Physics Manometer Question?
I've established P1 at the lower end of the 69.9cm mark. (0.669m) and calculated
P1= Pgas + roh*g*h
= 93,643.5 Pa
Established P2 directly horizontal at the point where the Water and Oil meet,
P2=Patm + ro*g*h
solving for h, i get -0.7498...m,
What am i doing wrong here?
- NCSLv 75 years agoFavorite Answer
On the LHS, P1 = Patm + rho*g*H.
where rho is the density of water.
From the same base mark on the RHS, you have
P1 = Pgas + RHO*g*0.699m
where RHO is the density of the oil.
Patm + rho*g*H = Pgas + roh*g*h
1.01e5Pa + 1000kg/m³ * 9.8m/s² * H = 0.88e5Pa + 823kg/m³ * 9.8m/s² * 0.699m
H = -0.751 m
I, too, get a negative number, and very close to yours.
Either the gas pressure is too low, or the head of oil is too small, or the oil's density is too low.
I believe that the gas pressure SHOULD have been 98 kPa, not 88 kPa. Then
1.01e5Pa + 1000kg/m³ * 9.8m/s² * H = 0.98e5Pa + 823kg/m³ * 9.8m/s² * 0.699m
H = 0.269 m
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