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유체 역학 8 판 솔루션 주제에 대한 동영상 보기
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d여기에서 제8강 소방유체역학 8 – 유체 역학 8 판 솔루션 주제에 대한 세부정보를 참조하세요
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Fundamentals of Thermodynamics 8th edition SI solution. thank you flu mechanics 8th edition white solutions manual full download: chapter pressure.
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[솔루션] 유체역학 8판 (Fluid Mechanics 8th Edition), White
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frank white 유체역학 8판 솔루션 * 레포트다운 검색결과
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유체역학 8판 (저자 Clayton T.Crowe 8th ed , Engineer – 완도군청
자료제목 : 유체역학 8판 솔루션 (저자 Clayton T.Crowe 8th ed , Engineering Flu Mechanics) 8판 솔루션 입니다! 파일이름 : 유체역학 8판 전체 …
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주제와 관련된 이미지 유체 역학 8 판 솔루션
주제와 관련된 더 많은 사진을 참조하십시오 제8강 소방유체역학 8. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.
주제에 대한 기사 평가 유체 역학 8 판 솔루션
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Fundamentals of Thermodynamics 8th edition SI solution. thank you
Fluid Mechanics 8th Edition White SOLUTIONS MANUAL
Full download:
testbanklive/download/fluid-mechanics-8th-edition-white-
solutions-manual/
Chapter 2 • Pressure Distribution in a Fluid
P2 For the two-dimensional stress field in Fig. P2, 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 element. This gives the freebody shown at right. Now sum forces normal and tangential to side AA. Denote side length AA as ―L.‖
Fig. P2.
P2 For the stress field of Fig. P2, 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 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
In like manner, solve for the shear stress on plane AA, using our result for σxy:
This problem and Prob. P2 can also be solved using Mohr‘s circle.
P2 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 N/m, contact angle θ = 0°, and γ = 9790 N/m 3. The capillary rise in the tube, from Example 1 of the text, is
Then the rise due to applied pressure is less by that amount: h press = 0 m − 0 m = 0 m. The applied pressure is estimated to be p = γ h press = (9790 N/m 3 )(0 m) ≈ 2160 Pa Ans.
P2 Pressure gages, such as the Bourdon gage
in Fig. P2, are calibrated with a deadweight piston.
W θ?
Bourdon gage
If the Bourdon gage is designed to rotate the pointer
10 degrees for every 2 psig of internal pressure, how
many degrees does the pointer rotate if the piston and
weight together total 44 newtons?
2 cm diameter
Oil
Fig. P2.
Solution : The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon gage. Stay in SI units for the moment:
2-4 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
From Table A, methanol has ρ = 791 kg/m 3 and a large vapor pressure of 13,400 Pa. Then the manometer rise h is given by
P2 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the surface of Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.
Solution : We need some data from the Internet: Mars gravity is 3 m/s 2 , surface pressure is 700 Pa, and surface temperature is -10ºF (above the freezing temperature of ethanol). Then the bottom pressure is given by the hydrostatic formula, with ethanol density equal to 789 kg/m 3 from Table A. Convert ½ mile = ½(5280) ft = 2640 ft * 0 m/ft = 804 m. Then
P2 A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20°C. ( a ) What is the gage pressure, in lbf/in 2 , at the bottom of the tank? ( b ) How does your result in ( a ) change if the tank diameter is reduced to 15 ft? ( c ) Repeat ( a ) if leakage has caused a layer of 5 ft of water to rest at the bottom of the (full) tank.
Solution : This is a straightforward problem in hydrostatic pressure. From Table A, the density of SAE 30W oil is 891 kg/m 3 ÷ 515 = 1 slug/ft 3. ( a ) Thus the bottom pressure is
pbottom = ρ oil g h = (1. slug ft 3
)(32.
ft
s 2
)(36 ft ) = 2005 lbf ft 2
= 13.
lbf in 2
gage Ans .( a )
( b ) The tank diameter has nothing to do with it, just the depth: p bottom = 13 psig. Ans .( b )
( c ) If we have 31 ft of oil and 5 ft of water (ρ = 1 slug/ft 3 ), the bottom pressure is
2-5 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
P2 A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a depth of 50 ft. The absolute pressure at the bottom of the tank is approximately 221 kPa. From Table A, what might this liquid be?
