A u-tube manometer is used to measure the pressure of a vessel under vacuum. Ass

Question

A u-tube manometer is used to measure the pressure of a vessel under vacuum. Assume the manometer is open to atmospheric pressure (101.325 kPa) and the density of                 the fluid in the manometer is 12.0 g/cm3. When the vessel is under 14.5 psi of vacuum, the height of the fluid column is 50 cm. If the vessel partially loses                 vacuum and is now under 7.5 psi of vacuum, what will be the new height of fluid column (answer in cm)?

All previous answers to similar u-tube manometer problems on Chegg are either incorrect or do not show an in-depth step-by-step solution. I want to learn how to solve this type of problem, so please                 use step-by-step instructions without skipping any steps.

Explanation / Answer

this is a fairly easy problem

firstly u need to understand that in a u tube manometer the height differencce in the two colums of the manometer gives the pressure difference of outside and inside the container

and this pressure difference is given by density of liwuid in the manometer * accleration due to gravity * height of the liquid column

here tlet thet height of the liquid column = h m

p outside = 101.325 kpa = 101325 pa

p inside = 7.5 psi = 51711pa

density of liquid = 12 g/cm^3 = 12000 kg/m^3

so

pout - pin = 12000*9.81*(h)

49614=58860 *h

h=0.4215 m

h=42.15 cm

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