CE Board Exam Randomizer

Question 1

Pressure Variation and Measurement Basics

Pressure measurement problems are solved by moving through fluid columns. Moving downward in a fluid increases pressure by $\gamma h$, while moving upward decreases pressure by $\gamma h$.

Variation in pressure:

$$\Delta P=P_2-P_1$$ $$P_2=P_1+\Delta P$$ $$\Delta P=\sum \gamma h=\sum SG\,\gamma_w h$$ $$P_{gage}=\sum \gamma h$$

Pressure head:

$$h=\frac{P}{\gamma}$$

Pressure may be measured relative to absolute zero, relative to the atmosphere, or as a vacuum below atmospheric pressure.

$$p_{abs} = p_{atm} + p_{gage}$$ $$p_{abs} = p_{atm} - p_{vac}$$

At the same elevation in the same continuous static liquid, pressures are equal. This is the main balancing idea in manometer equations.

Answer

$$p = \gamma h$$ $$p = 10070(3200) = 32224000 \text{ N/m}^2$$ $$p = 32224 \text{ kPa}$$ $$1 \text{ bar} = 100 \text{ kPa}$$ $$p = 322.24 \text{ bars}$$

Answer: The pressure is 322.24 bars.

Answer

$$p_A + 9.81(0.15 + 0.45) - 9.81(10)(0.45) = 0$$ $$p_A = 9.81(10)(0.45) - 9.81(0.60)$$ $$p_A = 38.25 \text{ kPa}$$

Answer: The pressure at $A$ is 38.25 kPa.

Answer

$$p = 9.81(0.80)(6) = 47.09 \text{ kPa}$$ $$p = \gamma_w(SG)h$$ $$47.09 = 9.81(2.90)h$$ $$h = 1.66 \text{ m}$$

Answer: The required height is 1.66 m.

Answer

Start at A, work through each column to B, adding when going down and subtracting when going up:

$$p_A + \gamma_w(0.60) - \gamma_{Hg}(0.25) - \gamma_w(0.30) = p_B$$ $$p_A - p_B = \gamma_{Hg}(0.25) - \gamma_w(0.60) + \gamma_w(0.30)$$ $$p_A - p_B = 9.81(13.6)(0.25) - 9.81(0.30)$$ $$p_A - p_B = 33.35 - 2.94 = 30.41 \text{ kPa}$$

Answer: $p_A - p_B = 30.41 \text{ kPa}$. Point A is at higher pressure. This is the principle used in venturi and orifice meter readings.

Answer

Traverse from A downward through oil, across mercury, then up through water to the open atmosphere (gage = 0):

$$p_A + \gamma_{oil}(1.20) - \gamma_{Hg}(0.55) + \gamma_w(0.40) = 0$$ $$p_A = \gamma_{Hg}(0.55) - \gamma_{oil}(1.20) - \gamma_w(0.40)$$ $$p_A = 9.81(13.6)(0.55) - 9.81(0.90)(1.20) - 9.81(0.40)$$ $$p_A = 73.39 - 10.60 - 3.92 = 58.87 \text{ kPa}$$

Answer: The gage pressure at the pipe connection is 58.87 kPa. The heavy mercury reading amplifies small pressure differences, making it ideal for high-pressure pipe measurements.

Answer

$$p_{sea} = \gamma_{Hg} h_{sea} = 133420(0.760) = 101399 \text{ Pa}$$ $$\Delta p = \gamma_{air} \cdot z = 12(1800) = 21600 \text{ Pa}$$ $$p_{elev} = 101399 - 21600 = 79799 \text{ Pa}$$ $$h_{Hg} = \frac{79799}{133420} = 0.598 \text{ m} = 598 \text{ mm}$$

Answer: The mercury barometer reads approximately 598 mm at 1800 m. Atmospheric pressure decreases with altitude, which is why altimeters can be calibrated from barometric readings.

