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

Answer:

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:

$A*x = a*L$

$\dfrac{\pi}{4}D^2x = \dfrac{\pi}{4}d^2 L$

$x = ( \dfrac{d}{D})^2L$

Replacing x into (1) ; we have;

$\Delta \ P = \rho _{oil} g ( ( \dfrac{d}{D})^2L+L)$

$\Delta \ P = \rho _{oil} g \ L ( ( \dfrac{d}{D})^2+1)$

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

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

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

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\rho_{water} \g \Delta \ h}{\rho _{water} SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\g \Delta \ h}{SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{25}{0.827 ( ( \dfrac{6}{18})^2+1)}$

L = 27.21  mm