\pi (0.075 m)^2]
Energy Equation. Establishing the gravitational datum at A, and applying the energy equation between A (in) and B (out), we have
\frac{p_A}{\gamma} + \frac{V_A^2}{2g} + z_A + h_{pump} = \frac{p_B}{\gamma} + \frac{V_B^2}{2g} + z_B + h_{turbine} + h_{L}0 + 0 + 0 + h_{pump} = 0 + \frac{(\frac{16}{\pi} m/s)}{2(9.81 m/s^2)} + 4 m + 0 + (0.1 m/m)(7.5 m)
h_{pump} = 6.072 m
This positive result indicates the pump head of water is transferred into the system by the pump.
Power. Using Eq. 5–17, the pump must therefore have a power of
\dot{W}_s=\dot{m} g h_s=Q \gamma h_s (5–17)
\dot{W_s} = Q\gamma_w h_{pump} = (0.09 m^3/s)[9.81 (10^3) N/m^3](6.072 m)= 5.361(10^3) W = 5.36 kW
Of this amount, the power needed to compensate for the frictional head loss is
\dot{W_L} = Q\gamma_w h_{L} = (0.09 m^3/s)[9.81 (10^3) N/m^3][(0.1 m/m)(7.5 m)]= 662.18 W
EGL and HGL. Recall the EGL is a plot of the total head H = p/\gamma + V^2/2g + z along the pipe. The HGL lies V^2/2g below and parallel to the EGL. Here the velocity head is
\frac{V^2}{2g} = \frac{(\frac{16}{\pi} m/s)}{2(9.81 m/s^2)} = 1.322 mIt remains constant along the pipe, since the pipe has the same diameter throughout its length. Point A′ in Fig. 5–27 is where the water reaches its velocity V, having been accelerated from rest at the entrance of the pipe. At A′, C, and D, the pressure head is
\frac{p_A}{\gamma} + \frac{V_A^2}{2g} + z_A + h_{pump} = \frac{p_{A′}}{\gamma} + \frac{V_{A′}^2}{2g} + z_{A′} + h_{turb} + h_{L}0 + 0 + 0 + 0 = \frac{p_{A′}}{\gamma} + 1.322 m + 0 + 0 + 0
\frac{p_{A′}}{\gamma} = – 1.322 m
\frac{p_A}{\gamma} + \frac{V_A^2}{2g} + z_A + h_{pump} = \frac{p_{C}}{\gamma} + \frac{V_{C}^2}{2g} + z_{C} + h_{turb} + h_{L}
0 + 0 + 0 + 0 = \frac{p_C}{\gamma} + 1.322 m + 0 + 0 + (0.1 m/m)(3.5 m)
\frac{p_C}{\gamma} = 1.672 m
\frac{p_A}{\gamma} + \frac{V_A^2}{2g} + z_A + h_{pump} = \frac{p_{D}}{\gamma} + \frac{V_{D}^2}{2g} + z_{D} + h_{turb} + h_{L}
0 + 0 + 0 + 0 = \frac{p_D}{\gamma} + 1.322 m + 4 m + 0 + (0.1 m/m)(7.5 m)
\frac{p_D}{\gamma} = – 6.072 m
The signs indicate a negative pressure caused by the suction of the pump. The total head at A′, C, D, and B is therefore
H = \frac{p}{\gamma} + \frac{V^2}{2g} + z
H_A′ = -1.322 m + 1.322 m + 0 = 0
H_C = -1.672 m + 1.322 m + 0 = -0.35 m
H_D = -6.072 m + 1.322 m + 4 m = -0.75 m
H_B = 0 + 1.322 m + 4 m = 5.32 m
Stretching the pipe out, these values are plotted in Fig. 5–27b to produce the EGL. The HGL lies 1.322 m below and parallel to the EGL.