Answer

Starting at A and traversing to the open mercury surface (right leg exposed to atmosphere):

$$p_A + \gamma_w(0.40) - \gamma_{Hg}(0.30) = 0 \text{ (gage)}$$ $$p_{A,gage} = 9.81(13.6)(0.30) - 9.81(0.40)$$ $$p_{A,gage} = 40.02 - 3.92 = 36.10 \text{ kPa}$$ $$p_{abs} = 36.10 + 101.3 = 137.4 \text{ kPa}$$

Answer: Gage pressure at A is 36.10 kPa; absolute pressure is 137.4 kPa.

Answer

$$A_L = \frac{\pi(0.200)^2}{4} = 0.031416 \text{ m}^2$$ $$p = \frac{W}{A_L} = \frac{15{,}000}{0.031416} = 477.5 \text{ kPa}$$ $$A_S = \frac{\pi(0.030)^2}{4} = 7.069 \times 10^{-4} \text{ m}^2$$ $$F_S = p \cdot A_S = 477.5(7.069 \times 10^{-4}) = 0.3376 \text{ kN} = 337.6 \text{ N}$$ $$MA = \frac{15000}{337.6} = 44.4$$

Answer: Oil pressure is 477.5 kPa. A force of only 337.6 N on the small piston supports the 15 kN load, giving a mechanical advantage of 44.4.

Answer

The tank is open to the atmosphere. Convert the specified bottom absolute pressure to gage pressure, then sum the pressure contributions of the four liquid layers.

$$P_{bottom,gage}=231.3-101.325=129.975\ \mathrm{kPa}$$ $$P_{bottom,gage}=\gamma_w\left[(0.89)(1.5)+(1.00)(2.5)+S_{olive}(2.9)+(13.6)(0.4)\right]$$ $$129.975=9.81\left[1.335+2.5+2.9S_{olive}+5.44\right]$$ $$13.249=9.275+2.9S_{olive}$$ $$\boxed{S_{olive}=1.37}$$

Answer

Start at the water-pipe pressure and traverse the connected columns to the oil pipe. Pressure increases while moving downward and decreases while moving upward.

$$P_{oil}=P_{water}+\gamma_w(0.15)-\gamma_{Hg}(0.10)-\gamma_{0.68}(0.20)+\gamma_{0.86}(0.15)$$ $$P_{oil}-P_{water}=9.81\left[(0.15)-(13.6)(0.10)-(0.68)(0.20)+(0.86)(0.15)\right]$$ $$P_{oil}-P_{water}=9.81(-1.217)$$ $$\boxed{P_{oil}-P_{water}=-11.94\ \mathrm{kPa}}$$

Thus, the water-pipe pressure is 11.94 kPa greater than the oil-pipe pressure.

Answer

The mercury is higher in the leg connected to the tank, so the tank air pressure is below atmospheric pressure.

$$P_{air,gage}=-\gamma_{Hg}H=-(13.6)(9.81)(0.16)=-21.35\ \mathrm{kPa}$$ $$P_{gage}=P_{air,gage}+\gamma_w(4)$$ $$P_{gage}=-21.35+(9.81)(4)$$ $$\boxed{P_{gage}=17.89\ \mathrm{kPa}}$$

Answer

Move down 7 cm through water from A, rise through the mercury by the vertical component of the 9 cm inclined length, then rise 10 cm through oil to B.

$$\Delta z_{Hg}=0.09\sin40^{\circ}=0.05785\ \mathrm{m}$$ $$P_B=P_A+\gamma_w(0.07)-\gamma_{Hg}(0.05785)-\gamma_{oil}(0.10)$$ $$P_B=10+(9.81)(0.07)-(13.6)(9.81)(0.05785)-(0.87)(9.81)(0.10)$$ $$P_B=10+0.6867-7.719-0.8535$$ $$\boxed{P_B=2.115\ \mathrm{kPa}}$$

Answer

With piston weights neglected and the pressure transferred by Pascal's law, the force ratio equals the piston-area ratio. The oil SG does not affect the result because no elevation difference is specified.

$$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$ $$F_1=F_2\frac{A_1}{A_2}=20.95\left(\frac{D_1}{D_2}\right)^2$$ $$F_1=20.95\left(\frac{30}{75}\right)^2$$ $$\boxed{F_1=3.352\ \mathrm{kN}}$$

