SOLUTIONS MANUAL
FOR FLUID MECHANICS 8TH EDITION WHITE
(AUTHOR) GRADED A+
LATEST UPDATE.
, CHAPTER 1 • PRESSURE DISTRIBUTION IN A
FLUID
P2.1 For The Two-Dimensional Stress
Field In Fig. P2.1, Let
Find The Shear And Normal Stresses On
Plane Aa Cutting Through At 30°.
Solution: Make Cut —Aa‖ So That It Just
Hits The Bottom Right Corner Of The
Fig. P2.1
Element. This Gives The Freebody Shown
At Right. Now Sum Forces Normal And
Tangential To Side Aa. Denote Side Length
Aa As —L.‖
P2.2 For The Stress Field Of Fig. P2.1, Change The Known Data To Σxx = 2000 Psf, Σyy
= 3000 Psf, And Σn(Aa) = 2500 Psf. Compute Σxy And The Shear Stress On Plane Aa.
Solution: Sum Forces Normal To And Tangential To Aa In The Element Freebody
Above, With Σn(Aa) Known And Σxy Unknown:
,2-2 2-2
In Like Manner, Solve For The Shear Stress On Plane Aa, Using Our Result For Σxy:
This Problem And Prob. P2.1 Can Also Be Solved Using Mohr‘S Circle.
P2.3 A Vertical Clean Glass Piezometer Tube Has An Inside Diameter Of 1 Mm. When
A Pressure Is Applied, Water At 20°C Rises Into The Tube To A Height Of 25 Cm. After
Correcting For Surface Tension, Estimate The Applied Pressure In Pa.
Solution: For Water, Let Y = 0.073 N/M, Contact Angle Θ = 0°, And Γ = 9790 N/M3.
The Capillary Rise In The Tube, From Example 1.9 Of The Text, Is
Then The Rise Due To Applied Pressure Is Less By That Amount: Hpress = 0.25 M − 0.03 M =
0.22 M. The Applied Pressure Is Estimated To Be P = Γhpress = (9790 N/M3)(0.22 M) ≈
2160 Pa Ans.
θ? Bourdon
P2.4 Pressure Gages, Such As The Bourdon Gage W gage
In Fig. P2.4, Are Calibrated With A Deadweight
Piston. If The Bourdon Gage Is Designed To Rotate
The Pointer
2 cm Oil
10 Degrees For Every 2 Psig Of Internal Pressure, How diameter
Many Degrees Does The Pointer Rotate If The Piston
And Weight Together Total 44 Newtons?
Fig. P2.4
Solution: The Deadweight, Divided By The Piston Area, Should Equal The Pressure
Applied To The Bourdon Gage. Stay In Si Units For The Moment:
2-3
At 10 Degrees For Every 2 Psig, The Pointer Should Move Approximately 100 Degrees. Ans.
, P2.5 Quito, Ecuador Has An Average Altitude Of 9,350 Ft. On A Standard Day,
Pressure Gage A In A Laboratory Experiment Reads 63 Kpa And Gage B Reads 105 Kpa.
Express These Readings In Gage Pressure Or Vacuum Pressure, Whichever Is
Appropriate.
Solution: Convert 9,350 Ft X 0.3048 = 2,850 M. We Can Interpolate In The Standard
Altitude Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
Good Interpolating! Then Pa = 71500-63000 = 8500 Pa (Vacuum Pressure) Ans.(A),
And Pb = 105000 - 71500 = 33500 Pa (Gage Pressure) Ans.(B)
P2.6 Express Standard Atmospheric Pressure As A Head, H = P/Ρg, In (A) Feet Of Glycerin;
(B) Inches Of Mercury; (C) Meters Of Water; And (D) Mm Of Ethanol.
Solution: Take The Specific Weights, Γ = Ρg, From Table A.3, Divide Patm By Γ :
(a) Glycerin: H = (2116 Lbf/Ft2)/(78.7 Lbf/Ft3) ≈ 26.9 Ft Ans. (A)
(b) Mercury: H = (2116 Lbf/Ft2)/(846 Lbf/Ft3) = 2.50 Ft ≈ 30.0 Inches Ans. (B)
(c) Water: H = (101350 N/M2)/(9790 N/M3) ≈ 10.35 M Ans. (C)
(d) Ethanol: H = (101350 N/M2)/(7740 N/M3) = 13.1 M ≈ 13100 Mm Ans. (D)
P2.7 La Paz, Bolivia Is At An Altitude Of Approximately 12,000 Ft. Assume A
Standard Atmosphere. How High Would The Liquid Rise In A Methanol Barometer,
Assumed At 20°C? [Hint: Don‘T Forget The Vapor Pressure.]
