# 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,

P1=P2

P2=Patm + ro*g*h

solving for h, i get -0.7498...m,

What am i doing wrong here?

### 1 Answer

- 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.

So

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

Et voila!

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