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1. Homework Statement
2. Homework Equations
##P=\rho gh##
3. The Attempt at a Solution
So this is the first time I'm doing fluid mechanics, and I'm trying my best to understand it. This was the first question, and truth to be told, I'm not very confident.
I know what gauge pressure is: the total pressure minus the atmospheric pressure. I don't know what it means in this question though, and how I'm supposed to use it.
This was my attempt at finding the elevation of the petrol:
##Pressure~due~to~air = \rho_{air} gh##
##=1.225 \times 9.81 \times 2##
##=24.0345~kPa##
##Pressure~due~to~petrol = \rho_{petrol} gh##
(The height I'll be using is the height considering everything above the tube)
##=750 \times 9.81 \times 0.75##
##=5518.125~kPa##
##Atmospheric~pressure=101.325~kPa##
##Total~pressure=24.0345+5518.125+101.325=5643.4845~kPa##
##5643.4845=\rho_{petrol} gh##
##5643.4845=750\times 9.81\times h##
##h=0.767~m~(to~3~s.f.)##
This would be the height from the base of the tube (just a little higher at the dotted line).
Like I said before, I don't know how to use that gauge value, or even if my calculations without it are correct.
Any help would be appreciated.
Thank you
2. Homework Equations
##P=\rho gh##
3. The Attempt at a Solution
So this is the first time I'm doing fluid mechanics, and I'm trying my best to understand it. This was the first question, and truth to be told, I'm not very confident.
I know what gauge pressure is: the total pressure minus the atmospheric pressure. I don't know what it means in this question though, and how I'm supposed to use it.
This was my attempt at finding the elevation of the petrol:
##Pressure~due~to~air = \rho_{air} gh##
##=1.225 \times 9.81 \times 2##
##=24.0345~kPa##
##Pressure~due~to~petrol = \rho_{petrol} gh##
(The height I'll be using is the height considering everything above the tube)
##=750 \times 9.81 \times 0.75##
##=5518.125~kPa##
##Atmospheric~pressure=101.325~kPa##
##Total~pressure=24.0345+5518.125+101.325=5643.4845~kPa##
##5643.4845=\rho_{petrol} gh##
##5643.4845=750\times 9.81\times h##
##h=0.767~m~(to~3~s.f.)##
This would be the height from the base of the tube (just a little higher at the dotted line).
Like I said before, I don't know how to use that gauge value, or even if my calculations without it are correct.
Any help would be appreciated.
Thank you
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