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below figure shows an inclined U-tube mercury manometer. The vertical end of the tube is exposed to a gas of gauge pressure 50 kPa and the inclined end is exposed to the atmosphere. The inclined part of the tube is at an angle of 30o with the horizontal. Find the value of h (in cm) (take g = 9.8 m/s2, ρmercury = 13600 kg/m3)

Correct Answer: 75
Pressure along the dotted line will be 50 kPa. Gauge pressure in an inclined manometer is given by P = ρ.g.h.sin (Ɵ). Substituting P, ρ and Ɵ, we get the value of h as 0.75 m.

In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).

A student wants to find the absolute pressure of water at a point below the surface of water. He has a barometer and a manometer pressure gauge. The barometer reads 1.3152 bar where as the manometer pressure gauge reads 0.3152 bar. What is the absolute pressure? (Assume that pressure at one end of the manometer is atmospheric.)

In a U-tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).

A mercury (specific gravity = 13.6) manometer connected across an orificemeter fitted in a pipe shows a manometer reading of 2 cms. If the manometer liquid is changed to carbon tetrachloride (specific gravity = 1.6), then for the same flow rate of water the new manometer reading will be __________ cms.

In the manometer given above, 2 immiscible fluids mercury (ρ = 13600 kg/m3) and water (ρ = 1000 kg/m3) are used as manometric fluids. The water end is exposed to atmosphere (100 kP

The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the lift limb is connected to a pipe in which a fluid of specific gravity 0.85 is flowing. The centre of the pipe is 14 cm below the level of mercury in the right limb.Evaluate the pressure of fluid flowing in the pipe if the difference of mercury level in the two limbs is 22 cm.

A U-tube manometer is open at one end and has a gas at the other end if the level of the mercury in both legs of the manometer is equal then what is the temperature of the gas?

A student wishes to find the velocity of air flowing through a pipe. He has a pressure gauge which displays only the dynamic pressure. The pressure gauge reads 0.018 mm Hg. Assume density of air to be 1.225 kg/m3, find the velocity V of air (ρHg = 13600 kg/m3).

A Differential manometer is connected at the points A and Bat the centre of two pipes. The pipe A(left limb) contains a liquid of specific gravity = 1.5 while pipe B (right limb)contains a liquid of specific gravity 0.85. The pressure at A and B are .5 kgf/cm2 and 1.2 kgf/cm2 respectively. Find the difference in level of mercuru in the differential manometer. A is 2.5m above B and 5 m above the mercury in its own limb. B is 2.5 m above the mercury level in limb A.

Pressure difference between two points in vessels, pipelines or in two different pipelines can be measured by a differential manometer. The pressure difference measured as the mm of water coloumn in case of mercury-water, differential manometer is equal to (where, H = difference in height of mercury column in mm).