CE Board Exam Randomizer

⬅ Back to Subject Topics

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.

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.

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

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.

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

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 $SG_{Hg} = 13.6$.

$$h_w = 13.6h_{Hg}$$

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.

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

Answer: The pressure is 322.24 bars.

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.

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

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

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 $SG = 0.80$?

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

Answer: The required height is 1.66 m.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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

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.

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

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.

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

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

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.

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

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

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

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

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.

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

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

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.

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

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?

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 $(500/50)^2 = 100$.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: longquiz11

HGE - Hydraulics / Hydrostatic Pressure / Engr. Anne

A closed cylindrical tank having a height of 6 m and a radius of 2 m is 2/3 full of oil having a sp. gr. of 0.80. The pressure in the air space above the oil surface is 120 kPa absolute. Atmospheric pressure is 101.3 kPa.

longquiz11

What is the gage pressure at the bottom of the tank?

  1. 50.09 kPa
  2. 31.39 kPa
  3. 151.70 kPa
  4. 18.70 kPa
Gage pressure of the air space:
$P_{abs}=P_{atm}+P_{gage}$
$120=101.3+P_{gage}$
$P_{gage}=18.7\text{ kPa}$

Pressure at the bottom:
The oil depth is $\frac{2}{3}(6)=4$ m, so
$P_A=18.7+9.81(0.80)(4)$
$\boxed{P_A=50.09\text{ kPa}}$

Question Bank: longquiz12

HGE - Hydraulics / Hydrostatic Pressure / Engr. Anne

A pressure gauge at elevation 8 m at the side of a tank containing a liquid reads 80 kPa. Another gauge at elevation 3 m reads 120 kPa.

longquiz12

Compute the specific weight of the liquid.

  1. 8.00 kN/m3
  2. 9.81 kN/m3
  3. 6.40 kN/m3
  4. 10.00 kN/m3

Compute the density of the liquid.

  1. 815.49 kg/m3
  2. 1000.00 kg/m3
  3. 784.80 kg/m3
  4. 652.39 kg/m3

Compute the specific gravity of the liquid.

  1. 0.82
  2. 0.78
  3. 1.00
  4. 0.64
Specific weight:
The two gauges are $8-3=5$ m apart vertically, so
$120=80+5\gamma$
$\boxed{\gamma=8\text{ kN/m}^3}$

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

Specific gravity:
$s=\dfrac{\rho}{\rho_w}=\dfrac{815.49}{1000}$
$\boxed{s=0.82}$

Question Bank: longquiz13

HGE - Hydraulics / Manometers / Engr. Anne

From the figure shown, the gage reading at the top of a closed tank is −18 kPa. Below the air space the tank contains, in order, a liquid of sp. gr. 0.7 (8 m deep), water (6 m deep), and a liquid of sp. gr. 1.6 (4 m deep). A U-tube mercury manometer is connected at point B, which is 3 m below the bottom of the sp. gr. 1.6 layer. Neglect the weight of air.

longquiz13

Determine the deflection h of the mercury in the U-tube manometer.

  1. 1.32 m
  2. 1.14 m
  3. 1.51 m
  4. 0.97 m
Sum the pressure heads from the gage down to the mercury surface and back up, taking $\gamma_w=9.81$ kN/m3 and $s_{Hg}=13.6$:
$P_B=P_A+9.81(0.7)(8)+9.81(6)+9.81(1.6)(4)+(3-h)(9.81)-9.81(13.6)h$
$0=-18+54.94+58.86+62.78+29.43-9.81h-133.42h$
$143.23h=189.01$
$\boxed{h=1.32\text{ m}}$

Question Bank: longquiz16

HGE - Hydraulics / Manometers / Engr. Anne

A manometer is attached to a pipe carrying water to measure pressure as shown. The mercury (sp. gr. 13.6) leg starts 450 mm below point A on the pipe.

longquiz16

Compute the deflection h of the mercury if the pressure at A is 16 kPa.

