# Manometry¶

## Learning outcomes¶

• Apply knowledge of hydrostatic pressure to pressure measurement
• Understand the manometer instrument
• Learn how an inclined manometer increases measurement sensitivity

# Introduction¶

Manometers are a class of instruments used to measure fluid pressure which make use of the Hydrostatic Equation:

\begin{align} p_2 - p_1 = -\rho g \left(z_2 - z_1 \right) \end{align}

For a fluid of known density in a constant gravity well, a change in elevation $(z_2 - z_1)$ linearly corresponds to a change in pressure.

Consider the section of pipe shown below. A liquid flows along the length of the pipe (we can assume it is much longer than shown and the flow moves entirely parallel to the pipe) and at some point along its length a hole is drilled and a small open tube is inserted normal to the upper surface. This arrangement is called a piezometer tube.

Shown below is a cross sectional view of the geometry. The arrows indicate the flow direction. The small vertical tube is also shown.

What do think the value of $h$ is?

Will the flow in the main pipe squirt out of this smaller tube? Will the flow continue along its path and ignore the perpendicular tube completely?

The answer is that for some constant velocity in the main pipe is that the flow will rise up the perpendicular tube some finite $h$ and stop once it's reached an equilibrium. If we change the velocity in the pipe the level $h$ will change accordingly. Why? In order to increase the velocity in the pipe we have too apply more pressure.

If we are applying a positive gauge pressure at the pipe inlet and then let the pipe exit to the atmosphere (zero gauge pressure) then there will be a pressure gradient along the length of the pipe and the liquid will flow from high to low pressure — at a constant velocity (it's a little more complicated than that but more on that later). This result is that the hydrostatic pressure is constant perpendicular to the direction of flow (assuming a sufficiently small diameter). In fact if we place a series of these 'piezometer tubes' along the length of the pipe we could visualise the pressure gradient along the pipe.

The pressure at each location $n$ can be measured by measuring the local height $h_n$:

\begin{align} p_{abs}(x) = p_{atm} + \rho g h_{n} \end{align}

or

\begin{align} p_{gauge}(x) = \rho g h_{n} \end{align}

It is the static pressure that pushes (piezo is greek for push) the fluid laterally into the piezometer tube. The velocity of flow moving along the pipe does not directly play a role in the height $h$. In fact in the pipe was instead a closed pressure vessel containing a static fluid as observed in our Bourdon gauge example we would observe the same behaviour. However the Piezometer tube may need to be extremely long for any practical usage!

We can use the height of a column of liquid as a unit of pressure. We see this used in the Mercury Barometer where atmospheric pressure is often recorded in $mmHg$ or $inHg$. Since the height of the liquid column is dependent on the density and acceleration due to gravity (the specific weight, $\gamma = \rho g$):

\begin{align} \text{Pressure Head} = \frac{\Delta P}{\gamma} \end{align}

When working with a liquid in a complex flow system such as the pipe network of a chemical plant it is often convenient to work in pressure head.

## The U-tube Manometer¶

We've seen various arrangements of tubes used to measure fluids so far; from Boyle's 'J-tube' to the Piezometer tube. Now we will consider another, the 'U-tube Manometer'

If we take a U-shaped tube, open to atmosphere at both ends, and partially fill it form one side with a liquid, what happens? The fluid flows down and around and eventually reaches an equilibrium with the free surfaces in each arm perfectly level with one another.

We can perhaps understand this equilibrium better if we consider the toy example shown in the video below. Owing to friction, the spheres that fall down the right side of the tube eventually come to rest. Each sphere has an identical finite mass and therefore gravity pulls on each with the same force which results in the distribution of the spheres balanced being between either side. In the case of the fluid we can, for now, think of the fluid as consisting of many many tiny spheres.

Gravity $g$ pulls the fluid (with density $\rho$) down evenly and atmospheric pressure which acts at each open end (which we will denote as 1 and 2) is equal. The Hydrostatic Equation is satisfied so that $p_1 = p_2$ and the height of the fluid relative to a common datum is $h_1 = h_2$.

But what if we connect one end of our U-tube to a pressure vessel as shown on the left side of the image below?

On the right is shown an illustration of the U-tube from the previous video and on the left is it after the left side is connected to the pressure vessel containing a fluid ($\gamma_1 = \rho_1 g$) at some unknown pressure $P_A$.

Is the pressure at point A greater or less than atmospheric? Which is the same as asking is the gauge pressure greater or less than zero?

