Question 1

A rectangular tank $80 \mathrm {~cm}$ high and $180 \mathrm {~cm}$ long is filled with water and pressurized to $20 \mathrm { kPa }$ at the top. If it is accelerated at the rate of $5 \mathrm {~m} / \mathrm { s } ^ { 2 }$ on a horizontal plane in the direction of its longer side, what is the highest pressure in the tank?

A) 49.1 kPa
B) 44.9 kPa
C) 41.2 kPa
D) 36.8 kPa

36.8 kPa

Accepted Answer

$36.8 \mathrm { kPa }$
With $a _ { z } = 0$, Eq. 2.5.3 gives $h$ as
$$h = L \tan \alpha = 1.8 \times \frac { a _ { x } } { g } . \quad \therefore h = \frac { 1.8 \times 5 } { 9.81 } = 0.917 \mathrm {~m}$$ 11edd9e4_a535_7ca4_bca1_fdc70641c1ef_TB8827_11
The pressure at the bottom left corner is then
$$p = \gamma H = 9810 \times ( 0.917 + 0.8 ) = 16840 \mathrm {~Pa}$$
Since it was pressurized to $20 \mathrm { kPa }$, we simply add $20 \mathrm { kPa }$ and obtain
$$p _ { \text {max } } = 16.84 + 20 = \underline { 36.84 \mathrm { kPa } }$$

Question 2

If the U-tube of No. 4 is rotated about the right leg, what maximum rpm will result in the $20 ^ { \circ } \mathrm { C }$ water being thrown from the tube?&#10;A) 741 rpm&#10;B) 672 rpm&#10;C) 569 rpm&#10;D) 432 rpm

Accepted Answer

$672 \mathrm { rpm }$
The water will leave the tube when the lowest pressure in the tube, the pressure at the top of the right leg, reaches the vapor pressure which is $2.34 \mathrm { kPa }$ abs (see Table B.1 in the Appendix). The pressure there is given by Eq. 2.6.4:
11edd9e4_deb4_6b87_bca1_53aa40265cc8_TB8827_11
$$\begin{aligned}
101300 - 2340 & = \frac { 1000 } { 2 } \omega ^ { 2 } \times 0.2 ^ { 2 } . \quad \therefore \omega = 70.34 \mathrm { rad } / \mathrm { s } \text { or } 672 \mathrm { rpm }
\end{aligned}$$
We used $p _ { 1 } = 101300 \mathrm {~Pa}$ since the pressure at point 2 is in absolute pressure. The pressure must be in Pa not $\mathrm { kPa }$.

Question 3

The U-tube shown is rotated abou (A) $5.6 \mathrm { kPa }$&#10;(B) $4.2 \mathrm { kPa }$&#10;(C) $2.2 \mathrm { kPa }$&#10;(D) $0.8 \mathrm { kPa }$ t the left vertical leg at 100

Accepted Answer

$4.2 \mathrm { kPa }$
The highest pressure occurs at the right-hand corner of the bottom where $r _ { 2 } = 0.2 \mathrm {~m}$. Select $p _ { 1 } = 0$ at the top of the left leg. Using $\omega = 100 \times 2 \pi / 60 = 10.47 \mathrm { rad } / \mathrm { s }$, Eq.2.6.4 provides
11edd9e4_be42_c186_bca1_c324ae564811_TB8827_11
$p_{2}=\frac{1000}{2} 10.47^{2} \times 0.2^{2}-9810 \times(-0.2)=\underline{4154 \mathrm{~Pa}}$ 
Careful: Point 2 is at $( 0.2 , - 0.2 ) \mathrm { m }$. (The units will work out if we use $\mathrm { kg } , \mathrm { N } , \mathrm { m }$, and $\mathrm { s }$ as the units. Check them if you're uncertain. Angular velocity must be in rad/s, never rpm).

Question 4

The tank in Problem 1 has a small hole positioned in its top at the very rear of the tank which is accelerated at $a _ { x } = 5 \mathrm {~m} / \mathrm { s } ^ { 2 }$ in the direction of the longer side. The lowest pressure in the tank, which is filled with water, is nearest:

A)

12 \mathrm { kPa }

B)

9 \mathrm { kPa }

C)

- 9 \mathrm { kPa }

D)

- 18 \mathrm { kPa }

Accepted Answer

The answer of The tank in Problem 1 has a...

Question 5

The force acting on the bottom of the cylinder of Example 2.12 is nearest:&#10;A) 55.6 N&#10;B) 50.4 N&#10;C) 45.2 N&#10;D) 38.4 N rpm. The highest pressure in the U-tube is nearest:

Accepted Answer

$55.6 \mathrm {~N}$
The pressure distribution on the bottom of the cylinder, using $p = 0$ at $r = 0$, is (see Eq. 2.6.4)
$$p - 0 = \frac { \rho \omega ^ { 2 } } { 2 } \left( r ^ { 2 } - 0 \right) \quad \text { or } \quad p = \frac { 1000 } { 2 } 26.6 ^ { 2 } r ^ { 2 } = 353800 r ^ { 2 }$$
Integrate over the bottom area:
$$F _ { \text {bottom } } = \int _ { 0 } ^ { 0.1 } 353800 r ^ { 2 } \times 2 \pi r d r = 2 \pi \times \frac { 353800 \times 0.1 ^ { 4 } } { 4 } = \underline { 55.6 \mathrm {~N} }$$

Deck 5: The Differential Forms of the Fundamental Laws