Answer

Collinear hydraulic system: 50 mm piston, 500 mm cylinder, and a third 50 mm piston joined by a rod

The water pressure produced by the small piston acts on the large piston, giving the force $F_2$ carried by the rod. That same rod force must be balanced by the pressure in the third pipe acting on the third 50 mm piston.

$$p_1=\frac{F_1}{A_1}=\frac{90}{\frac{\pi}{4}(0.05)^2}=45836\ \mathrm{Pa}$$ $$F_2=p_1A_2=45836\left[\frac{\pi}{4}(0.50)^2\right]=9000\ \mathrm{N}$$ $$p_3=\frac{F_3}{A_3}=\frac{9000}{\frac{\pi}{4}(0.05)^2}$$ $$\boxed{p_3=4\,583\,662\ \mathrm{Pa}\approx 4.58\ \mathrm{MPa}}$$

Answer: The required intensity of pressure in the third pipe is approximately 4.58 MPa. Because the third piston has the same diameter as the first, the pressure is amplified by the area ratio $(500/50)^2 = 100$.

Answer

The pressure difference between the two barometer readings equals the pressure of the air column:
$\Delta p=\gamma_l\Delta h=\gamma_{air}H$
The liquid weighs $0.735$ times mercury, so:
$H=\frac{0.735(13{,}600)(9.81)(0.850-0.600)}{12}$
$H=2042.9$ m
$\boxed{H\approx 2.04\text{ km}}$

Answer

Convert the barometric pressure, then add the oil pressure at 9 m depth:
$p_{atm}=\frac{720}{760}(101.356)=96.08$ kPa
$p_{oil}=0.85(9.81)(9)=75.04$ kPa
$p_{abs}=96.08+75.04=171.07$ kPa
$\boxed{p_{abs}=171.07\text{ kPa}}$

Answer

Pressure balance through the mercury gives the mercury-level difference:
$\Delta h_{Hg}=\frac{0.760-0.80(0.450)}{13.6}=0.0294$ m
The difference between the free surfaces is:
$0.760-0.450-0.0294=0.2806$ m
$\boxed{\Delta h\approx281\text{ mm}}$

Answer

Oil depth $= \frac{2}{3}(6) = 4$ m.
$P_{bottom,abs} = P_{air} + \gamma_{oil}h = 120 + (0.80\times9.81)(4) = 120 + 31.39 = 151.39$ kPa abs.
$P_{gage} = 151.39 - 101.3$
$\boxed{= 50.09 \text{ kPa}}$

Answer

Part 1.

$\gamma = \frac{\Delta P}{\Delta h} = \frac{120 - 80}{8 - 3} = \frac{40}{5}$
$\boxed{= 8 \text{ kN/m}^3}$

Part 2.

$\rho = \frac{\gamma}{g} = \frac{8000}{9.81}$
$\boxed{= 815.49 \text{ kg/m}^3}$

Part 3.

$s = \frac{\gamma}{\gamma_w} = \frac{8}{9.81}$
$\boxed{= 0.82}$

Answer

Manometer balance: $P_A + \gamma_w(0.45) = \gamma_{Hg}\,h$
$16 + 9.81(0.45) = 13.6(9.81)h$
$20.41 = 133.42h$
$\boxed{h = 0.153 \text{ m}}$

Answer

Because the unit weight changes with depth, integrate $dp=\gamma\,dh$: $$p=\int_0^2.8(14+0.4h)\,dh=14(2.8)+\frac{0.4}{2}(2.8)^2.$$ Thus $p=40.768$ kPa.
Computed answer: 40.77 kPa

Answer

Pressure head is pressure divided by the unit weight of water: $$h_w=\frac{p}{\gamma_w}=\frac{450}{9.81}=45.871559633\text{ m}.$$
Computed answer: 45.9 m

Answer

The pressure difference indicated by mercury balances the air column: $$H=\Delta h_{Hg}\frac{\gamma_{Hg}}{\gamma_{air}}=\frac{(720-620)}{1000}\frac{9.81(13.6)}{9/1000}=1482.4\text{ m}.$$
Computed answer: 1482 m