We can use the hydrostatic equation to reason that the pressure $p_2 = p_3$ since any pressure acting on the interface at point 2 will be balanced at point 3 by the atmosphere and the weight of fluid in the column $h_2$. Working in gauge pressure we offset by atmospheric pressure so that the forces on the right side are equal to $\gamma_2 h_2$. On the left side the pressure $p_A$ is invariant along the dash-dot line so the forces acting on the left side are equal to $p_{A_\text{gauge}} + \gamma_1 h_1$. Since the sum of the forces is zero:

\begin{align} p_{A_\text{gauge}} + \gamma_1 h_1 = \gamma_2 h_2 \end{align}

which is easily rearranged: \begin{align} p_{A_\text{gauge}} = \gamma_2 h_2 - \gamma_1 h_1 \end{align}

This tells us that by measuring the difference in height between the two sides for two fluids of differing specific weights we can compute the pressure at point $A$ — which, remember, is at the same vertical location as point 1 and therefore at the same hydrodynamic pressure.

To answer the question, the pressure at point A is higher than atmospheric unless the density of the fluid in the tank is greater than the fluid we are using as our manometer gauge fluid — which would simply not work as the blue gauge fluid would float up through the red fluid.

### Gauge fluids¶

Dense liquids like water and mercury are popular gauge fluids. See here for a list of commercial gauge fluids with various specific gravity ranges

http://www.dwyer-inst.com/PDF_files/GageFluids_i.pdf

Remember, ($SG = \frac{\rho}{\rho_{H_2O}}$)

## An Example¶

A tank containing air and oil (SG 0.9) is connected to a mercury U-tube manometer (SG 13.8).

When $h_1 = 914~mm$, $h_2 = 152~mm$ and $h_3 = 228~mm$ determine the pressure of the air inside the tank.

We can start by listing the forces on each side. On the left we have the air pressure and the hydrostaic pressure of the height of oil from the air/oil interface down to point 1. We will consider any hydrostatic pressure in the air volume as negligible since air has a very low density. Since $p_1 = p_2$ we can ignore the mercury below this level. On the right we simply have the weight of $h_3~mm$ of mercury.

This gives:

$p_{air} + \gamma_{oil} (h_2 + h_1) = \gamma_{Hg} h_3$

Rearranging:

$p_{air} = \gamma_{Hg} h_3 - \gamma_{oil} (h_2 + h_1)$

and expanding for $\gamma = \rho g = SG~\rho_{H_2O}~g$:

$p_{air} = [SG_{Hg}~\rho_{H_2O}~g] h_3 - [SG_{oil}~\rho_{H_2O}~g] (h_2 + h_1)$

In [1]:
# Physical properties
rho_w = 1000  # kg/m^3
SG_HG = 13.6
SG_oil = 0.9
g = 9.81      # m/s/s
h_1 = 0.914   # m
h_2 = 0.152   # m
h_3 = 0.228   # m

p_air = (SG_HG * rho_w  * g * h_3) - (SG_oil * rho_w * g * (h_1 + h_2))

# print result using an f-string
print(f"Air presure is {p_air/1000:.2f} kPa")

Air presure is 21.01 kPa


## Differential Manometer¶

We can also connect both ends of our u-tube manometer to different pressure vessels and measure the differential pressure between them; the atmosphere is just a pressure vessel after all!

Again we can equate the pressures at the bottom $p_2 = p_3$

$p_{A_{gauge}} + \gamma_1 h_1 = p_{B_{gauge}} + \gamma_3 h_3 + \gamma_2 h_2$

$p_{A_{gauge}} - p_{B_{gauge}} = \gamma_3 h_3 + \gamma_2 h_2 - \gamma_1 h_1$

## Inclined-Tube Manometer¶

The inclined manometer is variant of the U-tube manometer used for increased sensitivity when measuring small pressure differences. This greater sensitivity is due to the fact that the gauge fluid is moved over a greater length of tubing on the inclined side for the same vertical displacement on the vertical side. Once again we can gain some intuition by considering a toy model with some spheres in place of the fluid. Here our manometer tube is vertical on the right and inclined at an angle of $30^\circ$ on the right.

The image below shows the final frame of the video when the system has reached equilibrium. The vertical height of the spheres is equal on each side. Two red laser beams marking the topmost spheres are used to illustrate this.

