Fluid Dynamics - Using the Manometer Equation

[dylanwalt]

Homework Statement

The two water reservoirs shown in the figure are open to the atmosphere. The water has a density of 1.00 Mg/m°.
A u-shaped tube between the reservoirs contains incompressible mercury with a density of 13.60 Mg/m'.
The difference in height, d, of the mercury on either side of the u-shaped tube is 25.0 cm as shown.
What is the difference in elevation h?

Relevant Equations

(delta)P=pg(delta)h

I tried to use this equation, so I isolated the delta h because that is what im solving for and then I thought because the pressure on both ends of the reservoir is both atmospheric pressure the change in pressure is 0. This makes my entire equation 0 and thus height is 0 which is definitely not the case.

 

[osilmag]

Think of the two reservoirs as the right hand side and left hand side of an equation. In this instance the equation for pressure. You would put mercury's pressure on the right hand side.

P_h + Px + P25 = Pmerc + Px

 

[erobz]

Homework Statement: The two water reservoirs shown in the figure are open to the atmosphere. The water has a density of 1.00 Mg/m°.
A u-shaped tube between the reservoirs contains incompressible mercury with a density of 13.60 Mg/m'.
The difference in height, d, of the mercury on either side of the u-shaped tube is 25.0 cm as shown.
What is the difference in elevation h?
Relevant Equations: (delta)P=pg(delta)h

I tried to use this equation, so I isolated the delta h because that is what im solving for and then I thought because the pressure on both ends of the reservoir is both atmospheric pressure the change in pressure is 0. This makes my entire equation 0 and thus height is 0 which is definitely not the case.

You appear to be using Bernoulli's between 1 and 3? While not wrong...it's not going to get you anywhere. You have to think about the pressures at the various points in the manometer. You can relate these from surface $\enclose{circle}{\text{1}}$ to surface $\enclose{circle}{\text{3}}$.

 

[Chestermiller]

Is it clear to you that, at the bottom of the blue section in either leg of the manometer, the pressures are equal (since there is only mercury of equal heights in the two legs below this level)?

 

[dylanwalt]

Yes it is clear to me, I worked it out. I used the idea that the pressure on the left is equal to the pressure on the right because the system is in equilibrium and then placed all the values in and got an answer of 3.15 meters.

 

[Lnewqban]

Yes it is clear to me, I worked it out. I used the idea that the pressure on the left is equal to the pressure on the right because the system is in equilibrium and then placed all the values in and got an answer of 3.15 meters.

Welcome, @dylanwalt !

For homework questions, we require you to show your attempt to resolve the problem.

Another way to see it:
A hydrostatic pressure able to create a 0.25-meter column of mercury, will be able to induce a [(13.60 / 1.00) x 0.25]-meter column of water (which is not equal to answer C).

 

[Steve4Physics]

A hydrostatic pressure able to create a 0.25-meter column of mercury, will be able to induce a [(13.60 / 1.00) x 0.25]-meter column of water (which is not equal to answer C).

To avoid possible confusion, it's worth noting that the question asks for the value of $h$. But $h$ is not the length of a "[(13.60 / 1.00) x 0.25]-meter column of water".

So, unless I'm misunderstanding something, the OP has the correct answer (C, 3.15m in Post #5).

Minor edit made.