An inverted U-tube manometer containing oil (S.G. = 0.8) is located between two

Question

An inverted U-tube manometer containing oil (S.G. = 0.8) is located between two tanks. The tank on the left contains carbon tetrachloride and is pressurized to 9 psig. The tank on the right contains water and is open to the atmosphere. Given the measurements shown below, what is the height, h, of water in the tank on the right? Note that the water level in the drawing might not be shown "to scale". The specific weight of carbon tetrachloride and the density of water are equal to 15.6 kN/m^3 and 1000 kg/m^3, respectively. (1 psi = 6.894757 kPa 1 m = 3.2808 ft 1 kPa = 1000 N/m^2)

Explanation / Answer

Solution:

Let's Start with the pressure at point 1 at the air-carbon tetrachloride interface, and moving downwards by adding (as we go down) or subtracting (as we go up) the gh terms until we reach point 2 (water-air interface open to the atmosphere), and setting the result equal to Patm since the tube is open to the atmosphere gives:

P1 + carbon tetrachloride gh1 + oil gh2 water gh3 = Patm

Here, P1 = 9 psig = 9x6.8947 = 62 kPa, h1 = 2 ft = 2x0.3 = 0.6 m, h2 = 0.7 ft = 0.7x0.3 = 0.21 m, h3 = ?

Patm - P1 = carbon tetrachloride gh1 + oil gh2 water gh3

or, 101.32 kPa - 62 kpa = g (carbon tetrachloride h1 + oil h2 water h3)

or, (39.32x100)/g. N/m2 = 15600 N/m3 x 0.6 m + 0.8 kg/m3 x 0.21 m – 1000 kg/m3 x h3

or, 401.22 N / s2m3 = 1590.73 kg/m3 x 0.6 m + 0.8 kg/m3x0.21 m – 1000 kg/m3 x h3              [g = 9.8 m/s2]

or, 401.22 N s2/m3 = 954.6 kg/m2 – 1000 kg/m3 x h3

or, 401.22 kg/m2 = 954.6 kg/m2 – 1000 kg/m3 x h3

or, 533.38 kg/m2 = 1000 kg/m3 x h3

or, 0.553 m = h3

or, h3 = 0.553 x 3.2 ft = 1.77 ft.

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