Now lets consider the case where there is a force sufficient to move the right side down by one sphere diameter. Our laser beams remain fixed and clearly demonstrate that on the left side the spheres are also displaced one diameter vertically. However the topmost sphere on the left is also displaced significantly to the left. Note also that since the top level of the spheres is now one diameter lower the balance of equal force is also one diameter lower.

The result of all of this is apparent from the measurement checker board placed next to each 'surface'. For the current angle, a 4 unit (1 diameter) vertical reduction on the right corresponds to a 12 unit (3 diameters) displacement along the inclined tube a 3:1 ratio. A regular u-tube manometer gives us a 2:1 ratio.

### Lets look at this a bit more rigorously¶

In the image below, measurements (in this case differential pressure between $p_A$ and $p_B$) are read along a scale on the inclined tube by reading the level of the gauge fluid ($\gamma_3$)

In the upper image the pressure difference between $p_A$ and $p_B$ results in the gauge fluid moving a distance $l$ along the incline relative to $h_B$. The sum of the forces is:

\begin{equation*} P_A + \gamma_{1}~h_A + \gamma_{3}~l \sin{\theta} = P_B + \gamma_{2}~(h_B) \end{equation*}

If, as shown in the lower part of the image, the pressure $p_A$ is reduced by $\Delta P$ the length of gauge fluid $l$ will increase by distance $\alpha$. Accordingly the height $h_A$ will have to decrease by $\alpha \sin{\theta}$ and the the height $h_B$ will increase by $\Delta h_B = \alpha$. The sum of the forces becomes:

\begin{equation*} [P_A - \Delta P] + \underbrace{\gamma_{1}~[h_A - \alpha \sin{\theta}]}_\text{fluid 1} + \underbrace{\gamma_{3}~[(l+\alpha) \sin{\theta} + \alpha]}_\text{gauge fluid} = P_B + \underbrace{\gamma_{2}~[h_B + \alpha]}_\text{fluid 2} \end{equation*}

It is very important to observe that the measured distance along the inclined portion of the manometer is given by the distance $\alpha + l + \alpha/\sin{\theta}$ as the increase in $h_B$ to $h_B + \alpha$ shifts the dash-dot line of equal pressure in the gauge fluid down. This is clearly apparent in the toy example above where there is a three fold increase in the displacement of the fluid on the inclined side of the manometer.

### Example¶

Determine the new differential reading along the inclined leg of the mercury manometer if the pressure in pipe A is decreased 10 kPa and the pressure in pipe B remains unchanged. The fluid in A has a specific gravity of 0.9 and the fluid in B is water.

We can write the force balance for both sides of the manometer in its initial state:

$$P_A + \gamma_{A}~(0.1) + \gamma_{Hg}~(0.05) \sin{30^\circ} = P_B + \gamma_{B}~(0.08) .$$

Rewriting the balance with a $10kPa$ pressure reduction in $P_A$:

\begin{align} [P_A - 10\times 10^{3}] + \gamma_{A}[({0.1-\alpha \sin{30^\circ}})] + \gamma_{Hg}[({\alpha \sin{30^\circ} + l\sin{30^\circ} + \alpha})] = P_B + \gamma_{B}[({0.08 + \alpha})] \end{align}

Subtracting $2$ from $1$:

\begin{equation*} -10\times 10^{3} - \gamma_A({~\alpha \sin{30^\circ}}) + \gamma_{Hg}({\alpha\sin{30^\circ} + \alpha}) = \gamma_{B}({\alpha}) \end{equation*}

and solving for $\alpha$ \begin{equation*} \alpha = \frac{-10\times10^{3}}{\gamma_B + \gamma_A ({\sin{30^\circ}}) + \gamma_{Hg}({\sin{30^\circ}+1})} = 0.054~m \end{equation*}

The differential pressure is read along the inclined portion of the manometer so we need to calculate the total change in length $L$ of mercury along it.

\begin{equation*} L \sin{\theta} = [\alpha + l\sin(\theta) + \alpha\sin(\theta)] \end{equation*}\begin{equation*} L = \left[\frac{\alpha}{\sin{\theta}} + l + \alpha\right] \end{equation*}

The resulting differential reading is then:

\begin{equation*} \frac{\alpha}{\sin{30^\circ}} + \alpha + 0.05 = \frac{0.054}{\sin{30^\circ}}+0.054+0.05=0.215m \end{equation*}

We can also solve this graphically, accounting for positive or negative changes in $\Delta P$.