$\text {The pump inputs $2 \mathrm { hp }$ into the flow.}$
$\text { The cast-iron pipe has a total length}$
$\text {of $100 \mathrm {~m}$. Estimate the flow rate.}$
A) $0.018 \mathrm {~m} ^ { 3 } / \mathrm { s }$
B) $0.024 \mathrm {~m} ^ { 3 } / \mathrm { s }$
C) $0.034 \mathrm {~m} ^ { 3 } / \mathrm { s }$
D) $0.042 \mathrm {~m} ^ { 3 } / \mathrm { s }$

Question

$\text {The pump inputs $2 \mathrm { hp }$ into the flow.}$
$\text { The cast-iron pipe has a total length}$
 $\text {of $100 \mathrm {~m}$. Estimate the flow rate.}$
A) $0.018 \mathrm {~m} ^ { 3 } / \mathrm { s }$ 
B) $0.024 \mathrm {~m} ^ { 3 } / \mathrm { s }$ 
C) $0.034 \mathrm {~m} ^ { 3 } / \mathrm { s }$ 
D) $0.042 \mathrm {~m} ^ { 3 } / \mathrm { s }$

Quizplus · Accepted Answer

11edd9db_26f8_388d_bca1_e17c35a88ba0_TB8827_11
$\text {Assume that the friction factor is in the "completely turbulent regime" in Fig. 7.13.}$
$\text { With $e / D = 0.26 / 80 = 0.00325$. Then $f \cong 0.027$. The energy equation from the surface}$ $\text {of the water to the nozzle exit is}$
$$\begin{aligned}
\frac { Y _ { 1 } ^ { \prime } } { 2 g } + \frac { y _ { 1 } } { \gamma } + z _ { 1 } + H _ { p } & = \frac { V _ { 2 } ^ { 2 } } { 2 g } + \frac { y _ { 2 } } { \gamma } + \not f _ { 2 } + h _ { L } \
8 + \frac { 2 \times 746 } { 9810 \times \pi \times 0.04 ^ { 2 } V } & = \frac { ( 4 V ) ^ { 2 } } { 2 \times 9.81 } + 0.027 \frac { 100 } { 0.08 } \frac { V ^ { 2 } } { 2 \times 9.81 } + 0.8 V ^ { 2 } \
8 + \frac { 303.3 } { V } & = 3.34 V ^ { 2 }
\end{aligned}$$
$\text {where we used $V _ { 2 } = 4 \mathrm {~V}$, the velocity in the pipe being $\mathrm { V }$. We also used $K = 0.8$ for the }$$\text {reentrant entrance. The loss thru a nozzle can be neglected. A trial-and-error solution}$$\text { is used as follows $( V = 5$ appears close $)$ :}$
Let $V = 5 : \quad 68.7 \neq 83.5$
Let $V = 4 : \quad 83.8 \neq 53.4$
Let $V = 4.7 : \quad 72.5 \neq 73.8$
$\text {results in $V \cong 4.7 \mathrm {~m} / \mathrm { s }$. Check $\operatorname { Re } = V D / \nu = 4.7 \times 0.8 / 10 ^ { - 6 } = 3.8 \times 10 ^ { 6 } . \therefore$ OK.}$
$$Q = A V = 4.7 \times \pi \times 0.04 ^ { 2 } = \underline { 0.0236 \mathrm {~m} ^ { 3 } / \mathrm { s } }$$
$\text {Note: If $2 \mathrm { hp }$ is input into the flow, no efficiency is used. An $80 \%$ efficient pump}$ $\text {would require an energy input to the pump of $2 / 0.8 = 2.5 \mathrm { hp }$.}$

The pump inputs 2hp into the flow.\text {The pump inputs \(2 \mathrm { hp }\) into the flow.}The pump inputs 2hp into the flow.

$\text {The pump inputs \(2 \mathrm { hp }\) into the flow.}$