Answer

For uniform air unit weight, $p=\gamma h$: $$H=\frac{p_{atm}}{\gamma_{air}}=\frac{103}{12/1000}=8583.33333333\text{ m}.$$
Computed answer: 8583 m

Answer

Use $pA=W$ and $A=\pi D^2/4$: $$D=1000\sqrt{\frac{4W}{\pi p}}=1000\sqrt{\frac{4(95)}{\pi(1.35)(1000)}}=299.329815309\text{ mm}.$$
Computed answer: 299 mm

Question 2

Manometers, Barometers, and Hydraulic Pistons

Manometer — Any device that measures pressure. Unless otherwise qualified, the term most often refers to a U-shaped tube partially filled with fluid.

Piezometer — The simplest type of manometer.

Barometer — A scientific instrument used to measure atmospheric pressure.

Hydraulic pistons use Pascal's principle to transmit pressure through a confined fluid.

$$p_{atm} = \gamma_{Hg}h_{Hg}$$ $$p_1 = p_2$$ $$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

Question 3

Manometer Sign Convention

In manometer problems, begin at a known pressure and move point by point through the connected fluids. The sign depends on whether the path goes down or up through a fluid column.

$$p_{lower} = p_{upper} + \gamma h$$ $$p_{upper} = p_{lower} - \gamma h$$

For several fluids, add or subtract each column separately using its own unit weight.

$$p_2 = p_1 \pm \gamma_1h_1 \pm \gamma_2h_2 \pm \gamma_3h_3$$ $$\gamma_i = SG_i\gamma_w$$

Question 4

Common Pressure Conversions

Atmospheric-pressure equivalents:

$$P_{atm}=101.325\ \mathrm{kPa}=2116.22\ \mathrm{psf}=760\ \mathrm{mmHg}$$ $$P_{atm}=760\ \mathrm{Torr}=14.70\ \mathrm{psi}=29.92\ \mathrm{inHg}$$

Many board questions ask for the same pressure expressed as another liquid column. Equate pressures and solve for the unknown height.

$$\gamma_1h_1 = \gamma_2h_2$$ $$SG_1h_1 = SG_2h_2$$ $$h_2 = \frac{SG_1h_1}{SG_2}$$

For mercury and water, a mercury column is commonly converted using $SG_{Hg} = 13.6$.

$$h_w = 13.6h_{Hg}$$

Question 5

Problem: Deep Sea Pressure

Assuming sea water to be incompressible with $\gamma = 10070 \text{ N/m}^3$, determine the pressure in bars at 3200 m below the ocean surface.

Question 6

Problem: Simple Manometer Reading

A manometer is attached to a conduit. The specific gravity of the manometer liquid is $10$, with water column components $0.15 \text{ m}$ and $0.45 \text{ m}$, and heavy liquid height $0.45 \text{ m}$. Compute pressure at $A$ in kPa.

Question 7

Problem: Equivalent Liquid Columns

What height of a special gage liquid with $SG = 2.90$ exerts the same pressure as a 6 m column of oil with $SG = 0.80$?

Question 8

Problem: Differential Manometer Between Two Pipe Sections

Two horizontal pipes at the same elevation carry water and are connected by a mercury-filled U-tube differential manometer. Point A is 0.60 m above the left mercury meniscus and point B is 0.30 m above the right mercury meniscus. The mercury deflection between the two legs is 0.25 m (left leg lower). Mercury SG = 13.6. Determine the pressure difference $p_A - p_B$ in kPa.

Question 9

Problem: Three-Fluid Manometer — Pipe to Atmosphere

A manometer connects a pressurized pipe carrying oil (SG = 0.90) to the open atmosphere. The oil in the left leg stands 1.20 m above the mercury surface in the left leg. A mercury column 0.55 m tall separates the left and right legs. Water occupies the right leg from the mercury surface to the open top, 0.40 m above the mercury. Determine the gage pressure at the oil pipe connection point A in kPa.

