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help me with some manometers here please. 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?

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I think alcohol and water are miscible (never tried though!). So we have to use average density of liquid.
that is what I am thinking as well. but my classmates insist that there is a height difference here and we can assume the two fluids to be immiscible.
if they are miscible, their height will be the same at both sides right?

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It would be complicated if they were immiscible. Also in that case we will have to know on which side pressure was applied - where there is alcohol on top or where there is water on top.
we can say that the water is going to at the bottom since the water is more dense than alcohol right?
\[avg \ density, \ d = (\rho _{w} * n _{w}+\rho_{a}*n_{a}) / 100\] where \( \rho_{w}\) is density of water = 1000 kg/\(m^{3}\) \(n_{w}\) is % of water present = 40% \(\rho_{a}\) = density of alcohol = 0.8 \(\rho_{w}\) = 800kg/m\(^{3}\) \(n_{a}\) = % of alcohol = 60% You will be able to calculate h from the formula i showed in the drawing above after using this value of d.
ok, ill try this one. lots of thanks.
i cant see the whole version of the picture, can u type the written words here?
Strange ,i can see it ! I have written (exact words) : "See that 'h' height of liquid will have to balance 50kPa pressure So 50 kPa= hdg where 'd' i average density of 40% water and 60% alcohol"
maybe my screen is not wide enough. thanks a lot again. :)

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