Question

A reservoir manometer has vertical tubes of diameter D518 mm and d56 mm. The manometer liquid is Meriam red oil. Develop an algebraic expression for liquid deflection L in the small tube when gage pressure Δp is applied to the reservoir. Evaluate the liquid deflection when the applied pressure is equivalent to 25 mm of water (gage).

Answer 1

Step-by-step explanation:

Given that :

the diameter of the reservoir D = 18 mm

the diameter of the manometer d = 6 mm

For an equilibrium condition ; the pressure on both sides are said to be equal

∴

`\Delta \ P = \rho _(water) g \Delta h_(water) = \rho _(oil) g \Delta h_(oil)`

`\Delta \ P = \rho _(oil) g (x+L) ----- (1)`

According to conservation of volume:

`A*x = a*L`

`(\pi)/(4)D^2x = (\pi)/(4)d^2 L`

`x = ( (d)/(D))^2L`

Replacing x into (1) ; we have;

`\Delta \ P = \rho _(oil) g ( ( (d)/(D))^2L+L)`

`\Delta \ P = \rho _(oil) g \ L ( ( (d)/(D))^2+1)`

`L = (\Delta \ P)/(\rho _(oil) g \ ( ( (d)/(D))^2+1))`

Thus; the liquid deflection is :
`L = (\Delta \ P)/(\rho _(oil) g \ ( ( (d)/(D))^2+1))`

when the applied pressure is equivalent to 25 mm of water (gage); the liquid deflection is:

`L = (\Delta \ P)/(\rho _(oil) g \ ( ( (d)/(D))^2+1))`

`L = (\rho_(water) \g \Delta \ h)/(\rho _(water) SG_(oil)g \ ( ( (d)/(D))^2+1))`

`L = (\g \Delta \ h)/(SG_(oil)g \ ( ( (d)/(D))^2+1))`

`L = (25)/(0.827 ( ( (6)/(18))^2+1))`

L = 27.21 mm