  1. 0.153 m
  2. 0.120 m
  3. 0.186 m
  4. 0.220 m
Sum pressure heads from A to the open end of the manometer, in metres of water:
$\dfrac{P_A}{\gamma_w}+0.45-h(13.6)=0$
$\dfrac{16}{9.81}+0.45=13.6h$
$1.631+0.45=13.6h$
$\boxed{h=0.153\text{ m}}$

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:

  1. 2.04km
  2. 3.02km
  3. 1.84km
  4. 2.46km
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
$\boxed{H\approx 2.04\text{ km}}$

Question Bank: q353

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

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

q353

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

  1. 86.328kPa
  2. 85.734kPa
  3. 87.628kPa
  4. 88.234kPa

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

  1. 8.0m
  2. 8.2m
  3. 8.4m
  4. 8.6m

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

  1. 7.6m
  2. 5.6m
  3. 7.8m
  4. 8.0m

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

  1. 5.5m
  2. 7.6m
  3. 6.7m
  4. 5.8m
### Multi-fluid pressure and piezometer levels The pressure at the tank bottom is the sum of the pressure contributions of the three liquid layers: $$p=\gamma_w[0.80(2)+1.00(4)+1.60(2)].$$ Thus $$p=9.81(8.8)=\boxed{86.328\ \text{kPa}}.$$ A piezometer contains the same liquid as its connected layer, so its liquid level is the piezometric head for that liquid. From the layer interfaces and the corresponding specific gravities, $$\boxed{h_A=8.0\ \text{m}},\qquad \boxed{h_B=7.6\ \text{m}},\qquad \boxed{h_C=5.5\ \text{m}}.$$

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:

  1. 171.07kPa
  2. 172.70kPa
  3. 168.54kPa
  4. 164.58kPa
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
$\boxed{p_{abs}=171.07\text{ kPa}}$

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?

q372

Answer:

  1. 281mm
  2. 276mm
  3. 339mm
  4. 326mm
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
$\boxed{\Delta h\approx281\text{ mm}}$

Question Bank: q516

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

For the system shown:

q516

Determine the resultant of the force system.

  1. 380
  2. 700
  3. 320
  4. 300

Determine the z-coordinate of the resultant force.

  1. 1.474m below the x-y plane
  2. 1.474m above the x-y plane
  3. 1.105m below the x-y plane
  4. 1.105m above the x-y plane

Determine the y-coordinate of the resultant force.

  1. 1.105m left of the x-z plane
  2. 1.105m right of the x-z plane
  3. 1.474m left of the x-z plane
  4. 1.474m right of the x-z plane
All shown forces are parallel to the $x$-axis. Add their signed magnitudes to obtain the resultant:
$$R_x=380\text{ kN}$$
Take moments of every force about the coordinate planes. From the force locations in the figure:
$$\sum M_y=-560.12\text{ kN}\cdot\text{m},\qquad \sum M_z=419.90\text{ kN}\cdot\text{m}$$
For a resultant $R_x$ acting at $(y_R,z_R)$:
$$M_y=z_RR_x,\qquad M_z=-y_RR_x$$
Therefore:
$$z_R=\frac{-560.12}{380}=-1.474\text{ m}$$
$$y_R=-\frac{419.90}{380}=-1.105\text{ m}$$
Thus the resultant is 380 kN, acting 1.474 m below the $x$-$y$ plane and 1.105 m left of the $x$-$z$ plane.

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 (Patm = 101.3 kPa).

  1. 50.09 kPa
  2. 65.80 kPa
  3. 18.70 kPa
  4. 47.10 kPa
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.
$P_{gage} = 151.39 - 101.3$
$\boxed{= 50.09 \text{ kPa}}$

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.

  1. 8 kN/m3
  2. 9.81 kN/m3
  3. 7.5 kN/m3
  4. 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.

  1. 815.49 kg/m3
  2. 1000 kg/m3
  3. 764.53 kg/m3
  4. 800 kg/m3

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

  1. 0.82
  2. 0.75
  3. 0.80
  4. 1.00

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.

