Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 2, Problem 2.19P
To determine

The free-surface height in the right leg.

The free-surface height in the left leg.

Expert Solution & Answer
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Explanation of Solution

Write the expression for the volume of the right end leg of U-tube.

Vr=π4d2h …… (I)

Here, the volume is Vr, the diameter of the leg is d and the height of the water column is h.

The extra water added to the U-tube will displace the mercury, so the height of the column will be changed.

Write the expression for the final pressure in the left leg of U-tube.

pL=pa+γm(hl+lbL) …… (II)

Here, the pressure in the left leg is pL, the atmospheric pressure is pa, the specific weight of mercury is γm, the length of the displaced mercury column is L, the height of the mercury column in the left leg is hl and the length of the mercury column at the base is lb.

Write the expression for the final pressure in the right leg of U-tube.

pr=pa+γwh+γmL …… (III)

Here, the pressure in the right leg is pr, the specific weight of water is γw.

Since all the liquid column of the U-tube is in equilibrium, the pressure in the right and left leg are same

Equate Equation (II) and Equation (III).

pa+γm(hl+lbL)=pa+γwh+γmLγm(hl+lbL)=γwh+γmL …… (IV)

Write the expression for the final height of the right leg.

zr=L+h …… (V)

Write the expression for the final height of the left leg.

zl=hl+lbL …… (VI)

Conclusion:

Refer to the Table 2.1 “Specific Weight of Some Common Fluids” to obtain the specific weight of the water as 9790N/m3.

Refer to the Table 2.1 “Specific Weight of Some Common Fluids” to obtain the specific weight of the mercury as 133100N/m3.

Substitute 20cm3 for Vr and 1cm for h in Equation (I).

(20cm3)=π4(1cm)2hh=4π20cm31cm2=25.46cm

Substitute 133100N/m3 for γm, 9790N/m3 for γw, 10cm for hl, 10cm for lb and 25.46cm for h in Equation (II) and solve for L in Equation (IV).

(133100N/m3)[(10cm)+(10cm)L]=[(9790N/m3)(25.46cm)+(133100N/m3)L](133100N/m3)[(10cm)(1m100cm)+(10cm)(1m100cm)L]=[{(9790N/m3)×(25.46cm)(1m100cm)}+(133100N/m3)L]L=0.0906m

Substitute 0.0906m for L and 25.46cm for h in Equation (V).

zr=(0.0906m)+(25.46cm)=(0.0906m)(100cm1m)+(25.46cm)=34.52cm

Thus, the free-surface height in the right leg is 34.52cm.

Substitute 0.0906m for L, 10cm for lb and 10cm for hl in Equation (VI).

zl=(10cm)+(10cm)(0.0906m)=20cm(0.0906m)(100cm1m)=10.94cm

Thus, the free-surface height in the left leg is 10.94cm.

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Chapter 2 Solutions

Fluid Mechanics

Ch. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - The system in Fig. P2.18 is at 20°C. If...Ch. 2 - Prob. 2.19PCh. 2 - The hydraulic jack in Fig. P2.20 is filled with...Ch. 2 - At 20°C gage A reads 350 kPa absolute. What is the...Ch. 2 - The fuel gage for a gasoline tank in a car reads...Ch. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - P2.27 Conduct an experiment to illustrate...Ch. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - Prob. 2.30PCh. 2 - In Fig. P2.31 all fluids arc at 20°C. Determine...Ch. 2 - For the inverted manometer of Fig. P2.32, all...Ch. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Water flows upward in a pipe slanted at 30°, as in...Ch. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - If the pressure in container A in Fig. P2.38 is...Ch. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - P2.41 The system in Fig. P2.41 is at 20°C....Ch. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Prob. 2.45PCh. 2 - In Fig. P2.46 both ends of the manometer are open...Ch. 2 - Prob. 2.47PCh. 2 - The system in Fig. P2.4H is open to 1 atm on the...Ch. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into...Ch. 2 - Example 2.5 calculated the force on plate AB and...Ch. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Gate AB in Fig. P2.55 is 5 ft wide into the paper,...Ch. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Gate AB has length L and width b into the paper,...Ch. 2 - Prob. 2.60PCh. 2 - Gale AB in Fig. P2.61 is homogeneous mass of 180...Ch. 2 - Gale AB in Fig. P2.62 is 15 ft long and 8 ft wide...Ch. 2 - The tank in Fig. P2.63 has a 4-cm-diameter plug at...Ch. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - P2.68 Isosceles triangle gate AB in Fig. P2.68 is...Ch. 2 - P2.69 Consider the slanted plate AB of length L in...Ch. 2 - Prob. 2.70PCh. 2 - Prob. 2.71PCh. 2 - Prob. 2.72PCh. 2 - P2.73 Gate AB is 5 ft wide into the paper and...Ch. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. 2.76PCh. 2 - P2.77 The circular gate ABC in Fig. P2.77 has l-m...Ch. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. 2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. 2.82PCh. 2 - Prob. 2.83PCh. 2 - Prob. 2.84PCh. 2 - P2.85 Compute the horizontal and vertical...Ch. 2 - Prob. 2.86PCh. 2 - The bottle of champagne (SG = 0.96) in Fig. P2.87...Ch. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - The lank in Fig. P2.90 is 120 cm long into the...Ch. 2 - The hemispherical dome in Fig. P2.91 weighs 30 kN...Ch. 2 - A 4-m-diameter water lank consists of two half...Ch. 2 - Prob. 2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - The mega-magnum cylinder in Fig. P2.99 has a...Ch. 2 - Pressurized water fills the tank in Fig, P2.100....Ch. 2 - Prob. 2.101PCh. 2 - Prob. 2.102PCh. 2 - Prob. 2.103PCh. 2 - Prob. 2.104PCh. 2 - P2.105 it is said that Archimedes discovered the...Ch. 2 - Prob. 2.106PCh. 2 - Prob. 2.107PCh. 2 - P2.108 A 7-cm-diameter solid aluminum ball (SG =...Ch. 2 - Prob. 2.109PCh. 2 - Prob. 2.110PCh. 2 - P2.111 A solid wooden cone (SG = 0.729) floats in...Ch. 2 - The uniform 5-m-long round wooden rod in Fig....Ch. 2 - Prob. 2.113PCh. 2 - Prob. 2.114PCh. 2 - P2.115 The 2-in by 2-in by 12-ft spar buoy from...Ch. 2 - Prob. 2.116PCh. 2 - The solid sphere in Fig. P2.117 is iron ( SG7.9 )....Ch. 2 - Prob. 2.118PCh. 2 - Prob. 2.119PCh. 2 - Prob. 2.120PCh. 2 - Prob. 2.121PCh. 2 - Prob. 2.122PCh. 2 - Prob. 2.123PCh. 2 - Prob. 2.124PCh. 2 - Prob. 2.125PCh. 2 - Prob. 2.126PCh. 2 - Prob. 2.127PCh. 2 - Prob. 2.128PCh. 2 - Prob. 2.129PCh. 2 - Prob. 2.130PCh. 2 - Prob. 2.131PCh. 2 - Prob. 2.132PCh. 2 - Prob. 2.133PCh. 2 - Prob. 2.134PCh. 2 - P2.135 Consider a homogeneous right circular...Ch. 2 - Prob. 2.136PCh. 2 - Prob. 2.137PCh. 2 - Prob. 2.138PCh. 2 - P2.139 The tank of liquid in Kg. 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