Solution: Convert 12,000 Ft To 3658 Meters, And Table A.6, Or Eq. (2.20), Give
FOR FLUID MECHANICS 8TH EDITION WHITE
(AUTHOR) GRADED A+
LATEST UPDATE.
, CHAPTER 1 • PRESSURE DISTRIBUTION IN A
FLUID
P2.1 For The Two-Dimensional Stress
Field In Fig. P2.1, Let
Find The Shear And Normal Stresses On
Plane Aa Cutting Through At 30°.
Solution: Make Cut —Aa‖ So That It Just
Hits The Bottom Right Corner Of The
Fig. P2.1
Element. This Gives The Freebody Shown
At Right. Now Sum Forces Normal And
Tangential To Side Aa. Denote Side Length
Aa As —L.‖
P2.2 For The Stress Field Of Fig. P2.1, Change The Known Data To Σxx = 2000 Psf, Σyy
= 3000 Psf, And Σn(Aa) = 2500 Psf. Compute Σxy And The Shear Stress On Plane Aa.
Solution: Sum Forces Normal To And Tangential To Aa In The Element Freebody
Above, With Σn(Aa) Known And Σxy Unknown:
,2-2 2-2
In Like Manner, Solve For The Shear Stress On Plane Aa, Using Our Result For Σxy:
This Problem And Prob. P2.1 Can Also Be Solved Using Mohr‘S Circle.
P2.3 A Vertical Clean Glass Piezometer Tube Has An Inside Diameter Of 1 Mm. When
A Pressure Is Applied, Water At 20°C Rises Into The Tube To A Height Of 25 Cm. After
Correcting For Surface Tension, Estimate The Applied Pressure In Pa.
Solution: For Water, Let Y = 0.073 N/M, Contact Angle Θ = 0°, And Γ = 9790 N/M3.
The Capillary Rise In The Tube, From Example 1.9 Of The Text, Is
Then The Rise Due To Applied Pressure Is Less By That Amount: Hpress = 0.25 M − 0.03 M =
0.22 M. The Applied Pressure Is Estimated To Be P = Γhpress = (9790 N/M3)(0.22 M) ≈
2160 Pa Ans.
θ? Bourdon
P2.4 Pressure Gages, Such As The Bourdon Gage W gage
In Fig. P2.4, Are Calibrated With A Deadweight
Piston. If The Bourdon Gage Is Designed To Rotate
The Pointer
2 cm Oil
10 Degrees For Every 2 Psig Of Internal Pressure, How diameter
Many Degrees Does The Pointer Rotate If The Piston
And Weight Together Total 44 Newtons?
Fig. P2.4
Solution: The Deadweight, Divided By The Piston Area, Should Equal The Pressure
Applied To The Bourdon Gage. Stay In Si Units For The Moment:
2-3
At 10 Degrees For Every 2 Psig, The Pointer Should Move Approximately 100 Degrees. Ans.
, P2.5 Quito, Ecuador Has An Average Altitude Of 9,350 Ft. On A Standard Day,
Pressure Gage A In A Laboratory Experiment Reads 63 Kpa And Gage B Reads 105 Kpa.
Express These Readings In Gage Pressure Or Vacuum Pressure, Whichever Is
Appropriate.
Solution: Convert 9,350 Ft X 0.3048 = 2,850 M. We Can Interpolate In The Standard
Altitude Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
Good Interpolating! Then Pa = 71500-63000 = 8500 Pa (Vacuum Pressure) Ans.(A),
And Pb = 105000 - 71500 = 33500 Pa (Gage Pressure) Ans.(B)
P2.6 Express Standard Atmospheric Pressure As A Head, H = P/Ρg, In (A) Feet Of Glycerin;
(B) Inches Of Mercury; (C) Meters Of Water; And (D) Mm Of Ethanol.
Solution: Take The Specific Weights, Γ = Ρg, From Table A.3, Divide Patm By Γ :
(a) Glycerin: H = (2116 Lbf/Ft2)/(78.7 Lbf/Ft3) ≈ 26.9 Ft Ans. (A)
(b) Mercury: H = (2116 Lbf/Ft2)/(846 Lbf/Ft3) = 2.50 Ft ≈ 30.0 Inches Ans. (B)
(c) Water: H = (101350 N/M2)/(9790 N/M3) ≈ 10.35 M Ans. (C)
(d) Ethanol: H = (101350 N/M2)/(7740 N/M3) = 13.1 M ≈ 13100 Mm Ans. (D)
P2.7 La Paz, Bolivia Is At An Altitude Of Approximately 12,000 Ft. Assume A
Standard Atmosphere. How High Would The Liquid Rise In A Methanol Barometer,
Assumed At 20°C? [Hint: Don‘T Forget The Vapor Pressure.]
Solution: Convert 12,000 Ft To 3658 Meters, And Table A.6, Or Eq. (2.20), Give