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

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. 1.32 m
  2. 1.15 m
  3. 1.50 m
  4. 0.95 m

Solution pending in psadquestions/t23.json.

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.

  1. 58.76 cm
  2. 45.20 cm
  3. 62.10 cm
  4. 50.00 cm

Solution pending in psadquestions/t24.json.

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).

  1. 0.153 m
  2. 0.211 m
  3. 0.115 m
  4. 0.120 m
Manometer balance: $P_A + \gamma_w(0.45) = \gamma_{Hg}\,h$
$16 + 9.81(0.45) = 13.6(9.81)h$
$20.41 = 133.42h$
$\boxed{h = 0.153 \text{ m}}$

Question Bank: v48

HGE - Hydraulics / Hydrostatic Pressure / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

The weight-density of mud varies with depth according to $\gamma=14+0.4h$, where $\gamma$ is in kN/m3 and $h$ is in meters. Determine the pressure at a depth of 2.8 m in kPa.

  1. 44.03 kPa
  2. 38.73 kPa
  3. 40.77 kPa
  4. 42.81 kPa
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

Question Bank: v49

HGE - Hydraulics / Hydrostatic Pressure / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A liquid column produces a pressure of 450 kPa. What is the equivalent height of water column in meters?

  1. 45.9 m
  2. 47.7 m
  3. 3.4 m
  4. 4.7 m
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

Question Bank: v50

HGE - Hydraulics / Hydrostatic Pressure / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Mercury barometers at the base and top of a mountain read 720 mm and 620 mm, respectively. Assume the unit weight of air is 9 N/m3. Approximate the mountain height.

  1. 1616 m
  2. 1868 m
  3. 1482 m
  4. 1171 m
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

Question Bank: v51

HGE - Hydraulics / Hydrostatic Pressure / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

If air had a uniform unit weight of 12 N/m3, determine the height of the atmosphere for an atmospheric pressure of 103 kPa.

  1. 8583 m
  2. 9785 m
  3. 7639 m
  4. 6523 m
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

Question Bank: v52

HGE - Hydraulics / Measurements of Pressure / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A hydraulic jack supports a load of 95 kN. If the pressure below the piston is 1.35 MPa, obtain the piston diameter in mm.

  1. 299 mm
  2. 263 mm
  3. 470 mm
  4. 332 mm
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 Bank: v93

HGE - Hydraulics / Manometers / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

A manometer is attached to a conduit. Liquid B has specific gravity 10.5. The water level difference is h = 0.11 m and the liquid-B column difference is d = 0.35 m. Compute the pressure at A in kPa.

  1. 20.50 kPa
  2. 25.86 kPa
  3. 31.54 kPa
  4. 37.22 kPa
Summing pressure from the conduit point A to the atmospheric free surface gives $$p_A+\gamma_w(h+d)-\gamma_Bd=0.$$ Therefore $$p_A=\gamma_w[SG_Bd-(h+d)]=31.53915 ext{ kPa}.$$
Computed answer: 31.54 kPa

Question Bank: v99

HGE - Hydraulics / Measurements of Pressure / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

At an elevation above sea level, a mercury barometer reads 640 mm. Mercury has specific gravity 13.3, and air weighs 12 N/m^3. Determine the corresponding water-barometer reading, atmospheric pressure, and approximate elevation above sea level under normal atmospheric conditions.

What is the corresponding water-barometer reading in millimeters?

  1. 6895 mm
  2. 7661 mm
  3. 8512 mm
  4. 9533 mm

Evaluate the atmospheric pressure in kPa at that level.

  1. 76.8 kPa
  2. 90.2 kPa
  3. 83.5 kPa
  4. 61.8 kPa

What is the approximate elevation in meters above sea level?

