Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 2, Problem 2.34P

(a)

To determine

The formula for the pressure difference papb when the oil-water interface on the right rises Δh<h, for dD.

(a)

Expert Solution
Check Mark

Explanation of Solution

The following figure shows the two containers with water and oil.

Fluid Mechanics, Chapter 2, Problem 2.34P , additional homework tip  1

Figure-(1)

The height of oil in container is H, the height of water in left container is L, the height of oil in limb is h and the XX is the datum for analysis.

Write the hydrostatic equation for the left limb.

pl=pa+γw(L+h)

Here, the pressure at a is pa, and the specific weight of water is γw.

Write the hydrostatic equation for the right limb.

pr=pb+γoil(H+h)

Here, the pressure at b is pb and the specific weight of oil is γoil.

The pressure of the right limb and the left limb is same at the datum XX.

pl=pr …… (I)

Substitute pb+γoil(H+h) for pr and pa+γw(L+h) for pl in Equation (I).

pb+γoil(H+h)=pa+γw(L+h)papb=γoil(H+h)γw(L+h) …… (II)

The figure below shows the rise in limb ΔH.

Fluid Mechanics, Chapter 2, Problem 2.34P , additional homework tip  2

Figure-(2)

Both the containers are of equal diameter hence, the change in height of limb is Δh, the depression in container a is equal to increase in other container b.

The volume rise and fall in both the containers is same.

Write the expression for balance the volume in both containers.

π4D2ΔH=π4d2×ΔhΔH=(dD)2Δh

Here, the diameter of the limb is d.

Write the hydrostatic equation for the left limb at the datum XX.

pl=pa+γw(LΔH+hΔh)

Write the hydrostatic equation for the right limb at the datum XX.

pr=pb+γoil(H+ΔH+hΔh)

The pressure at both the limbs is same at the datum.

pl=pr …… (III)

Substitute (pa+γw(LΔH+hΔh)) for pl and (pb+γoil(H+ΔH+hΔh)) for pr in Equation (III).

pa+γw(LΔH+hΔh)=pb+γoil(H+ΔH+hΔh)papb=γoil(H+ΔH+hΔh)γw(LΔH+hΔh)papb=γoil(H+h)+γoil(ΔHΔh)γw(L+h)+γw(ΔH+Δh) …. (IV)

From Equation (IV) and Equation (II).

papb=γoil(H+h)+γoil(HΔh)γoil(H+h)+γw(H+Δh)=γoil(ΔHΔh)+γw(ΔH+Δh) …… (V)

Substitute (dD)2Δh for ΔH in Equation (V).

papb=γoil[(dD)2ΔhΔh]+γw[(dD)2Δh+Δh]=Δh{γoil[(dD)21]+γw[(dD)2+1]} …… (VI)

When dD then dD0.

Substitute 0 for dD in Equation (VI).

papb=Δh{γoil[(0)21]+γw[(0)2+1]}=Δh[γwγoil] …… (VII)

Substitute 9790N/m3 for γw and 8740.71N/m3 in Equation (VII).

papb=Δh[9790N/m38740.71N/m3]=1049.2ΔhN/m2

Thus, the formula for the pressure difference papb when the oil-water interface on the right rises Δh<h, for dD is 1049.2ΔhN/m2.

(b)

To determine

The formula for the pressure difference papb when the oil-water interface on the right rises Δh<h, for d=0.15D.

The percentage change in pressure difference.

(b)

Expert Solution
Check Mark

Answer to Problem 2.34P

The formula for the pressure difference papb when the oil-water interface on the right rises Δh<h, for d=0.15D is 1466.231ΔhN/m2.

The percentage change in pressure difference is 40%.

Explanation of Solution

Write the expression for diameter of limb.

d=0.15DdD=0.15

Substitute 0.15 for dD in Equation (VI).

Δp=Δh{γoil[(0.15)21]+γw[(0.15)2+1]}=Δh{γoil[0.02251]+γw[0.0225+1]}=Δh{γw[1.0225]γoil[0.9775]} …… (VIII)

Write the percentage change in pressure.

Δp%=Δp2Δp1Δp1×100 …… (IX)

Here, the change in pressure in case 1 is Δp1 and the change in pressure in case 2 is Δp2.

Conclusion:

Substitute 9790N/m3 for γw and 8740.71N/m3 in Equation (VIII).

Δp=Δh{(9790N/m3)[1.0225](8740.71N/m3)[0.9775]}=1466.231ΔhN/m2

Thus, the formula for the pressure difference papb when the oil-water interface on the right rises Δh<h, for d=0.15D is 1466.231ΔhN/m2.

Substitute 1466.231ΔhN/m2 for Δp2 and 1049.2ΔhN/m2 for Δp1 in Equation (IX).

Δp%=1466.231ΔhN/m21049.2ΔhN/m21049.2ΔhN/m2×100=417.031049.2×10040%

Thus, the percentage change in pressure difference is 40%.

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Chapter 2 Solutions

Fluid Mechanics

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