Fluid Mechanics: Differential Hg Manometer

[nicmat42]

Homework Statement

A differential mercury manometer is connected to pipe A containing gasoline (SG=0.65), and to pipe B containing water. Determine the differential reading, h, corresponding to a pressure in A of 20kPa and a vacuum of 150 mm Hg in B.

Homework Equations

Distance from pipe A to fill line is 0.3 m and distance from pipe B to other fill line is 03 m.

The Attempt at a Solution

I have done a few manometer problems before, however, I am not sure what to make of the "vacuum of 150 mm Hg in B". I do not know what this is implying to the problem?
If someone could kindly explain I would be very grateful and will be able to complete the question.
Thanks

 

[Redbelly98]

It sounds like they are talking about the pressure relative to 1 atm (760 mm Hg).

So the "vacuum of 150 mm Hg" means an absolute pressure of (760-150) or 610 mm Hg.

This probably means the 20 kPa pressure is really 1 atm + 20 kPa.

 

[tomcue]

I would approach this problem by first clarifying the meaning of "vacuum of 150 mm Hg in B". In this context, vacuum refers to a region of low pressure. So, in B, there is a pressure of -150 mm Hg (negative because it is lower than atmospheric pressure). This information is important because it affects the calculation of the differential reading in the manometer.

To solve this problem, we can use the equation for pressure difference in a manometer: P1-P2 = ρgh, where P1 and P2 are the pressures in the two pipes, ρ is the density of the fluid (mercury in this case), g is the acceleration due to gravity, and h is the height difference between the two arms of the manometer.

First, we need to calculate the pressure in pipe A, which is given as 20 kPa. We can convert this to mm Hg by multiplying by 760 (since 1 mm Hg = 1 torr = 1/760 atm). So, P1 = 20 kPa * 760 mm Hg/kPa = 15200 mm Hg.

Next, we need to calculate the pressure in pipe B. Since there is a vacuum of -150 mm Hg, the pressure in B is equal to atmospheric pressure (760 mm Hg) minus the vacuum (-150 mm Hg). So, P2 = 760 mm Hg - (-150 mm Hg) = 910 mm Hg.

Now, we can plug these values into the manometer equation: 15200 mm Hg - 910 mm Hg = ρgh. We know the density of mercury (ρ) and the height difference (h) between the two arms of the manometer (0.3 m - 0.3 m = 0), so we can solve for the differential reading, h.

h = (15200 mm Hg - 910 mm Hg) / (13600 kg/m^3 * 9.8 m/s^2) = 1.11 meters.

Therefore, the differential reading in the manometer is 1.11 meters. This means that the level of mercury in the arm connected to pipe A will be 1.11 meters higher than the level in the arm connected to pipe B.