# Pressure on manometer with different fluids

## Homework Statement

The fluid in a manometer tube is 40% water and 60% alcohol (specific gravity = 0.8). What is the manometer fluid height difference if a 50kPa pressure difference is applied across the two ends of manometer?

## Homework Equations

P1-P2=50 kPa

P3-P2=(ρgh)alc
P3-P1=(ρgh)h2o

then,
P2=P3-(ρgh)alc
P1=P3-(ρgh)h2o

equating
(ρgh)alc-(ρgh)h2o = 50 kPa

## The Attempt at a Solution

P1-P2=50 kPa

P3-P2=(ρgh)alc
P3-P1=(ρgh)h2o

then,
P2=P3-(ρgh)alc
P1=P3-(ρgh)h2o

equating
(ρgh)alc-(ρgh)h2o = 50 kPa

Now, i'm stuck with getting h. since i don't have any idea how to get it using my attempts.

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Are the liquids separated somehow, or are they allowed to mix freely? In the latter case, you need to look up the density of the resultant mixture in a table, you cannot treat it as two separate liquids.

i think they are separated since the problem is asking for a height difference.

voko, since the specific gravity is given, why would you have to look up the densities in a table? assuming you know the density of water?

Like we can treat the alcohol as oil or any immiscible fluid

i think they are separated since the problem is asking for a height difference.
Even if the liquid was some pure substance, the problem would still be asking for a height difference. That's how manometer tubes work.

voko, since the specific gravity is given, why would you have to look up the densities in a table? assuming you know the density of water?
Ethanol mixes with water so eagerly that the volume of a mixture is less than the volume of pure ingredients, but how exactly less is complicated. So a naive computation of density will be misleading.

SteamKing
Staff Emeritus
Homework Helper
The OP doesn't specify the type of alcohol. I'm sure from experience that we all know that certain types of alcohol (ethanol, I'm talking to you) have no problem mixing with water.

But, to get back to the poster's original problem, the reason she was having difficulty in solving for the height was because she never introduced any information about the densities of water and alcohol into her calculations, and was thus doomed to keep chasing her tail while manipulating various formulas.

I think that in order to proceed with a solution, the first order of business is to find the density ρ of the mixture of water and alcohol. You are given the SG of the alcohol and you know the proportions of the mixture. Care to take a guess at the mixture SG, Miss? Here, have a drink while you do.

ok, they are supposed to mix. then will the SG of the alcohol be the SG of the mixture?

qwer

i changed my mind, the SG of the solution is 0.92? am i right?

You are given the SG of the alcohol and you know the proportions of the mixture. Care to take a guess at the mixture SG, Miss? Here, have a drink while you do.
That makes me thirsty :)

Yet, as I said, an arithmetic computation might easily be a few percent off. Besides, it is not even known whether the stated proportions are by mass or by volume.

SteamKing
Staff Emeritus
Homework Helper
i changed my mind, the SG of the solution is 0.92? am i right?
Show us the detailed calculation for this result.

0.6 (SG of alcohol) + 0.4 (SG of water) = SG of mixture
0.6 (0.8) + 0.4 (1) = SG of mixture
SG of mixture = 0.92

Last edited:
SteamKing
Staff Emeritus