# Pressure in a manometer - Absolute pressure

286
1. The problem statement, all variables and given/known data

Consider the manometer in the figure. If the specific weight of fluid A is 100kN/m3, what is the absolute pressure, in kPa, indicated by the manometer when the local atmospheric pressure is 90kPa?

2. Relevant equations

Specific weight = $\gamma _{s}=\rho g$

Pressure = $\rho gh$

3. The attempt at a solution

$$\rho_{a}=\left (100,000\frac{N}{m^{3}} \right )/\left (9.81\frac{m}{s^{2}} \right )=10,194\frac{kg}{m^{3}}$$

$$\rho_{b}=\left (8,000\frac{N}{m^{3}} \right )/\left (9.81\frac{m}{s^{2}} \right )=815.49\frac{kg}{m^{3}}$$

Find absolute pressure P1.

$$P_{1}+\rho_{a}g(.05m)=\rho_{b}g(.12m)+P_{atm}$$

$$P_{1}=\rho_{b}g(.12m)+P_{atm}-\rho_{a}g(.05m)$$

$$P_{1}=\left (815.49\frac{kg}{m^{3}} \right )\left (9.81\frac{m}{s^{2}} \right )+\left (90,000Pa \right )-\left (10,194\frac{kg}{m^{3}} \right )\left (9.81\frac{m}{s^{2}} \right )\left (.05m \right )$$

$$P_{1}=86kPa$$

I'm approaching this problem as a force balance equation. The answer I came up with looks reasonable, but I am unsure if I used the correct height (.12m) for fluid B. Also, I know the image shows 10kN/m3 for the SG of fluid A but the problem statement uses 100kN/m3 which is what I'm basing my calculations on.

Last edited: Jan 24, 2012
2. ### Curious3141

2,970
No need to calculate these. They've already given you the respective ρg terms in kN/m3, just multiply by the heights in m to get kPa.

Why? P1 is actually higher in pressure than the interface between fluids A and B by 5cm (0.05m) of Fluid A. Remember, with an unbroken column of a single fluid in equilibrium, the *lower* point is at the higher pressure. So why are you adding 0.05m of fluid A to P1, that makes no sense.

Also, forget about the 12cm and the 30cm measurements, they're red herrings. All you need is the 15cm (0.15m) measurement. The interface between fluids A and B is *higher* than atmospheric pressure by 15cm of fluid B.

Now you can easily calculate the gauge pressure of P1, then add it to the atmospheric pressure to get the absolute pressure. Since the question is ambiguous, present two calculations with the 2 different values for specific weight of fluid A.

286
Now I see that $\gamma_{s}=\rho g$ so the comutation of $\rho_{a}$ and $\rho_{b}$ were unnecessary.

I still believe I wrote the force-balance equation correctly. I'll re-write using SG:

$$P_{1}+\gamma_{A}(.05m)=\gamma_{B}(.15m)+P_{atm}$$

$$P_{1}=\left (8\frac{kN}{m^3} \right )(.15m)-\left (100\frac{kN}{m^3} \right )(.05m)+90kPa$$

$P_{1}=86.2kPa$ (guage)

$$P_{1abs}=P_{1guage}-P_{atm}=86.2kPa-90kPa=-3.8kPa$$

I'm unclear as to whether P1 in the graphic indicates gauge or absolute pressure. Should one or the other be assumed?

It seems like my new equation is the same as the last, except that you helped me realize to use 15cm instead of 12cm. I'm stuck!

4. ### Curious3141

2,970
Why do you feel you've written the force-balance equation correctly? It's wrong, the sign on the 5cm term should have a minus sign on it!

Imagine a horizontal plane cutting through the fluid A level on which P1 acts (on the leftmost tubing) - this plane denotes points at equal pressure within fluid A. The level at which that plane cuts the next tubing on the right lies 5cm *below* the interface between fluids A and B. That means that there's an additional pressure of 5cm of fluid A pressing down on that point compared to the A/B interface. So the pressure at that point is 5cm of A higher than that of the interface. Consequently, the pressure at the fluid A level in the leftmost tube (P1) is *also* 5cm of A higher than that at the interface. P1 is the highest pressure here, so don't add 5cm of A to P1, subtract it!

You've specified the correct sign for the other term (the 15cm of fluid B).

And you've already taken the atmospheric pressure into account in the first calculation, so you already have the absolute pressure, hence there is no need to do the last calculation. At any rate, the equation is wrong, it's P(absolute) = P(gauge) + P(atm). Positive gauge pressures indicate absolute pressures above atmospheric and negative gauge pressures indicate absolute pressures below atmospheric.

Don't get confused between gauge and absolute pressure. The manometer only indicates relative pressures - the pressure at one end of the tubing compared to the other. The pressure at one end of the tubing (the rightmost side here) is considered the reference pressure. Here, the reference pressure is the atmospheric pressure.

P1 (leftmost tubing) just represents an unknown pressure that you have to determine. You can determine whether it's higher or lower (and how much higher or lower) than the reference pressure (on the other side of the manometer) by looking at the fluid levels.

Here P1 is higher than the reference (atmospheric) pressure. The gauge pressure indicates *how much* higher P1 is compared to the reference pressure. In other words, P(gauge) = P1 - P(reference) = P1(absolute) - P(atm). Rearrange the equation, and you can see that to find the absolute P1 value, you must *add* the gauge pressure to the atmospheric pressure.

If you've calculated the gauge pressure for P1 initially (which should be positive here) by considering the fluid levels, then just add it to 90kPa (the reference atmospheric pressure) to get the absolute pressure in the last step.

If you've already taken the atmospheric pressure (90kPa) into account in the calculation, there's no need to do this, since you've already calculated the absolute pressure. The answer should be more than 90kPa.

Last edited: Jan 24, 2012

286
Thanks for your persistence and explanations. I read through your post several times and got the "aha" moment.

$$P_{1}=\gamma_{A}(.05m)+\gamma_{B}(.15m)+P_{atm}$$

$$P_{1}=\left (100,000\frac{N}{m^{3}} \right )\left (.05m \right )+\left (8,000\frac{N}{m^{3}} \right )\left (.15m \right )+90,000Pa$$

$$P_{1}=96.2kPa$$

The image in my head is:

Consider that the column of liquid A and the column of liquid B are added to the Patm on the right hand side of the manometer in order to balance out the P1 term on the left hand side of the manometer. The extra bends in the manometer actually cancel out due to the common fluid in them.

And as you said, lower point = higher pressure.