CE Board Exam Randomizer

Question 1

Hydraulics and Fluid Properties

Hydraulics is the branch of mechanics that studies the behavior of water and other liquids at rest or in motion. The usual branches are hydrostatics, hydrokinetics, and hydrodynamics.

$$\rho = \frac{m}{V}$$ $$\gamma = \frac{W}{V} = \rho g$$ $$v_s = \frac{1}{\rho}$$ $$SG = \frac{\gamma}{\gamma_w}$$

For board problems, the most frequent first move is converting a pressure into an equivalent head or converting a fluid head into pressure.

$$p = \gamma h$$ $$h = \frac{p}{\gamma}$$ $$\gamma = SG\,\gamma_w$$

Answer

$$p_{gage} = p_{abs} - p_{atm}$$ $$p_{gage} = (2.75 - 1)(101.325) = 177.32 \text{ kPa}$$ $$p = \gamma_w h$$ $$h = \frac{177.32}{9.81} = 18.1 \text{ m}$$

Answer: The pressure head is approximately 18.1 m of water.

Answer

$$p = \gamma_w h$$ $$101.3 = 9.81h$$ $$h = 10.32 \text{ m}$$

Answer: Atmospheric pressure is equivalent to about 10.32 m of water.

Answer

$$dp = \gamma\,dh$$ $$p = \int_0^5 (10 + 0.5h)\,dh$$ $$p = \left[10h + 0.25h^2\right]_0^5$$ $$p = 10(5) + 0.25(5)^2 = 56.25 \text{ kPa}$$

Answer: The gage pressure is 56.25 kPa.

Answer

$$h_w = -0.250(13.6) = -3.40 \text{ m of water}$$ $$p = \gamma_w h = 9.81(-3.40) = -33.35 \text{ kPa}$$

Answer: Pressure head is -3.40 m of water and gage pressure is -33.35 kPa.

Answer

$$p_{bottom} = p_{top} + \gamma_{oil}h_{oil} + \gamma_w h_w + \gamma_{heavy}h_{heavy}$$ $$p_{bottom} = 20 + 9.81(0.80)(0.60) + 9.81(1.00)(0.90) + 9.81(1.60)(0.30)$$ $$p_{bottom} = 20 + 4.709 + 8.829 + 4.709$$ $$p_{bottom} = 38.25 \text{ kPa}$$

Answer: The gage pressure at the bottom of the tank is 38.25 kPa.

Answer

$$BF_{water} = 450 - 300 = 150 \text{ N}$$ $$V = \frac{BF}{\gamma_w} = \frac{150}{9810} = 0.01529 \text{ m}^3$$ $$\rho_{obj} = \frac{W}{gV} = \frac{450}{9.81(0.01529)} = 3000 \text{ kg/m}^3 \Rightarrow SG_{obj} = 3.0$$

$$BF_{unknown} = 450 - 270 = 180 \text{ N}$$ $$\gamma_{unknown} = \frac{BF}{V} = \frac{180}{0.01529} = 11774 \text{ N/m}^3$$ $$SG_{unknown} = \frac{11774}{9810} = 1.20$$

Answer: The object has volume 0.01529 m³, density 3000 kg/m³ (SG = 3.0). The unknown liquid has SG = 1.20.

Answer

$$h = \frac{2\sigma\cos\theta}{\rho g r}$$ $$r = \frac{0.0015}{2} = 0.00075 \text{ m}$$ $$h = \frac{2(0.0728)\cos 0°}{1000(9.81)(0.00075)} = \frac{0.1456}{7.358} = 19.8 \text{ mm}$$

For the 3.0 mm tube, $r = 0.0015$ m:

$$h = \frac{2(0.0728)}{1000(9.81)(0.0015)} = 9.9 \text{ mm}$$

Answer: Water rises 19.8 mm in the 1.5 mm tube and 9.9 mm in the 3.0 mm tube. Yes — doubling the diameter exactly halves the capillary rise because $h \propto 1/r$.

Answer

$$c = \sqrt{\frac{E_v}{\rho}} = \sqrt{\frac{2.34 \times 10^9}{1025}} = \sqrt{2{,}283{,}000} = 1511 \text{ m/s}$$ $$t = \frac{L}{c} = \frac{1800}{1511} = 1.19 \text{ s}$$

Answer: Sound travels at 1511 m/s in sea water, and the pulse reaches 1.80 km in 1.19 s. This celerity value directly enters the water hammer pressure formula $\Delta p = \rho c \Delta V$.