Solution : Convert 50 ft to 15 m. Use the hydrostatic formula to calculate the bottom pressure:
P2 In Fig. P2, sensor A reads 1. kPa (gage). All fluids are at 20°C. Determine the elevations Z in meters of the liquid levels in the open piezometer tubes B and C.
Solution: (B) Let piezometer tube B be an arbitrary distance H above the gasoline- glycerin interface. The specific weights are γair ≈ 12 N/m 3 , γgasoline = 6670 N/m 3 , and γglycerin = 12360 N/m 3. Then apply the hydrostatic formula from point A to point B:
Fig. P2. 11
Solve for ZB = 2 m (23 cm above the gasoline-air interface) Ans. (b)
Solution (C): Let piezometer tube C be an arbitrary distance Y above the bottom. Then
1500 + 12(2) + 6670(1) + 12360(1 − Y) − 12360(ZC − Y) = pC = 0 (gage)
Solve for ZC = 1 m (93 cm above the gasoline-glycerin interface) Ans. (c)
2-7 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2-
oil water mercury
P2 For the three-liquid system
shown, compute h 1 and h 2. water
SG= 0.
mercury h 2
27 cm
Solution : The pressures at
the three top surfaces must all be
8 cm h 1 5 cm
atmospheric, or zero gage pressure. Compute γ = (0)(9790) = 7636 N/m 3. Also, from Table 2, γ = 9790 N/m 3 equality is
and γ = 133100 N/m 3. The surface pressure
P2 In Fig. P2 all fluids are at 20°C. Gage A reads 15 lbf/in 2 absolute and gage B reads 1 lbf/in 2 less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual reading of gage C in lbf/in 2 absolute.
Fig. P2. 15 Solution: First evaluate γair = (pA/RT)g = 15 × 144/(1717 × 528) ≈ 0 lbf/ft 3. Take γwater = 62 lbf/ft 3. Then apply the hydrostatic formula from point B to point C:
2-8 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2-
With the oil weight known, we can now apply hydrostatics from point A to point C:
P2 If the absolute pressure at the interface
between water and mercury in Fig. P2 is 93 kPa, what, in lbf/ft 2 , is ( a ) the pressure at the
surface, and ( b ) the pressure at the bottom
of the container?
Fig. P2.
Water
75° 75°
Mercury
28 cm
8 cm
32 cm
Solution : Do the whole problem in SI units and then convert to BG at the end. The bottom width and the slanted 75-degree walls are irrelevant red herrings. Just go up and down:
2- 10 Solutions Chapter Ma 2 nu • al Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2- 10
P2 The U-tube at right has a 1-cm ID and contains mercury as shown. If 20 cm 3 of water is poured into the right-hand leg, what will be the free surface height in each leg after the sloshing has died down?
Solution: First figure the height of water added:
Then, at equilibrium, the new system must have 25 cm of water on the right, and a 30-cm length of mercury is somewhat displaced so that ―L‖ is on the right, 0 m on the bottom, and ―0 − L‖ on the left side, as shown at right. The bottom pressure is constant:
Thus right-leg-height = 9 + 25 = 34 cm Ans. left-leg-height = 20 − 9 = 10 cm Ans.
P2 The hydraulic jack in Fig. P2. is filled with oil at 56 lbf/ft 3. Neglecting piston weights, what force F on the handle is required to support the 2000-lbf weight shown?
Fig. P2. 20
Solution: First sum moments clockwise about the hinge A of the handle:
or: F = P/16, where P is the force in the small (1 in) piston. Meanwhile figure the pressure in the oil from the weight on the large piston:
2- 11 Solutions Chapter Ma 2 nu • al Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2- 11
Therefore the handle force required is F = P/16 = 222/16 ≈ 14 lbf Ans.
P2 In Fig. P2 all fluids are at 20°C. Gage A reads 350 kPa absolute. Determine (a) the height h in cm; and (b) the reading of gage B in kPa absolute.
Solution: Apply the hydrostatic formula from the air to gage A:
Fig. P2. 21
Then, with h known, we can evaluate the pressure at gage B:
P2 The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. P2. If the tank accidentally contains 2 cm of water plus gasoline, how many centimeters ―h‖ of air remain when the gage reads ―full‖ in error?
