The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations differ by h = 20.0 mm. determine the elevation difference when a pressure of 98.0 pa (gage) is applied to the right tube. (hint: when the gage pressure is applied to the right tube, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward)

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7 May, 00:58

The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations differ by h = 20.0 mm. determine the elevation difference when a pressure of 98.0 pa (gage) is applied to the right tube. (hint: when the gage pressure is applied to the right tube, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward)

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Castillo · Answer 1

The solution for this problem is: In the figure, you now know that total length of the kerosene column

So at x - xPatm + Pkg (H0 th) = Pa + Pwgh

Now H0 + h = 20 + 91.1 mm = 111.1 mm

Therefore = Pkg 0.1111 - P2g = h = 56 x 0.111 - 98 / 1000 x 9.81 = 0.081 m or 81 mn

Therefore H0 = 111.1 - 81 = 30.1 mm