Answer

$$W_{oil} = \gamma_w \cdot SG_{oil} \cdot V_{oil} = 9.81(0.85)(0.80) = 6.671 \text{ kN}$$ $$W_{water} = 9.81(1.00)(0.20) = 1.962 \text{ kN}$$ $$W_{total} = 8.633 \text{ kN}$$ $$V_{total} = 1.00 \text{ m}^3$$ $$SG_{avg} = \frac{W_{total}}{\gamma_w V_{total}} = \frac{8.633}{9.81} = 0.88$$

Answer: Total weight is 8.633 kN. The average specific gravity of the liquid-filled pipeline is 0.88.

Question 2

Bulk Modulus, Compressibility, and Capillarity

These properties describe how fluids respond to pressure change and surface interaction.

$$E_v = \frac{\Delta p}{\Delta V/V}$$ $$\beta = \frac{1}{E_v}$$ $$c = \sqrt{\frac{E_v}{\rho}}$$ $$h = \frac{2\sigma \cos\theta}{\rho g r}$$

Use consistent SI units. In most CE board computations, $\gamma_w = 9.81 \text{ kN/m}^3$, $1 \text{ atm} \approx 101.3 \text{ kPa}$, and mercury has $SG \approx 13.6$.

Question 3

Pressure Classifications

Pressure may be reported as absolute pressure, gage pressure, or vacuum pressure. Always identify which one is being used before substituting values.

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

A positive gage pressure means the pressure is above atmospheric pressure. A vacuum reading means the pressure is below atmospheric pressure.

Question 4

Problem: Pressure Head from Gas Tank Pressure

The pressure in a gas tank is 2.75 atmospheres absolute. Compute the pressure head in meters of water.

Question 5

Problem: Atmospheric Pressure as Water Column

If water were used to measure atmospheric pressure, determine the height of water column in meters.

Question 6

Problem: Variable Unit Weight

The unit weight of a liquid is variable and given by $\gamma = 10 + 0.5h$, where $\gamma$ is in $\text{kN/m}^3$ and $h$ is the depth from the free surface in meters. Determine the gage pressure at a depth of 5 m.

Question 7

Problem: Vacuum Reading in Mercury

A gage on the suction side of a pump shows a vacuum of 250 mm of mercury. Compute the pressure head in meters of water and the gage pressure in kPa.

Question 8

Problem: Pressure at the Bottom of a Layered Tank

A closed tank contains three stratified liquids stacked vertically. At the top is 0.60 m of oil with SG = 0.80. Below it is 0.90 m of fresh water. At the bottom is 0.30 m of a heavy liquid with SG = 1.60. The pressure at the very top (oil surface) is 20 kPa gage. Find the gage pressure in kPa at the bottom of the tank.

Question 9

Problem: Specific Gravity by Submerged Weighing

An unknown solid object weighs 450 N in air. When fully submerged in fresh water, it weighs 300 N. When submerged in an unknown liquid, it weighs 270 N. Determine: (a) the volume and density of the object, and (b) the specific gravity of the unknown liquid.

Question 10

Problem: Capillary Rise in a Glass Tube

A clean glass tube with inside diameter 1.5 mm is inserted vertically into water at 20°C. The surface tension of water is σ = 0.0728 N/m and the contact angle with glass is 0°. Determine the capillary rise in millimeters. Would a tube of diameter 3.0 mm result in twice less rise?

Question 11

Problem: Speed of Sound Using Bulk Modulus

The bulk modulus of elasticity of sea water is $E_v = 2.34 \text{ GPa}$ and its mass density is 1025 kg/m³. Compute: (a) the speed of sound in sea water, and (b) the time for a pressure pulse to travel 1.80 km along a submarine pipeline.

Question 12

Problem: Average Unit Weight and SG of a Mixed Volume

A pipeline carries a mixture of 0.80 m³ of oil (SG = 0.85) and 0.20 m³ of fresh water. The two fluids are stratified (not emulsified). Determine the total weight of liquid in the pipe and the average specific gravity based on the combined volume.