Fig. P2. 22
Solution: Given γgasoline = 0(9790) = 6657 N/m 3 , compute the gage pressure when ―full‖:
Set this pressure equal to 2 cm of water plus ―Y‖ centimeters of gasoline:
Therefore the air gap h = 30 cm − 2 cm(water) − 27 cm(gasoline) ≈ 0 cm Ans.
2- 13 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 13
*P2 As measured by NASA‘s Viking landers, the atmosphere of Mars, where g = 3 m/s 2 , is almost entirely carbon dioxide, and the surface pressure averages 700 Pa. The temperature is cold and drops off exponentially: T ≈ T – Cz – o e , where C ≈ 1-5 m and T o ≈ 250 K. For example, at 20,000 m altitude, T ≈ 193 K. ( a ) Find an analytic formula for the variation of pressure with altitude. ( b ) Find the altitude where pressure on Mars has dropped to 1 pascal.
Solution : ( a ) The analytic formula is found by integrating Eq. (2) of the text:
( b ) From Table A for CO 2 2 , R = 189 m /(s -K). Substitute p = 1 Pa to find the altitude:
P2 For gases over large changes in height, the linear approximation, Eq. (2), is inaccurate. Expand the troposphere power-law, Eq. (2), into a power series and show that the linear approximation p ≈ p a – ρa g z is adequate when
Solution : The power-law term in Eq. (2) can be expanded into a series:
Multiply by p a, as in Eq. (2), and note that p a nB / T o = ( p a/ RT o) gz = ρa gz. Then the series may be rewritten as follows:
2- 14 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 14
For the linear law to be accurate, the 2nd term in parentheses must be much less than unity. If the starting point is not at z = 0, then replace z by δ z :
P2 This is an experimental problem: Put a card or thick sheet over a glass of water, hold it tight, and turn it over without leaking (a glossy postcard works best). Let go of the card. Will the card stay attached when the glass is upside down? Yes : This is essentially a water barometer and, in principle, could hold a column of water up to 10 ft high!
P2 A correlation of computational fluid dynamics results indicates that, all other things being equal, the distance traveled by a well-hit baseball varies inversely as the 0. power of the air density. If a home-run ball hit in NY Mets Citi Field Stadium travels 400 ft, estimate the distance it would travel in ( a ) Quito, Ecuador, and ( b ) Colorado Springs, CO.
Solution : Citi Field is in the Borough of Queens, NY, essentially at sea level. Hence the standard pressure is p o ≈ 101,350 Pa. Look up the altitude of the other two cities and calculate the pressure:
( a ) Quito : z ≈ 2850 m , pQ = po [1−
(0)(2850)
]5.
288.
= (101350)(0) = 71,500 Pa
( b ) Colorado Springs : z ≈ 1835 m , pCC = po [1−
(0)(1835)
]5.
288.
= (101350)(0) = 81,100 Pa
Then the estimated home-run distances are:
( a ) Quito : X = 40 0(
101350
)0 = 400(1) ≈ 454 ft Ans .( a ) 71500
( b ) Colorado Spri ngs : X = 40 0(
101350
)0 = 400(1) ≈ 433 ft Ans .( b ) 81100
The Colorado result is often confirmed by people who attend Rockies baseball games.
2- 16 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 16
P2 In Fig. P2 determine Δp between points A and B. All fluids are at 20°C.
Fig. P2. 31 Solution: Take the specific weights to be Benzene: 8640 N/m 3 Mercury: 133100 N/m 3 Kerosene: 7885 N/m 3 Water: 9790 N/m 3 and γair will be small, probably around 12 N/m 3. Work your way around from A to B:
P2 For the manometer of Fig. P2, all fluids are at 20°C. If pB − pA = 97 kPa, determine the height H in centimeters. Solution: Gamma = 9790 N/m 3 for water and 133100 N/m 3 for mercury and (0)(9790) = 8096 N/m 3 for Meriam red oil. Work your way around from point A to point B:
Fig. P2. 32
2- 17 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 17
P2 In Fig. P2 the pressure at point A is 25 psi. All fluids are at 20°C. What is the air pressure in the closed chamber B?
