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Pressure 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 $\gamma h$.

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

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

A manometer converts pressure into a measurable difference in liquid levels. A barometer measures atmospheric pressure. Hydraulic pistons use Pascal's principle to transmit pressure through a confined fluid.

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

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

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: Hydraulic Pistons

A 50 mm pipe is connected to a 500 mm cylinder, with water between pistons. If a force of 90 N is applied to the small piston, determine the pressure needed in a third 50 mm pipe to maintain equilibrium.

$$F_1 = p_1A_1$$ $$p_1 = \frac{90}{\frac{\pi}{4}(0.05)^2} = 45860 \text{ Pa}$$ $$F_2 = p_1A_2 = 45860\left[\frac{\pi}{4}(0.50)^2\right] = 9000 \text{ N}$$ $$9000 = p_3\left[\frac{\pi}{4}(0.05)^2\right]$$ $$p_3 = 4585987 \text{ N/m}^2 = 4.59 \text{ MPa}$$

Answer: The required pressure is approximately 4.59 MPa.

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.

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