Question 10

Problem: Barometric Pressure at Altitude

A mercury barometer reads 760 mm at sea level. The unit weight of air is taken as 12 N/m³ (assumed constant) and of mercury as 133,420 N/m³. Estimate the expected barometric reading in mm of mercury at an elevation of 1800 m above sea level.

Question 11

Problem: U-Tube Manometer on a Pressurized Pipe — Absolute Pressure

A pipe carries water under pressure. A simple U-tube mercury manometer is connected at point A, which is 1.50 m above a datum. The mercury in the left leg (connected to pipe) is 0.40 m below point A. The mercury column in the right (open) leg is 0.30 m higher than in the left leg. Atmospheric pressure is 101.3 kPa. Find the absolute pressure at point A.

Question 12

Problem: Hydraulic Press with Unequal Pistons

A hydraulic press has a small piston of diameter 30 mm and a large piston of diameter 200 mm. Both pistons are at the same horizontal level and connected by oil (SG = 0.90). A load of 15 kN is placed on the large piston. Find: (a) the oil pressure in kPa, (b) the force required on the small piston to maintain equilibrium, and (c) the mechanical advantage of the system.

Question 13

Specific Gravity of Olive Oil in a Layered Tank

Given the figure, if the atmospheric pressure is 101.325 kPa and the absolute pressure at the bottom of the tank is 231.3 kPa, determine the specific gravity of the olive oil.

See images:

Layered tank containing oil, water, olive oil, and mercury

Question 14

Pressure Difference Between Water and Oil Pipes

Find the difference in pressure between the water pipe and the oil pipe shown below.

See images:

Multifluid differential manometer between water and oil pipes

Question 15

Pressure-Gage Reading in a Tank with Vacuum Air Space

If the value of H = 16 cm, determine the pressure-gage reading.

See images:

Water tank connected to a mercury manometer with pressure gage four meters below the free surface

Question 16

Pressure in Pipe B Through an Inclined Manometer

If the pressure at Pipe A is 10 kPa, determine the pressure in Pipe B.

See images:

Inclined manometer joining water at Pipe A to oil at Pipe B through mercury

Question 17

Applied Force for a Hydraulic Car Jack

A hydraulic jack was used to lift a car. It is filled with oil at SG = 0.85. Neglecting the weight of the two pistons, find the force applied at Area 1 to lift a car with 20.95 kN weight. The diameter of Area 1 is 30 mm and the diameter of Area 2 is 75 mm.

See images:

Hydraulic jack lifting a car with small piston Area 1 and large piston Area 2

Question 18

Pressure in a Third Pipe of a Collinear Piston System

A 50 mm pipe is connected with the end of a cylinder having a diameter of 500 mm. There is a piston in the pipe and a piston in the cylinder, the space between being filled with water. The larger piston is connected by a rod with a 50 mm piston in a third pipe, the two pipes and cylinder having their axes horizontal and collinear. If a force of 90 N is applied to the small piston in the first pipe, what will be the necessary intensity of pressure in the third pipe to maintain equilibrium?

Question 19

Question Bank: q342

HGE - Hydraulics / Measurements of Pressure / Engr. Janclyde Espinosa (Clidez)

For lack of mercury, an improved barometer uses a liquid which was observed to weight 0.735 times that of mercury. At the base of the mountain, the barometer reads 850mm. Concurrently, another barometer of the same kind at the top of the mountain reads 600mm. Assuming the unit weight of air to be constant at 12N/m³, evaluate the height of the mountain in km.

Answer:

2.04km
3.02km
1.84km
2.46km

Question 20

Question Bank: q353

HGE - Hydraulics / Measurements of Pressure / Engr. Janclyde Espinosa (Clidez)

Liquids A, B, and C are inside the container shown.

What is the pressure in kPa at the bottom of the tank?

86.328kPa
85.734kPa
87.628kPa
88.234kPa

At what elevation will the liquid stand in the piezometer tube for liquid A?

8.0m
8.2m
8.4m
8.6m

At what elevation will the liquid stand in the piezometer tube for liquid B?