Solution: Take γ = 9790 N/m 3 for water, 8720 N/m 3 for SAE 30 oil, and (1)(9790) = 14196 N/m 3 for the third fluid. Convert the pressure at A from 25 lbf/in 2 to 172400 Pa. Compute hydrostatically from point A to point B:
Fig. P2. 33
P2 To show the effect of manometer dimensions, consider Fig. P2. The containers (a) and (b) are cylindrical and are such that pa = pb as shown. Suppose the oil-water interface on the right moves up a distance Δh < h. Derive a formula for the difference pa − pb when (a) and (b) d = 0. What is the % difference? Fig. P2. 34 Solution: Take γ = 9790 N/m 3 for water and 8720 N/m 3 for SAE 30 oil. Let ―H‖ be the height of the oil in reservoir (b). For the condition shown, pa = pb, therefore (1) Case (a), When the meniscus rises Δh, there will be no significant change in reservoir levels. Therefore we can write a simple hydrostatic relation from (a) to (b): 2- 19 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 19 P2 In Fig. P2 both the tank and the slanted tube are open to the atmosphere. If L = 2 m, what is the angle of tilt φ of the tube? Fig. P2. 36 Solution: Proceed hydrostatically from the oil surface to the slanted tube surface: P2 The inclined manometer in Fig. P2 contains Meriam red oil, SG = 0. Assume the reservoir is very large. If the inclined arm has graduations 1 inch apart, what should θ be if each graduation repre- sents 1 psf of the pressure pA? Fig. P2. 37 Solution: The specific weight of the oil is (0)(62) = 51 lbf/ft 3. If the reservoir level does not change and ΔL = 1 inch is the scale marking, then 2- 20 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 20 oil water mercury P2 If the pressure in container A is 200 kPa, compute the pressure in Fig. P2 B container B. A Water 18 cm Solution : The specific weights are γ = (0)(9790) = 7832 N/m 3 , γ = 133,100 N/m 3 , and γ = 9790 N/m 3. Solution : Begin at B and proceed around to A. 16 cm 8 cm Oil , SG = 0. Mercury 200,000+ (9790)(0 m )+ (133100)(0 − 0 m ) − ( 0 x 9790)(0 m ) = pA Solve for pA = 219,000 Pa = 219 kPa Ans.
유체 역학 8 판 솔루션 | 진샘미디어 기본 유체역학 2판 솔루션(1) 1장 점성 242 개의 새로운 답변이 업데이트되었습니다.
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Fundamentals of Thermodynamics 8th edition SI solution. thank you flu mechanics 8th edition white solutions manual full download: chapter pressure.
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유체역학 8판 솔루션 대학리포트 올레포트 무료표지 공업수학 솔루션 졸업논문 학업계획서 방송통신대학 자기소개서 이력서 사업계획서 기업분석.
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첨부파일 : 유체역학 8판 전체장(1장~15장) 저자 crowe, 김경호 등 공역 한티 … 자료제목 : 유체역학 8판 솔루션 (저자 Clayton T.Crowe 8th ed …
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오늘 소개할 자료는 바로바로 Munson 유체역학 8판 대학교재솔루션 (Munsons Flu Mechanics 8E Solutions Manual) 입니다. 문서자료 (유체역학 8판 …
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1) solution입니다. 19. 미시경제학 Pindyck 8th Edition Solutions Manual. 20. [솔루션] 유체역학 8판 (Fundamentals of Flu Mechanics 8th Edition) …
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유체역학8판솔루션.pdf. 유체역학 8th Edition (Engineering Flu Mechanics) Clayton T.Crowe, Donald F.Elger, John A.Roberson Wiley 한티미디어
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Fluid Mechanics 8th Edition White SOLUTIONS MANUAL
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Chapter 2 • Pressure Distribution in a Fluid
P2 For the two-dimensional stress field in Fig. P2, 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 element. This gives the freebody shown at right. Now sum forces normal and tangential to side AA. Denote side length AA as ―L.‖
Fig. P2.
P2 For the stress field of Fig. P2, 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 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
In like manner, solve for the shear stress on plane AA, using our result for σxy:
This problem and Prob. P2 can also be solved using Mohr‘s circle.
P2 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 N/m, contact angle θ = 0°, and γ = 9790 N/m 3. The capillary rise in the tube, from Example 1 of the text, is
Then the rise due to applied pressure is less by that amount: h press = 0 m − 0 m = 0 m. The applied pressure is estimated to be p = γ h press = (9790 N/m 3 )(0 m) ≈ 2160 Pa Ans.
