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Chapter 3: Problem 28

A reservoir manometer has vertical tubes of diameter \(D=18 \mathrm{mm}\) and \(d=6 \mathrm{mm} .\) The manometer liquid is Meriam red oil. Develop an algebraic expression for liquid deflection \(L\) in the small tube when gage pressure \(\Delta p\) is applied to the reservoir. Evaluate the liquid deflection when the applied pressure is equivalent to \(25 \mathrm{mm}\) of water (gage).

Short Answer

Expert verified
The algebraic expression for liquid deflection L in the small tube is \[ L = \frac{A_D}{A_d} \frac{\Delta p}{\rho g} \] When the applied pressure is equivalent to 25mm of water, the liquid deflection can be calculated by substituting in the specific values.

Step by step solution

01

Derive the Expression for Liquid Deflection

In a manometer, the difference in liquid levels reflects the applied pressure difference. The difference in pressure is given by: \(\Delta p = \rho g L\) Here, \(\rho\) is the density of the fluid, g is the gravitational acceleration, and L is the height difference. But, since volume is conserved, we have: \(A_D h = A_d L\) Here, A_D and A_d are the cross-sectional areas of the big and small tube respectively, h is the height in the large tube, and L is the height in the small tube. We substitute h = \(\rho\) g L / \(\rho\) g from the first equation into the second equation to obtain the deflection L.
02

Calculation of Area A_D and A_d

The cross-sectional area of a tube can be calculated by the formula: A = \(\pi r^2\) Here, r is the radius of the tube. Since, radius is half the diameter, we can replace r with D/2 or d/2 for the respective tubes. Hence, A_D = \(\pi (18/2 \mathrm{mm})^2\) and A_d = \(\pi (6/2 \mathrm{mm})^2\)
03

Calculate Deflection (L)

Substitute the known values in the derived equation to calculate the deflection (L). The density of Meriam red oil \(\rho\) can be found from standard tables and the value of g is approximately 9.8 m/s^2. Put these values in the equation to calculate L.
04

Calculate Liquid Deflection for Specific Pressure

Now, let鈥檚 calculate the liquid deflection when the applied pressure is equivalent to 25 mm of water. The pressure difference \(\Delta p\) is given by the height of the water column times the density of water times the gravitational acceleration. Substitute this into the expression for L derived previously to get the value of L for this specific pressure.

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