  1. 5845 m
  2. 6959 m
  3. 8211 m
  4. 4662 m
Convert the mercury column to an equivalent water column: $$h_w=SG_{Hg}h_{Hg}=8512 ext{ mm}.$$ Atmospheric pressure is $$p_{atm}=\gamma_wSG_{Hg}h_{Hg}=83.50272 ext{ kPa},$$ and the approximate elevation is $$z=p_{atm}/\gamma_{air}=6958.56 ext{ m}.$$


Computed answers:
1. 8512 mm
2. 83.5 kPa
3. 6959 m

Question Bank: v117

HGE - Hydraulics / Hydrostatic Pressure / HGE May 2022

Formula-mode item rendered with fixed values for lecture/PDF export.

An airplane climbs from an altitude of 8 km to 13 km. Assuming the unit weight of air remains constant at 10 N/m3, determine the magnitude of the pressure change.

  1. 39.1 kPa
  2. 50.0 kPa
  3. 56.9 kPa
  4. 63.1 kPa
For a constant unit weight, the magnitude of the pressure change is $$\Delta p=\gamma_{air}\Delta h.$$ Since 1 km = 1000 m and 1 kPa = 1000 Pa, $$\Delta p=10(13-8)=50\text{ kPa}.$$
Computed answer: 50.0 kPa

Question Bank: v140

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

Two mercury barometers are read simultaneously: 740 mm Hg at a low station and 500 mm Hg at a mountain station. Take the average unit weight of air as 0.012 kN/m3. Estimate the elevation difference.

Find the mountain height above the lower station.

  1. 2084 m
  2. 2668 m
  3. 3034 m
  4. 3365 m
Equate the pressure difference represented by the mercury columns to the pressure change through the air: γHgΔh = γairH.
Computed answers: 2668.32.

Question Bank: v152

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

What quantity is commonly obtained with a manometer connected to a pipe?

  1. Atmospheric temperature
  2. Only very low absolute pressure
  3. Fluid pressure in the pipe
  4. Mean channel velocity
A manometer converts a pressure difference into a measurable liquid-column difference.

Question Bank: v158

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

An aircraft changes altitude by 4250 m. Use an average air unit weight of 0.011 kN/m3.

Estimate the magnitude of the atmospheric pressure change.

  1. 36.5 kPa
  2. 46.8 kPa
  3. 53.2 kPa
  4. 59.0 kPa
Use Δp = γairΔz.
Computed answers: 46.75.

Question Bank: v179

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A column of liquid with specific gravity 3.2 and height 7 m produces the same pressure as a column of another liquid having specific gravity 0.75.

Determine the equivalent height of the second liquid.

  1. 23.33 m
  2. 33.96 m
  3. 29.87 m
  4. 37.66 m
Equate γ1h1 and γ2h2.
Computed answers: 29.8666666667.

Question Bank: v182

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A sealed rectangular tank 3 m long, 2 m wide, and 3 m high is half filled with water. The air pressure over the water is 8 kPa gage.

Find the bottom gage pressure.

  1. 17.74 kPa
  2. 25.83 kPa
  3. 28.64 kPa
  4. 22.72 kPa

Compute the total force on an end wall.

  1. 54.73 kN
  2. 70.07 kN
  3. 79.67 kN
  4. 88.36 kN

Locate the resultant below the water surface using the signed reference shown in the solution.

  1. 0.315 m
  2. 0.246 m
  3. 0.358 m
  4. 0.397 m
Superpose the uniform air-pressure loading and the triangular water-pressure loading; locate their combined resultant by moments.
Computed answers: 22.715, 70.0725, 0.31499518356.

Question Bank: v191

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

How is gage pressure related to absolute and atmospheric pressure?

  1. Absolute minus atmospheric
  2. Absolute plus atmospheric
  3. Atmospheric minus absolute
  4. Independent of atmospheric pressure
pgage = pabsolute - patmospheric.

Question Bank: v194

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A vertical piezometer of inside diameter 1.1 mm is connected to a liquid line. Capillary rise is estimated by hc = 4σcosθ/(γd). The indicated column is 0.26 m.

Find the corrected gage pressure.

