Worked examples · Sheet 2

U-tubes, barometers and manometers

Theory used: Communicating vessels and U-tube Torricelli's barometer Stevin's law Absolute pressure and gauge pressure

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A U-tube, open at both ends, contains water. Into one of the two arms oil of density $850 kg/ m^{3}$ , which does not mix with water, is poured until the oil column is $12.0 cm$ high. Compute the height of the water column, in the other arm, above the level of the interface between oil and water, and the difference in level between the two free surfaces.

The oil (

12 cm

) balances a water column of

10.2 cm

above the level of the interface

p_{A} = p_{B}, p_{A} = p_{0} + ρ_{o} g h_{o}, p_{B} = p_{0} + ρ_{a} g h_{a}

ρ_{o} h_{o} = ρ_{a} h_{a} ⟹ h_{a} = \frac{ρ _{o}}{ρ _{a}} h_{o}, Δ = h_{o} - h_{a}

h_{a} = \frac{850}{1000} \cdot 12.0 cm = 10.2 cm, Δ = 12.0 cm - 10.2 cm = 1.8 cm

We choose as reference the level of the interface between oil and water. The point $A$ , on this interface, and the point $B$ , at the same level in the other arm, both lie in the water, which fills the bottom of the tube continuously. By Stevin’s law points of the same liquid at rest at the same height have the same pressure, so $p_{A} = p_{B}$ . Above $A$ there is the oil column, above $B$ the water column, and both end with a free surface at pressure $p_{0}$ .

Equating the two pressures, $p_{0}$ and $g$ cancel and what remains is $ρ_{o} h_{o} = ρ_{a} h_{a}$ , so the heights of the columns are inversely proportional to the densities. Oil is less dense than water, therefore its column is higher and the free surface of the oil lies $1.8 cm$ higher than that of the water. The water column above the level of $A$ is $10.2 cm$ high.

The point $A$ is taken on the interface because below that level the two arms contain the same liquid; at a higher level, one arm contains oil and the other water, and the equality of the pressures does not hold.

Exercise 2

In a Torricelli barometer the mercury column ( $ρ = 13.6 \cdot 1 0^{3} kg/ m^{3}$ ) is $755 mm$ high above the surface of the mercury in the dish. Compute atmospheric pressure and the height of the column that a water barometer would have.

p_{0} = ρ_{Hg} g h, h_{a} = \frac{p _{0}}{ρ _{a} g}

p_{0} = 13.6 \cdot 1 0^{3} \cdot 9.81 \cdot 0.755 Pa = 1.01 \cdot 1 0^{5} Pa

h_{a} = \frac{1.007 \cdot 1 0 ^{5}}{1000 \cdot 9.81} m = 10.3 m

In the tube of the barometer, above the mercury, the pressure is negligible. The point of the tube at the level of the surface of the dish has the same pressure $p_{0}$ as the air pressing on the dish, because the two positions are at the same height in the same liquid. By Stevin’s law, between the top of the column, at zero pressure, and that point, $h$ lower, the pressure increases by $ρ_{Hg} g h$ , so $p_{0} = ρ_{Hg} g h$ . The height must be expressed in meters before substituting.

In a water barometer the same pressure supports a higher column, in the inverse ratio of the densities; the calculation gives a column of about ten meters. As a check, $\frac{13.6 \cdot 1 0 ^{3}}{1000} \cdot 0.755 m \approx 10.3 m$ . In the intermediate calculation one extra figure of $p_{0}$ is kept so as not to accumulate rounding errors.

Exercise 3

A gas tank is connected to one arm of a U-tube manometer containing mercury ( $ρ = 13.6 \cdot 1 0^{3} kg/ m^{3}$ ); the other arm is open to the atmosphere, at pressure $p_{0} = 1.013 \cdot 1 0^{5} Pa$ . The mercury in the open arm stands $18.0 cm$ higher than in the arm connected to the tank. Compute the gauge pressure and the absolute pressure of the gas.

p_{gas} = p_{0} + ρ g Δ h, p_{gas} - p_{0} = ρ g Δ h

p_{gas} - p_{0} = 13.6 \cdot 1 0^{3} \cdot 9.81 \cdot 0.180 Pa = 2.40 \cdot 1 0^{4} Pa

p_{gas} = 1.013 \cdot 1 0^{5} Pa + 0.240 \cdot 1 0^{5} Pa = 1.25 \cdot 1 0^{5} Pa

The pressure of the gas is considered uniform in the tank and in the connecting tube, because the density of the gas is negligible compared with that of mercury. The gas therefore presses directly on the surface of the mercury in the closed arm. At the level of this surface, in the open arm, there is a point of the same mercury, hence with the same pressure $p_{gas}$ ; above it the mercury column is $Δ h$ high and ends with the free surface at pressure $p_{0}$ .

By Stevin’s law we then have $p_{gas} = p_{0} + ρ g Δ h$ . The mercury is higher in the open arm, so the gas has a pressure greater than atmospheric pressure. The manometer directly measures the gauge pressure $ρ g Δ h$ , and adding $p_{0}$ gives the absolute pressure.

Exercise 4

A U-tube contains mercury ( $ρ_{m} = 13.6 \cdot 1 0^{3} kg/ m^{3}$ ). Into the left arm $20.0 cm$ of water is poured and into the right arm $10.0 cm$ of oil of density $800 kg/ m^{3}$ ; none of the liquids mixes with the others. Compute the difference in level between the two mercury surfaces and determine in which arm the mercury is higher.

p_{A} = p_{B}, p_{A} = p_{0} + ρ_{a} g h_{a}, p_{B} = p_{0} + ρ_{o} g h_{o} + ρ_{m} g Δ

ρ_{a} h_{a} = ρ_{o} h_{o} + ρ_{m} Δ ⟹ Δ = \frac{ρ _{a} h _{a} - ρ _{o} h _{o}}{ρ _{m}}

Δ = \frac{1000 \cdot 20.0 - 800 \cdot 10.0}{13.6 \cdot 1 0 ^{3}} cm = \frac{12 000}{13 600} cm = 0.882 cm

The water column, $ρ_{a} h_{a} = 1000 \cdot 20.0$ , weighs more than the oil column, $ρ_{o} h_{o} = 800 \cdot 10.0$ , for the same cross section. We therefore assume that the mercury is lower in the left arm, and we choose as reference the level $A$ of the interface between water and mercury. In the right arm, at the same level, there is a point $B$ of the mercury; above $B$ there are a stretch of mercury of height $Δ$ and then the oil column. The mercury fills the bottom of the tube continuously, so by Stevin’s law $p_{A} = p_{B}$ .

Equating the two expressions and canceling $p_{0}$ and $g$ we obtain $Δ$ , which turns out to be positive; hence the initial assumption is correct, and the mercury in the right arm stands $0.882 cm$ higher than in the left arm. If the calculation had given $Δ < 0$ , the mercury would have been higher in the left arm. The heights can be left in centimeters, because the units of the densities cancel in the ratio and $Δ$ comes out in the same unit as the heights.