$\text {Water at $20 ^ { \circ } \mathrm { C }$ is transported in a 10 -cm-diameter cast iron horizontal pipe at a flow rate of}$$\text { $0.02 \mathrm {~m} ^ { 3 } / \mathrm { s }$. The pressure drop over $100 \mathrm {~m}$ of the pipe is nearest:}$
A) $97 \mathrm { kPa }$
B) $84 \mathrm { kPa }$
C) $62 \mathrm { kPa }$
D) $51 \mathrm { kPa }$

Question

Quizplus · Accepted Answer

$84 \mathrm { kPa }$
$\text {The pressure drop is found using Eq. 7.6.23 using $\Delta p / \gamma = h _ { L }$. The average velocity }$$\text {is $V = Q / A = 0.02 / \left( \pi \times 0.05 ^ { 2 } \right) = 2.55 \mathrm {~m} / \mathrm { s }$. To find $f$ we need $e / D$ and Re:}$
$$\frac { e } { D } = \frac { 0.26 } { 100 } = 0.0026 \quad \text { and } \quad \operatorname { Re } = \frac { V D } { v } = \frac { 2.55 \times 0.1 } { 10 ^ { - 6 } } = 2.55 \times 10 ^ { 5 }$$
$\text {From the Moody diagram we find $f = 0.026$. The pressure drop is then}$
$$\frac { \Delta p } { 9810 } = 0.026 \frac { 100 } { 0.1 } \frac { 2.55 ^ { 2 } } { 2 \times 9.81 } . \quad \therefore \Delta p = \underline { 84500 \mathrm {~Pa} }$$
$\text {We could have used Eq. 7.6.29:}$
$$h _ { \mathrm { L } } = 1.07 \frac { 0.02 ^ { 2 } \times 100 } { 9.81 \times 0.1 ^ { 5 } } \left\{ \ln \left[ \frac { 0.26 } { 3.7 \times 100 } + 4.62 \left( \frac { 10 ^ { - 6 } \times 0.1 } { 0.02 } \right) ^ { 0.9 } \right] \right\} ^ { - 2 } = 8.52 \mathrm {~m}$$
$\text {The pressure drop is $\Delta p = \gamma h _ { L } = 9810 \times 8.52 = 83600 \mathrm {~Pa}$. Either answer gives $\underline { 84 \mathrm { kPa } \text {. } }$}$