  1. 1.782 kPa
  2. 2.282 kPa
  3. 2.594 kPa
  4. 2.877 kPa
Subtract capillary rise from the observed liquid-column height before converting head to pressure.
Computed answers: 2.28150909091.

Question Bank: v208

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Which scientist is associated with the principle that pressure applied to a confined fluid is transmitted undiminished?

  1. Archimedes
  2. Boyle
  3. Torricelli
  4. Pascal
This is Pascal's principle.

Question Bank: v213

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A liquid has depth-dependent unit weight γ = 14 + 0.6h kN/m3. Find the bottom pressure at depth 2.5 m.

Evaluate the bottom gage pressure.

  1. 28.80 kPa
  2. 41.93 kPa
  3. 36.88 kPa
  4. 46.50 kPa
Integrate p = ∫0Hγ(h)dh.
Computed answers: 36.875.

Question Bank: v219

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A mercury barometer reads 725 mm at an elevated location. Use mercury specific gravity 13.5, air unit weight 0.011 kN/m3, and sea-level pressure 101.5 kPa.

Convert the barometer reading to millimeters of water.

  1. 7644 mm water
  2. 11128 mm water
  3. 12342 mm water
  4. 9788 mm water

Find the local atmospheric pressure.

  1. 74.99 kPa
  2. 109.17 kPa
  3. 121.08 kPa
  4. 96.02 kPa

Estimate the elevation above sea level.

  1. 389 m
  2. 499 m
  3. 567 m
  4. 629 m
Convert the mercury column to equivalent water head, then use the pressure deficit divided by air unit weight.
Computed answers: 9787.5, 96.015375, 498.602272727.

Question Bank: v221

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Atmospheric pressure varies with which factors?

  1. Altitude
  2. Temperature
  3. Weather conditions
  4. All of these
All three affect local atmospheric pressure.

Question Bank: v229

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Hydrostatic pressure is the pressure at a point when which condition applies?

  1. The fluid accelerates.
  2. The fluid is incompressible.
  3. The fluid is at rest.
  4. Viscosity is high.
Hydrostatics concerns fluids at rest.

Question Bank: v240

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A liquid has unit weight γ=9.5+0.3h kN/m3. Evaluate pressure at depth 4.5 m.

Find gage pressure.

  1. 35.76 kPa
  2. 52.06 kPa
  3. 45.79 kPa
  4. 57.74 kPa
Integrate the variable unit weight over depth.
Computed answers: 45.7875.

Question Bank: v244

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

What is the line through piezometer water levels along a pipeline called?

  1. Hydraulic grade line
  2. Hydraulic head
  3. Energy head
  4. Energy gradient
The hydraulic grade line joins piezometric heads.

Question Bank: v245

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Which device measures static pressure head in a pipe with moving liquid?

  1. Manometer only
  2. Pitot tube
  3. Venturi meter
  4. Piezometer
A piezometer directly indicates pressure head for liquids.

Question Bank: v252

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A gas tank has absolute pressure 3.5 atmospheres. Take atmospheric pressure as 101.325 kPa.

Find gage pressure head in meters of water.

  1. 20.17 m
  2. 29.36 m
  3. 32.56 m
  4. 25.82 m
Subtract one atmosphere and divide by γw.
Computed answers: 25.8218654434.

Question Bank: v267

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A tank contains 2.5 m of a liquid having unit weight 7 kN/m3 over a water layer.

Find the gage pressure at the liquid interface.

  1. 17.50 kPa
  2. 13.67 kPa
  3. 19.90 kPa
  4. 22.07 kPa
At the interface, the upper liquid contributes $p=\gamma h$.
Computed answer: 17.5 kPa.

Question Bank: v272

HGE - Hydraulics / Measurements of Pressure / HGE Refresher Series

What is a space with pressure below atmospheric pressure commonly said to contain?

  1. Absolute pressure
  2. Vacuum
  3. Gage pressure
  4. Atmosphere
Pressure below atmospheric is described by its vacuum pressure.
Scroll to zoom