Question 1

The pressure on the lower surface of the cylinder is $20 \mathrm { kPa }$. If the variation of pressure in the vertical direction is given by $\partial p / \partial z = - 2400 \mathrm {~Pa} / \mathrm { m }$, the pressure on the upper surface is nearest:  &#10;(A) 19.52 kPa&#10;(B) 20.48 kPa&#10;(C) &#8722;28.0 kPa&#10;(D) 31.20 kPa

Accepted Answer

$19.52 \mathrm { kPa }$
Because the variation is constant, the change in pressure can be written as
$$\Delta p = \frac { \partial p } { \partial z } \Delta z = - 2400 \mathrm {~Pa} / \mathrm { m } \times 0.2 \mathrm {~m} = - 480 \mathrm {~Pa} . \quad \therefore p _ { \text {upper } } = 20 - 0.48 = \underline { 19.52 \mathrm { kPa } }$$

Question 2

The force shown is partially due to a pressure acting uniformly over the area of 100 cm2. The pressure is nearest:&#10; &#10;A) 6.02 kPa &#10;B) 6.91 N&#183;m&#10;C) 8.84 N&#183;m&#10;D) 10.1 N&#183;m

Accepted Answer

$346 \mathrm { kPa }$
The pressure is due to the normal component of the force acting on the area. We have
$$p = \frac { F } { A } = \frac { 4000 \mathrm {~N} \cos 30 ^ { \circ } } { 100 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 } } = 346000 \mathrm {~N} / \mathrm { m } ^ { 2 } \text { or } \underline { 346 \mathrm { kPa } }$$

Question 3

A water-well driller measures the water level in a well to be 20 ft below the surface. The point on the well is 250 ft below the surface. The pressure at the point, in SI units, is estimated to be:&#10;A) 632 kPa&#10;B) 652 kPa&#10;C) 688 kPa&#10;D) 698 kPa

Accepted Answer

$688 \mathrm { kPa }$
Using Eq. 2.4.4, the pressure is
$$p = \gamma h = 9810 \mathrm {~N} / \mathrm { m } ^ { 3 } \times ( 250 - 20 ) \mathrm { ft } \times 0.3048 \mathrm {~m} / \mathrm { ft } = 688000 \mathrm {~N} / \mathrm { m } ^ { 2 } \text { or } \underline { 688 \mathrm { kPa } }$$
Or, we could use English units as follows:
$$p = \gamma h = 62.4 \times ( 250 - 20 ) = 14,350 \mathrm { lb } / \mathrm { ft } ^ { 2 } . \quad \text { Then, } \frac { 14,350 \times 101.3 } { 14.7 \times 144 } = \underline { 685 \mathrm { kPa } }$$
The difference between the two numbers is due to the accuracy of the numbers used: $14.7,9810,62.4$, and $101.3$.

Question 4

The percentage error, considering the pressure from Table B.3 in the Appendix to be more accurate, is nearest:&#10;A) 0.08 %&#10;B) 0.13 %&#10;C) 0.24 %&#10;D) 0.31 %

Accepted Answer

$0.13 \%$
Equation 2.4.8 yields
$$\begin{aligned}
p = p _ { \text {atm } } \left( \frac { T _ { 0 } - \alpha z } { T _ { 0 } } \right) ^ { g / \alpha R } & = 101.3 \left( \frac { 288.2 - 0.0065 \times 6000 } { 288.2 } \right) ^ { 9.81 / 0.0065 \times 287 } = 47.15 \mathrm { kPa } \
\% \text { error } & = \frac { 47.21 - 47.15 } { 47.21 } \times 100 = \underline { 0.13 \% }
\end{aligned}$$

Deck 3: Introduction to Fluids in Motion