A square smooth conduit, 80 cm on a side, circulates 35 $$35 ^ { \circ } \mathrm { C }$$ C air in a large building. The pressure drop over a 50-m length of the horizontal conduit is not to exceed 800 Pa. Assume an average
Pressure in the conduit of 120 kPa. The maximum flow rate in the conduit is nearest:
A) $37 \mathrm {~m} ^ { 3 } / \mathrm { s }$
B) $29 \mathrm {~m} ^ { 3 } / \mathrm { s }$
C) $18 \mathrm {~m} ^ { 3 } / \mathrm { s }$
D) $11 \mathrm {~m} ^ { 3 } / \mathrm { s }$

Question

A square smooth conduit, 80 cm on a side, circulates 35 $$35 ^ { \circ } \mathrm { C }$$ C air in a large building. The pressure drop over a 50-m length of the horizontal conduit is not to exceed 800 Pa. Assume an average
Pressure in the conduit of 120 kPa. The maximum flow rate in the conduit is nearest: 
A) $37 \mathrm {~m} ^ { 3 } / \mathrm { s }$
B) $29 \mathrm {~m} ^ { 3 } / \mathrm { s }$
C) $18 \mathrm {~m} ^ { 3 } / \mathrm { s }$
D) $11 \mathrm {~m} ^ { 3 } / \mathrm { s }$

Quizplus · Accepted Answer

$29 \mathrm {~m} ^ { 3 } / \mathrm { s }$
$\text {The average density and hydraulic radius are}$
$$\rho = \frac { p } { R _ { \text {air } } T } = \frac { 120 } { 0.287 \times 308 } = 1.36 \mathrm {~kg} / \mathrm { m } ^ { 3 } . \quad R = \frac { A } { P } = \frac { 0.8 \times 0.8 } { 4 \times 0.8 } = 0.2 \mathrm {~m}$$
$\text {The head loss is}$
$$h _ { L } = \frac { \Delta p } { \rho g } = f \frac { L } { 4 R } \frac { V ^ { 2 } } { 2 \notin } \quad \text { or } \quad \frac { 800 } { 1.36 } = f \frac { 50 } { 4 \times 0.2 } \frac { V ^ { 2 } } { 2 } \quad \text { or } \quad f V ^ { 2 } = 19$$ The Moody diagram gives another relation between f and V (thru Re). So, a trial-and-error
technique follows using
 $$\operatorname { Re } = 4 \mathrm { VR } / \nu = 4 \times 0.2 \mathrm {~V} / 1.25 \times 10 ^ { - 5 } = 6.4 \times 10 ^ { 4 } \mathrm {~V}$$
$\text {where $v = \mu / \rho = 1.7 \times 10 ^ { - 5 } / 1.36 = 1.25 \times 10 ^ { - 5 } \mathrm {~m} ^ { 2 } / \mathrm { s }$ (find $\mu$ using Fig. B.1):}$
$\text {$f = 0.015$ : Eq. (1) gives $V = 36 \mathrm {~m} / \mathrm { s }$. Fig. $7.13$ gives $\operatorname { Re } = 2.5 \times 10 ^ { 5 }$ so from (2), $V = 3.9 \mathrm {~m} / \mathrm { s }$}$
$\text {$f = 0.01$ : Eq. (1) gives $V = 44 \mathrm {~m} / \mathrm { s }$. Fig. $7.13$ gives $\operatorname { Re } = 2.5 \times 10 ^ { 6 }$ so from (2), $V = 39 \mathrm {~m} / \mathrm { s }$}$
$\text {$f = 0.009$ : Eq. (1) gives $V = 46 \mathrm {~m} / \mathrm { s }$. Fig. $7.13$ gives $\operatorname { Re } = 5 \times 10 ^ { 6 }$ so from (2), $V = 78 \mathrm {~m} / \mathrm { s }$}$
$\text {Use $V = 45 \mathrm {~m} / \mathrm { s }$ so that the flow rate is $Q = A V = 0.8 \times 0.8 \times 45 = \underline { 28.8 \mathrm {~m} ^ { 3 } / \mathrm { s } }$.}$

A Square Smooth Conduit, 80 Cm on a Side, Circulates 35∘C35 ^ { \circ } \mathrm { C }35∘C

A Square Smooth Conduit, 80 Cm on a Side, Circulates $35 ^ { \circ } \mathrm { C }$