Question 1

The specific weight of air at 5 km is nearest: (A)

6.14 \mathrm {~N} / \mathrm { m } ^ { 3 }

(B)

6.66 \mathrm {~N} / \mathrm { m } ^ { 3 }

(C)

7.26 \mathrm {~N} / \mathrm { m } ^ { 3 }

(D)

8.22 \mathrm {~N} / \mathrm { m } ^ { 3 }

Accepted Answer

$7.26 \mathrm {~N} / \mathrm { m } ^ { 3 }$
The density is found in Table B.3 in the Appendix. Using a straight-line interpolation
$$\rho = 0.8194 - \frac { 0.8194 - 0.6602 } { 2 } = 0.7398 \text { or } 0.740 \mathrm {~kg} / \mathrm { m } ^ { 3 }$$
The interpolation will result in less accuracy so 4 significant digits is not appropriate. The specific weight is then, using Eq.1.5.1,
$$\gamma = \rho g = 0.740 \times 9.81 = \underline { 7.26 \mathrm {~N} / \mathrm { m } ^ { 3 } }$$

Question 2

Two clean, glass plates spaced

0.2 \mathrm {~mm}

apart are placed in

20 ^ { \circ } \mathrm { C }

water. How far up the plates will the water climb if the contact angle is

0 ^ { \circ }

?
(A) 75 mm
(B) 70 mm
(C) 65 mm
(D) 60 mm

Accepted Answer

$75 \mathrm {~mm}$
A free-body of the water between the plates would appear as in Fig. $1.10$ with $\beta = 0 ^ { \circ }$. Summing forces provides
$$\begin{array} { c } 
\sigma \times 2 L = \gamma \times V = \gamma t L h \quad \text { or } 0.0736 \times 2 L = 9810 \times 0.0002 \mathrm { Lh } \
\therefore h = 0.0750 \mathrm {~m} \text { or } \underline { 75 \mathrm {~mm} }
\end{array}$$

Question 3

Thunder is heard $2.2$ seconds after Lightning is observed in Arizona when the temperature is $50 ^ { \circ } \mathrm { C }$. The best estimate of how far away the lightning struck is nearest:&#10;A) 820 m&#10;B) 790 m&#10;C) 730 m&#10;D) 650 m

Accepted Answer

$790 \mathrm {~m}$
We assume the lightning is observed instantly. The speed with which a sound wave travels is given by
$$c = \sqrt { k R T } = \sqrt { 1.4 \times 287 \mathrm {~J} / ( \mathrm { kg } \cdot \mathrm { K } ) \times ( 273 + 50 ) \mathrm { K } } = 360 \mathrm {~m} / \mathrm { s }$$
where we have used
$$\frac { \mathrm { J } } { \mathrm { kg } \cdot \mathrm { K } } = \frac { \mathrm { N } \cdot \mathrm { m } } { \left( \mathrm { N } \cdot \mathrm { s } ^ { 2 } / \mathrm { m } \right) \cdot \mathrm { K } } = \frac { \mathrm { m } ^ { 2 } } { \mathrm {~s} ^ { 2 } \cdot \mathrm { K } }$$
The distance is then estimated to be the velocity multiplied by the time:
$$d = V t = 360 \times 2.2 = \underline { 792 \mathrm {~m} }$$

Question 4

The specific gravity of an unknown liquid is known to be 1.26. The liquid is most likely:&#10;A) Glycerin&#10;B) Carbon tetrachloride&#10;C) Mercury&#10;D) Water

Accepted Answer

Glycerin
$$\gamma _ { x } = S \gamma _ { \text {water } } = 1.26 \times 9810 = 12360 \mathrm {~N} / \mathrm { m } ^ { 3 }$$
Reviewing the specific weights of the various liquids in Table B.5 of the Appendix suggests the unknown liquid is glycerin.

Question 5

A farmer is applying compressed nitrogen to a field from a tank originally at

800 \mathrm { kPa }

. The temperature of the nitrogen as it leaves the tank is approximately:
(A)

- 140 ^ { \circ } \mathrm { C }

(B)

- 120 ^ { \circ } \mathrm { C }

(C)

- 100 ^ { \circ } \mathrm { C }

(D)

- 80 ^ { \circ } \mathrm { C }

Accepted Answer

$- 120 ^ { \circ } \mathrm { C }$
Assume the nitrogen is an ideal gas which undergoes an isentropic process so that
$$T _ { 2 } = T _ { 1 } \left( \frac { p _ { 2 } } { p _ { 1 } } \right) ^ { ( k - 1 ) / k } = ( 273 + 10 ) \left( \frac { 100 } { 800 } \right) ^ { 0.4 / 1.4 } = 156 \mathrm {~K} \text { or } \underline { - 117 ^ { \circ } \mathrm { C } }$$
The assumption of a constant entropy process as the nitrogen expands is quite acceptable since there is no heat transfer and the process is relatively frictionless.

Question 6

Estimate the torque needed to rotate the inner cylinder at $1000 \mathrm { rpm }$ if SAE-30 oil at $40 ^ { \circ } \mathrm { C }$ fills the gap. Assume a linear velocity distribution in the gap. Neglect end effects.&#10; &#10;&#10;(A) $16.0 \mathrm {~N} \cdot \mathrm { m }$&#10;(B) $16.9 \mathrm {~N} \cdot \mathrm { m }$&#10;(C) $18.8 \mathrm {~N} \cdot \mathrm { m }$&#10;(D) $20.2 \mathrm {~N} \cdot \mathrm { m }$&#10;$60 - \mathrm { cm }$ long concentric cylinders

Accepted Answer

$20.2 \mathrm {~N} \cdot \mathrm { m }$
The velocity of the outer cylinder is 0 and that of the inner cylinder is
$$V = r \omega = 0.08 \times \frac { 1000 \times 2 \pi } { 60 } = 8.38 \mathrm {~m} / \mathrm { s }$$
Using Eq.1.5.6, the shear stress on the inner cylinder is
$$\tau = \mu \frac { \Delta V } { \Delta r } = 0.1 \times \frac { 8.38 - 0 } { 0.001 } = 838 \mathrm {~N} / \mathrm { m } ^ { 2 }$$
The viscosity is found in the Appendix Fig. B.1. The torque is related to the shear stress as follows:
$$\begin{aligned}
T & = \tau A r = \tau \times ( 2 \pi r \times L ) \times r = 2 \pi r ^ { 2 } \tau L \
& = \left( 2 \pi \times 0.08 ^ { 2 } \right) \mathrm { m } ^ { 2 } \times 838 \mathrm {~N} / \mathrm { m } ^ { 2 } \times 0.6 \mathrm {~m} = \underline { 20.2 \mathrm {~N} \cdot \mathrm { m } }
\end{aligned}$$

Deck 2: Fluid Statics