Question 4.22: A pump is required to dewater a site as illustrated in Figur...

(a) Assuming the datum is at point 1, to determine the maximum height, z_{s} to which the fluid may be pumped (in order to establish a flow in the suction line), the energy equation is applied between points 1 and s, assuming a vapor pressure, P_{V} of the liquid at point s in order to avoid cavitation, and assuming no flow in the suction line as follows:

Z_{1}: = 0m                   P_{1}:= 0 \frac{N}{m^{2}}                   V_{1}: = 0 \frac{m}{sec}                   V_{s}: = 0 \frac{m}{sec}
P_{V}:= 2.34 \times 10^{3} \frac{N}{m^{2}} \times 10^{3} \frac{N}{m^{2}} = -9.899 \times 10^{4} \frac{N}{m^{2}}
P_{s}:= P_{V} = -9.899 \times 10^{4} \frac{N}{m^{2}}                   h_{ftotal1s}: = 0.7 m
\rho : = 998 \frac{kg}{m^{3}}                   g: = 9.81 \frac{m}{sec^{2}}                   \gamma : = \rho .g = 9.79 \times 10^{3} \frac{kg}{m^{2}s^{2}}

Guess value:                   Z_{s}: = 9 m

Given

\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} – h_{ftotal1s} = \frac{P_{s}}{\gamma } + Z_{s} + \frac{V^{2}_{s} }{2.g}
Z_{s}: Find ( Z_{s}) = 9.41 m

However, in order to establish a nonzero flow in the suction line, the maximum height, z_{s} to which the fluid may be pumped will be set at z_{s} = 9 m .
(b) In order to determine the maximum allowable velocity of the flow in the pipe without causing cavitation at the suction side, s of the pump, the energy equation is applied between points 1 and s, assuming that the pressure at point s is the vapor pressure of the water at 20^{◦} C as follows:

Guess value:                  V_{s}: = 1 \frac{m}{sec}

Given

\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} – h_{ftotal1s} = \frac{P_{s}}{\gamma } + Z_{s} + \frac{V^{2}_{s} }{2.g}
V_{s}: Find ( V_{s}) = 2.838 \frac{m}{s}

(c) To determine the minimum required design pipe size diameter in order to accommodate the maximum allowable velocity and the required flowrate, the continuity equation is applied as follows:

Guess value:                    D:= 1 m                       A: = 1 m^{2}

Given

Q = V_{s}.A
A= \frac{\pi .D^{2}}{4}
\left ( \begin{matrix} D \\ A \end{matrix} \right ) : = Find (D, A)
D = 0.449 m                         A: = 0.159 m^{2}

(d) The required head to be delivered by the pump is determined by applying the energy equation between points 1 and 2 as follows:

P_{2}: = 0 \frac{N}{m^{2}}                         V_{2}: = 0 \frac{m}{sec}                         Z_{2}: = 150 m
h_{ftotal1s}: = 0.7 m                         h_{ftotalds}: = 1.2 m

Guess value:                        h_{pump}: = 200 m

\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} – h_{ftotal1s} – h_{ftotalds} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2} }{2.g}
h_{pump}: Find ( h_{pump}) = 151.9 m

Thus, the required head to be delivered by the pump, h_{pump} = 151.9 m must overcome the difference in elevation between the two reservoirs, Δz= 150 m and the total head loss due to pipe friction and pipe components between points 1 and 2, h_{f,total} = 0.7 m + 1.2 m = 1.9 m.
(e) The required hydraulic power output from the pump is computed by applying Equation 4.186 h_{pump} = \frac{(P_{p})_{out}}{\gamma Q} = \frac{(\eta _{pump})(P_{p})_{in}}{\gamma Q} = \frac{(\eta _{pump})(w T_{shaft,in})}{\gamma Q} as follows:

(f) Assuming a typical pump efficiency of 80%, the required shaft power input to the pump by the motor is computed by applying Equation 4.165 \eta _{pump} = \frac{hydraulic power}{shaft power} = \frac{(\overset{\cdot }{W} _{pump})_{u}}{\overset{\cdot }{W} _{pump}} = \frac{(\overset{\cdot }{W} _{pump})_{u}}{\overset{\cdot }{W} _{shaft,in}} = \frac{(\overset{\cdot }{W} _{pump})_{u}}{wT _{shaft,in}} = \frac{(P_{p})_{out}}{(P_{p})_{in}} as follows:

Guess value:                      P_{pin}: = 700 kW

Given

\eta _{pump} = \frac{P_{pout}}{P_{pin}}
P_{pin}: = Find ( P_{pin})= 836.527 kW

(g) Assuming a typical motor efficiency of 85%, the required chemical or electric power input to the motor is computed by applying Equation 4.169 \eta _{motor} = \frac{shaft power output from motor}{electric power input from electical source} =\frac{W_{shaft,out}}{W_{elect,in}} = \frac{P_{out of motor}}{P_{in of motor}} = \frac{(P_{m})_{out}}{(P_{m})_{in}} as follows:

P_{mout}: = P_{pin} = 836.527 kW                     \eta _{motor}: = 0.85

Guess value:                    P_{min}: = 800 kW

\eta _{motor} = \frac{P_{mout}}{P_{min}}
P_{min}: = Find (P_{min}) = 984.149 kW

(h) The resulting overall pump–motor system efficiency is computed by applying Equation 4.172 \eta _{pump – motor}= \frac{hydraulic power}{electric power} = \frac{(\overset{\cdot }{W_{pump}} )_{u}}{\overset{\cdot }{W_{elect,in}}} = \frac{(P_{p})_{out}}{(P_{p})_{in}} = \eta _{pump}\eta _{motor} as follows:

\eta _{pumpmotor}: = \frac{P_{pout}}{P_{min}} = 0.68                             \eta _{pumpmotor}: = \eta _{pump}. \eta _{motor} = 0.68