Question 1

$\text {Water flows over a flat plate so that at }$$\text {the downstream end it has the velocity}$ $\text {profile shown. The average velocity$\bar { v }$}$ $\text {across the upper horizontal surface is }$$\text {nearest:}$&#10;A) 0.043 m/s  &#10;B) 0.067 m/s&#10;C) 0.082 m/s&#10;D) 0.097 m/s

Accepted Answer

$0.0667 \mathrm {~m} / \mathrm { s }$
$\text {The water enters on the left and leaves on the right but at a lower velocity (the}$$\text { velocity at the top of the profile on the right where $y = 0.1$ is $8 \mathrm {~m} / \mathrm { s }$ ), so there must}$ $\text {also be a velocity across the top surface. Assume unit depth (into the paper) and}$ $\text {continuity gives}$
$$\begin{aligned}
A _ { 1 } u _ { 1 } & = A _ { 2 } \bar { v } + \int _ { 0 } ^ { 0.1 } u ( y ) d y \
0.1 \times 8 & = 4 \times \bar { v } + \int _ { 0 } ^ { 0.1 } 160 \left( y - 5 y ^ { 2 } \right) d y . \quad \therefore \bar { v } = \underline { 0.0667 \mathrm {~m} / \mathrm { s } }
\end{aligned}$$

Question 2

The exit velocity is 12 m/s. The inlet velocity of the water is&#10;A) 6.31 m/s&#10;B) 4.32 m/s nearest:&#10; &#10;C) 3.95 m/s&#10;D) 3.24 m/s

Accepted Answer

$4.32 \mathrm {~m} / \mathrm { s }$
$\text {Eq. 4.4.6 can be used to find $V _ { 1 }$ :}$
$$\begin{array} { c } 
A _ { 1 } V _ { 1 } = A _ { 2 } V _ { 2 } \
\pi \times 0.02 ^ { 2 } \times V _ { 1 } = \pi \times 0.012 ^ { 2 } \times 12 . \quad \therefore V _ { 1 } = \underline { 4.32 \mathrm {~m} / \mathrm { s } }
\end{array}$$
$\text {We used the radius in meters: $r _ { 1 } = 0.2 \mathrm {~m}$ and $r _ { 2 } = 0.12 \mathrm {~m}$.}$

Question 3

$\text {Water flows into and out of the tank. If $V = 8 \mathrm {~m} / \mathrm { s }$ and}$ $Q = 0.02 \mathrm {~m} ^ { 3 } / \mathrm { s }$, $\text {the mass flux is nearest:}$&#10;A) 6 kg/s  &#10;B) 8 kg/s&#10;C) 10 kg/s&#10;D) 12 kg/s

Accepted Answer

$10 \mathrm {~kg} / \mathrm { s }$
$\text {The tank has two inlets and one outlet so the continuity equation (the conservation of}$ $\text {mass) demands the following:}$
$$\begin{array} { l } 
\rho A V + \dot { m } = \rho Q \
1000 \times \pi \times 0.02 ^ { 2 } \times 8 + \dot { m } = 1000 \times 0.02 . \quad \therefore \dot { m } = \underline { 9.95 \mathrm {~kg} / \mathrm { s } } \
\end{array}$$
$\text {The $60 - \mathrm { mm }$-diameter does not enter the solution. The density of water is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$}$ $\text {and the radius is $0.02 \mathrm {~m}$.}$

Question 4

\text {A stratified liquid flows in a \(80 - \mathrm { cm }\)-wide, 40 -cm-high channel with the density distribution }

\rho ( y ) = 800 ( 1 + 0.2 y ) \mathrm { kg } / \mathrm { m } ^ { 3 }

\text {and a uniform velocity distribution of \(1.2 \mathrm {~m} / \mathrm { s }\), with \(y\) measured}

\text { from the bottom of the channel. The mass flux is nearest:}

(A)

260 \mathrm {~kg} / \mathrm { s }

(B)

280 \mathrm {~kg} / \mathrm { s }

(C)

300 \mathrm {~kg} / \mathrm { s }

(D)

320 \mathrm {~kg} / \mathrm { s }

Accepted Answer

$320 \mathrm {~kg} / \mathrm { s }$
$$\begin{aligned}
\dot { m } = \int _ { A } \rho u d A & = \int _ { 0 } ^ { 0.4 } 800 ( 1 + 0.2 y ) 1.2 \times 0.8 d y \
& = 768 \int _ { 0 } ^ { 0.4 } ( 1 + 0.2 y ) d y = \underline { 319 \mathrm {~kg} / \mathrm { s } }
\end{aligned}$$

Deck 9: Compressible Flow