Question 1

The blade in Problem 3 is moving in the x-direction at 20 m/s. It is one of numerous blades on a rotor being energized by six jets each emitting water at 50 m/s. The entering blade angle is
The 45 $45 ^ { \circ }$ angle shown. If the six jets are oriented properly, the maximum power output of the
Rotor would be nearest:

A) 30 kW
B) 24 kW
C) 20 kW
D) 16 kW

16 kW

Accepted Answer

$16 \mathrm {~kW}$ 
$\text {In the two velocity polygons, it is the two relative velocities that are equal, i.e, $V _ { r 1 } =$}$$\text { $V _ { r 2 }$. The first polygon represents $\mathbf { V } _ { 1 } = \mathbf { V } _ { r 1 } + \mathbf { V } _ { B }$ and provides:}$
$$\left. \begin{array} { r l } 
V _ { r 1 } \sin 45 ^ { \circ } & = 50 \sin \beta _ { 1 } \
V _ { r 1 } \cos 45 ^ { \circ } + 20 & = 50 \cos \beta _ { 1 }
\end{array} \right\} \quad \text { Trial-and-error } \rightarrow \quad \beta _ { 1 } = 28.5 ^ { \circ } \text { and } V _ { r 1 } = 33.7 \mathrm {~m} / \mathrm { s }$$
$\text {The second polygon represents $\mathbf { V } _ { 2 } = \mathbf { V } _ { r 2 } + \mathbf { V } _ { B }$ and provides:}$
$$\left. \begin{array} { r } 
33.7 \sin 30 ^ { \circ } = V _ { 2 } \cos \beta _ { 2 } ^ { \prime } \
33.7 \cos 30 ^ { \circ } - 20 = V _ { 2 } \sin \beta _ { 2 } ^ { \prime }
\end{array} \right\} \quad \beta _ { 2 } ^ { \prime } = 28.6 ^ { \circ } \text { and } V _ { 2 } = 19.2 \mathrm {~m} / \mathrm { s }$$
$\text {The $x$-component of the force of the water acting on a blade (see the control volume}$$\text { in Problem 3) is}$
$$R _ { x } = \left[ 1000 \times \left( \pi \times 0.004 ^ { 2 } \right) \times 50 \right] \left( - 19.6 \sin 28.6 ^ { \circ } - 50 \cos 28.5 ^ { \circ } \right) = - 134 \mathrm {~N}$$
$\text {The maximum power output would be}$
$$\dot { W } = 6 R _ { x } V _ { B } = 6 \times 134 \times 20 = 16100 \mathrm {~N} \cdot \mathrm { m } / \mathrm { s } \text { or } \underline { 16.1 \mathrm {~kW} }$$

Question 2

$\text {Water flows over a 2-m-wide flat plate with}$$\text { the same velocities as in Problem 4 in the}$$\text { mini-exam for Section $4.4$ where we found}$ $\text {$\bar { v } = 0.0667 \mathrm {~m} / \mathrm { s }$. The frictional force $F$ }$$\text {acting on the horizontal surface is nearest:}$&#10; &#10;A) 250 N&#10;B) 1500 N&#10;C) 1700 N&#10;D) 1900 N

Accepted Answer

$1700 \mathrm {~N}$
$\text {The momentum equation (4.6.2) for this steady flow applied in the $x$-direction allow}$: $\text {us to calculate $F$ :}$
$\begin{aligned}
-F & =\int_{\text {c.s. }} \rho V_{x}(\mathbf{V} \cdot \hat{\mathbf{n}}) d A=\int_{B} \rho V_{x}^{2} d A-\int_{A} \rho V_{x}^{2} d A+\int_{\text {top }} \rho V_{x}(\mathbf{V} \cdot \hat{\mathbf{n}}) d A \
& =\rho \int_{0}^{0.1} 160^{2}\left(y-5 y^{2}\right)^{2} 2 d y-\rho \int_{0}^{0.1} 8^{2} \times 2 d y+\dot{m}_{\text {top }}\left(V_{x}\right)_{\text {top }} \
& =51.2 \times 10^{6} \int_{0}^{0.1}\left(y^{2}-10 y^{3}+25 y^{4}\right) d y-12800+[1000 \times(4 \times 2) \times 0.0667] \times 8 \
& =6830-12800+4270=-1700 . \quad \therefore F=\underline{1700 \mathrm{~N}}
\end{aligned}$