7.6m
5.6m
7.8m
8.0m

At what elevation will the liquid stand in the piezometer tube for liquid C?

5.5m
7.6m
6.7m
5.8m

Solution pending in psadquestions/q353.json.

Question 21

Question Bank: q370

HGE - Hydraulics / Measurements of Pressure / Engr. Janclyde Espinosa (Clidez)

What is the absolute pressure in KPa 9 m below the open surface in a tank of oil (sp.gr. = 0.85) if the barometric pressure is 720 mm of mercury? Use Pa=101.356kPa

Answer:

171.07kPa
172.70kPa
168.54kPa
164.58kPa

Question 22

Question Bank: q372

HGE - Hydraulics / Measurements of Pressure / Engr. Janclyde Espinosa (Clidez)

A U-tube with both ends open to the atmosphere contains contains mercury in the lower portion. In one leg, water stands 760 mm above the surface of the mercury. In the other leg, oil (sp.gr. = 0.80) stands 450 mm above the surface of the mercury. What is the difference in elevation between the surfaces of the oil and water columns?

Answer:

281mm
276mm
339mm
326mm

Question 23

Question Bank: q516

HGE - Hydraulics / Resultant of Parallel Force Systems / Engr. Deguma

For the system shown:

Determine the resultant of the force system.

380
700
320
300

Determine the z-coordinate of the resultant force.

1.474m below the x-y plane
1.474m above the x-y plane
1.105m below the x-y plane
1.105m above the x-y plane

Determine the y-coordinate of the resultant force.

1.105m left of the x-z plane
1.105m right of the x-z plane
1.474m left of the x-z plane
1.474m right of the x-z plane

Solution pending in psadquestions/q516.json.

Question 24

Question Bank: t20

HGE - Hydraulics / Hydrostatic Pressure / Civil Engineering Refresher

A closed cylindrical tank (h = 6 m, r = 2 m) is 2/3 full of oil (sp.gr. 0.80). If the air pressure above the oil is 120 kPa absolute, find the gage pressure at the bottom (P_atm = 101.3 kPa).

50.09 kPa
65.80 kPa
18.70 kPa
47.10 kPa

Question 25

Question Bank: t21

HGE - Hydraulics / Hydrostatic Pressure / Civil Engineering Refresher

A pressure gauge at elevation 8 m reads 80 kPa, and another at elevation 3 m reads 120 kPa.

Compute the specific weight of the liquid in kN/m³.

8 kN/m³
9.81 kN/m³
7.5 kN/m³
10 kN/m³

Using the gauge data from Question 21 (80 kPa at 8 m, 120 kPa at 3 m), compute the density of the liquid in kg/m³.

815.49 kg/m³
1000 kg/m³
764.53 kg/m³
800 kg/m³

Based on the data from Question 21, determine the specific gravity of the liquid.

0.82
0.75
0.80
1.00

Question 26

Question Bank: t23

HGE - Hydraulics / Manometers / Civil Engineering Refresher

A U-tube manometer contains mercury (sp.gr. 13.6). It is connected to a system with elevations at 8 m (sp.gr. 0.7), 6 m (sp.gr. 1.6), and 4 m. If the gage reads -18 kPa, find the mercury deflection h.

1.32 m
1.15 m
1.50 m
0.95 m

Solution pending in psadquestions/t23.json.

Question 27

Question Bank: t24

HGE - Hydraulics / Hydrostatic Pressure / Civil Engineering Refresher

A submerged bottle (30 cm dia, 30 cm high cylinder; 5 cm dia, 30 cm long neck) is inverted in water. If the water depth in the bottle is 30 cm, find the submerged depth of the open end.

58.76 cm
45.20 cm
62.10 cm
50.00 cm

Solution pending in psadquestions/t24.json.

Question 28

Question Bank: t26

HGE - Hydraulics / Manometers / Civil Engineering Refresher

A manometer is attached to a pipe at A. The pipe center is 0.45 m above the mercury-water interface. If pressure at A is 16 kPa, find the mercury deflection h (sp.gr. 13.6).

0.153 m
0.211 m
0.115 m
0.120 m

Question 29