Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 2, Problem 2.1CP
To determine

(a)

The expression for gage pressure in the reservoir.

Expert Solution
Check Mark

Answer to Problem 2.1CP

If the movement of the surface of the reservoir is considered, the expression for p1gage is ρmgh(1+d2D2h).

Explanation of Solution

Given:

Fluid Mechanics, Chapter 2, Problem 2.1CP , additional homework tip  1

Concept Used:

Force acting on the panel is given by

F=p.A

Where, p is the pressure in N/m2 and A is the area in m2.

Pressure p is given by

p=ρgh

Where, ρ is the density in kg/m3 and h is the height in m.

Calculation:

Let, d and D be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

π4d2h=π4D2HH=d2D2h

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

p1gage=ρmgh+ρmgH

Substituting the value of H, we get

p1gage=ρmgh(1+d2D2h)

Thus, the expression for p1gage is ρmgh(1+d2D2h).

Conclusion:

The expression for p1gage is ρmgh(1+d2D2h).

To determine

(b)

The expression for the gage pressure in the reservoir if the movement of the surface of the reservoir is neglected.

Expert Solution
Check Mark

Answer to Problem 2.1CP

If the movement of the surface of the reservoir is neglected, the expression for p1gage is ρmgh.

Explanation of Solution

Given:

Fluid Mechanics, Chapter 2, Problem 2.1CP , additional homework tip  2

Concept Used:

Force acting on the panel is given by

F=p.A

Where, p is the pressure in N/m2 and A is the area in m2.

Pressure p is given by

p=ρgh

Where, ρ is the density in kg/m3 and h is the height in m.

Calculation:

Let, d and D be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

π4d2h=π4D2HH=d2D2h

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

p1gage=ρmgh+ρmgH

Substituting the value of H, we get

p1gage=ρmgh(1+d2D2h)

If the movement of the surface of the reservoir is neglected, H=0.

Thus, the expression for p1gage is ρmgh.

Conclusion:

The expression for p1gage is ρmgh.

To determine

(c)

The ratio of dD to keep the error of the result in part (b) and part (a) within 1% and 0.1%.

Expert Solution
Check Mark

Answer to Problem 2.1CP

The ratio of dD for the error to be less than or equal to 1% and 0.1% is 9.95 and 31.6 respectively.

Explanation of Solution

Given:

Fluid Mechanics, Chapter 2, Problem 2.1CP , additional homework tip  3

Concept Used:

Force acting on the panel is given by

F=p.A

Where, p is the pressure in N/m2 and A is the area in m2.

Pressure p is given by

p=ρgh

Where, ρ is the density in kg/m3 and h is the height in m.

Calculation:

Let, d and D be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

π4d2h=π4D2HH=d2D2h

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

p1gage=ρmgh+ρmgH

Substituting the value of H, we get

p1gage=ρmgh(1+d2D2h)

In fact, the above expression is

(Δp)actual=p1pa=p1gage=ρmgh(1+d2D2h)

If the movement of the surface of the reservoir is neglected, H=0, the above expression becomes

(Δp)approx=p1pa=p1gage=ρmgh

It is clear from the above tow expressions, the values of ρm, g and h does not affect the % error. The % error is determined only by the ratio of dD.

The percentage error is given by

(%)E=(Δp)actual(Δp)approx(Δp)actual

(%)E=d2D21+d2D2

dD=1EE

If the error is less than or equal to 1% :

dD=10.010.019.94989.95

If the error is less than or equal to 0.1% :

dD=10.0010.00131.606931.6

Conclusion:

The ratio of dD for the error to be less than or equal to 1% and 0.1% is 9.95 and 31.6 respectively.

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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. P2.139...Ch. 2 - P2.140 The U-tube in Fig, P2.140 is moving to the...Ch. 2 - The same tank from Prob. P2.139 is now moving with...Ch. 2 - Prob. 2.142PCh. 2 - Prob. 2.143PCh. 2 - Prob. 2.144PCh. 2 - A fish tank 14 in deep by 16 by 27 in is to be...Ch. 2 - Prob. 2.146PCh. 2 - Prob. 2.147PCh. 2 - Prob. 2.148PCh. 2 - Prob. 2.149PCh. 2 - Prob. 2.150PCh. 2 - Prob. 2.151PCh. 2 - P2.152 A 16-cm-diamctcr open cylinder 27 cm high...Ch. 2 - Prob. 2.153PCh. 2 - Prob. 2.154PCh. 2 - Prob. 2.155PCh. 2 - Prob. 2.156PCh. 2 - Prob. 2.157PCh. 2 - Prob. 2.158PCh. 2 - Prob. 2.159PCh. 2 - Prob. 2.160PCh. 2 - Prob. 2.161PCh. 2 - Prob. 2.1WPCh. 2 - Prob. 2.2WPCh. 2 - W2.3 Consider a submerged curved surface that...Ch. 2 - Prob. 2.4WPCh. 2 - Prob. 2.5WPCh. 2 - W2.6 Consider a balloon of mass m floating...Ch. 2 - Prob. 2.7WPCh. 2 - W2.8 Repeat your analysis of Prob. W2.7 to let the...Ch. 2 - Prob. 2.9WPCh. 2 - Prob. 2.1FEEPCh. 2 - FE2.2 On a sea-level standard day, a pressure...Ch. 2 - Prob. 2.3FEEPCh. 2 - In Fig, FE2,3, if the oil in region B has SG = 0,8...Ch. 2 - Prob. 2.5FEEPCh. 2 - Prob. 2.6FEEPCh. 2 - Prob. 2.7FEEPCh. 2 - Prob. 2.8FEEPCh. 2 - Prob. 2.9FEEPCh. 2 - Prob. 2.10FEEPCh. 2 - Prob. 2.1CPCh. 2 - Prob. 2.2CPCh. 2 - Prob. 2.3CPCh. 2 - Prob. 2.4CPCh. 2 - Prob. 2.5CPCh. 2 - Prob. 2.6CPCh. 2 - Prob. 2.7CPCh. 2 - Prob. 2.8CPCh. 2 - Prob. 2.9CPCh. 2 - Prob. 2.1DPCh. 2 - Prob. 2.2DPCh. 2 - The Leary Engineering Company (see Popular...
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