Question 3

$\text {If $p _ { 1 } = 900 \mathrm { kPa }$, the force of the water on the}$$\text { nozzle is nearest:}$&#10;A) 6910 N  &#10;B) 6470 N&#10;C) 5120 N&#10;D) 4920 N

Accepted Answer

$5120 \mathrm {~N}$
11edd9df_0a1a_66de_bca1_7d7bea70f606_TB8827_11
$\text {The energy equation along with the continuity equation will allow the velocities to}$ $\text {be determined. Continuity demands that $V _ { 2 } = V _ { 1 } \left( 10 ^ { 2 } / 4 ^ { 2 } \right) = 6.25 V _ { 1 }$ so the energy}$ $\text {equation gives}$
$$\begin{aligned}
\frac { V _ { 1 } ^ { 2 } } { 2 g } + \frac { p _ { 1 } } { \gamma } + \not f _ { 1 } & = \frac { \left( 6.25 V _ { 1 } \right) ^ { 2 } } { 2 g } + \frac { p z ^ { 2 } } { / \gamma } + \not f _ { 2 } \
\frac { 900000 } { 9810 } & = \frac { 38.06 V _ { 1 } ^ { 2 } } { 2 \times 9.81 } . \quad \therefore V _ { 1 } = 6.88 \mathrm {~m} / \mathrm { s }
\end{aligned}$$
$\text {The momentum equation (4.6.6) applied to the control volume shown (if you don't}$ $\text {sketch the c.v., you could forget that pressure force) is used to find the force $F _ { x }$ of}$ $\text {the water acting on the contraction (actually, the force of the water on the contraction}$ $\text {is equal and opposite to the force shown which is the force of the contraction on the}$ $\text {water)}$
$$\begin{aligned}
- F _ { x } + p _ { 1 } A _ { 1 } = \dot { m } \left( V _ { 2 } - V _ { 1 } \right) & = \rho A _ { 1 } V _ { 1 } \left( V _ { 2 } - V _ { 1 } \right) \
- F _ { x } + 900000 \times \left( \pi \times 0.05 ^ { 2 } \right) & = \left[ 1000 \times \left( \pi \times 0.05 ^ { 2 } \right) \times 6.88 \right] ( 6.25 - 1 ) \times 6.88 \
\therefore F _ { x } & = \underline { 5120 \mathrm {~N} }
\end{aligned}$$
$\text {Make sure you check the units, especially on the right-hand side. They are}$
$$[ \dot { m } ] [ V ] = \frac { \mathrm { kg } } { \mathrm { s } } \times \frac { \mathrm { m } } { \mathrm { s } } = \frac { \mathrm { N } \cdot \mathrm { s } ^ { 2 } / \mathrm { m } } { \mathrm { s } } \times \frac { \mathrm { m } } { \mathrm { s } } = \mathrm { N }$$
$\text {Remember, if you use $\mathrm { kg } , \mathrm { m } , \mathrm { s }$, and $\mathrm { N }$ in the equations, the units will check out. }$$\text {There's no real need to always work two problems: one with the numbers and}$ $\text {another with the units. Check the units a couple times and then make sure you use }$$\text {$\mathrm { kg } , \mathrm { m } , \mathrm { s }$, and $\mathrm { N }$ when you input numbers into the equations.}$

Question 4

$\text {Water, issuing from an 8-mm-diameter nozzle, impacts the blade}$ $\text {with a velocity $V _ { 1 } = 50 \mathrm {~m} / \mathrm { s }$, as shown. The $x$-component of the }$$\text {force of the water on the blade is nearest}$  &#10;A) 1300 N &#10;B) 200 N&#10;C) 150 N&#10;D) 100 N

Accepted Answer

$198 \mathrm {~N}$
$\text {The velocity of the water as it moves over the blade}$ $\text {remains at $50 \mathrm {~m} / \mathrm { s }$ since the pressure remains constant and}$ $\text {there is negligible elevation change. We ignore any slight}$ $\text {slowing of the water due to friction. Apply the}$ $\text {momentum equation (4.6.6):}$
$$\begin{aligned}
R _ { x } = \dot { m } \left( V _ { 2 x } - V _ { 1 x } \right) & = \left[ 1000 \times \left( \pi \times 0.004 ^ { 2 } \right) \times 50 \right] \left( - 50 \cos 30 ^ { \circ } - 50 \cos 45 ^ { \circ } \right) \
& = - 198 \mathrm {~N}
\end{aligned}$$
The force component above represents the action of the blade on the water. The force of the water on the blade would be $ 198 \mathrm{~N} $.

Deck 11: Flows in Piping Systems