An engineer is installing an open-end mercury manometer attached to a pipe that

Question

An engineer is installing an open-end mercury manometer attached to a pipe that will eventually carry pressurized gas. While the pipe is open to the atmosphere, she adds enough mercury into the manometer that the mercury level in both arms of the manometer is 740.0 mm above the bottom of the manometer. (See Fig. 1) If the end of the open arm of the manometer is 1070.0 mm above the bottom of the manometer, what is the maximum pressure (gauge) of the pressurized gas that the pipe can carry before the mercury spills out the open end the manometer? (See Fig 2)

Explanation / Answer

Pressure due to a column of liquid can be written as follws

P = hg
Here
p= pressure
h= hight of the liquid coloumn
= density of the liquid

Let's apply this to the manometer under atmospheric pressure(Patm)

Patm = 740nm * g-----1

If the maximum pressure of the gas is P max

Pmax = 1070 * g------2

From 1, we can get g

g = Patm/740nm------3

By substituting 3 to eqation 2

Pmax = 1070nm *(Patm/740nm)---4

Normally the value of atmospheric pressure is 14.7 psia

But when we consider unit psig,
that means Pressure value - atmospheric pressure in psia units

By substituting Patm as 14.7 psia we can get Pmax from eqation 4

Pmax = 1070 *(14.7/740nm) = 21.2554 psia

In order to change this units to psig, we have to substract atmospheric pressure value 14.7 psia

So Pmax= 21.2554 - 14.7 = 6.5554 psig

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