P2 Pressure gages, such as the Bourdon gage
in Fig. P2, are calibrated with a deadweight piston.
W θ?
Bourdon gage
If the Bourdon gage is designed to rotate the pointer
10 degrees for every 2 psig of internal pressure, how
many degrees does the pointer rotate if the piston and
weight together total 44 newtons?
2 cm diameter
Oil
Fig. P2.
Solution : The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon gage. Stay in SI units for the moment:
2-4 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
From Table A, methanol has ρ = 791 kg/m 3 and a large vapor pressure of 13,400 Pa. Then the manometer rise h is given by
P2 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the surface of Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.
Solution : We need some data from the Internet: Mars gravity is 3 m/s 2 , surface pressure is 700 Pa, and surface temperature is -10ºF (above the freezing temperature of ethanol). Then the bottom pressure is given by the hydrostatic formula, with ethanol density equal to 789 kg/m 3 from Table A. Convert ½ mile = ½(5280) ft = 2640 ft * 0 m/ft = 804 m. Then
P2 A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20°C. ( a ) What is the gage pressure, in lbf/in 2 , at the bottom of the tank? ( b ) How does your result in ( a ) change if the tank diameter is reduced to 15 ft? ( c ) Repeat ( a ) if leakage has caused a layer of 5 ft of water to rest at the bottom of the (full) tank.
Solution : This is a straightforward problem in hydrostatic pressure. From Table A, the density of SAE 30W oil is 891 kg/m 3 ÷ 515 = 1 slug/ft 3. ( a ) Thus the bottom pressure is
pbottom = ρ oil g h = (1. slug ft 3
)(32.
ft
s 2
)(36 ft ) = 2005 lbf ft 2
= 13.
lbf in 2
gage Ans .( a )
( b ) The tank diameter has nothing to do with it, just the depth: p bottom = 13 psig. Ans .( b )
( c ) If we have 31 ft of oil and 5 ft of water (ρ = 1 slug/ft 3 ), the bottom pressure is
2-5 Solutions Chapter Manu 2 al • Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2-
P2 A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a depth of 50 ft. The absolute pressure at the bottom of the tank is approximately 221 kPa. From Table A, what might this liquid be?
Solution : Convert 50 ft to 15 m. Use the hydrostatic formula to calculate the bottom pressure:
P2 In Fig. P2, sensor A reads 1. kPa (gage). All fluids are at 20°C. Determine the elevations Z in meters of the liquid levels in the open piezometer tubes B and C.
Solution: (B) Let piezometer tube B be an arbitrary distance H above the gasoline- glycerin interface. The specific weights are γair ≈ 12 N/m 3 , γgasoline = 6670 N/m 3 , and γglycerin = 12360 N/m 3. Then apply the hydrostatic formula from point A to point B:
Fig. P2. 11
Solve for ZB = 2 m (23 cm above the gasoline-air interface) Ans. (b)
Solution (C): Let piezometer tube C be an arbitrary distance Y above the bottom. Then
1500 + 12(2) + 6670(1) + 12360(1 − Y) − 12360(ZC − Y) = pC = 0 (gage)
Solve for ZC = 1 m (93 cm above the gasoline-glycerin interface) Ans. (c)
2-7 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2-
oil water mercury
P2 For the three-liquid system
shown, compute h 1 and h 2. water
SG= 0.
mercury h 2
27 cm
Solution : The pressures at
the three top surfaces must all be
8 cm h 1 5 cm
atmospheric, or zero gage pressure. Compute γ = (0)(9790) = 7636 N/m 3. Also, from Table 2, γ = 9790 N/m 3 equality is
and γ = 133100 N/m 3. The surface pressure
P2 In Fig. P2 all fluids are at 20°C. Gage A reads 15 lbf/in 2 absolute and gage B reads 1 lbf/in 2 less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual reading of gage C in lbf/in 2 absolute.
Fig. P2. 15 Solution: First evaluate γair = (pA/RT)g = 15 × 144/(1717 × 528) ≈ 0 lbf/ft 3. Take γwater = 62 lbf/ft 3. Then apply the hydrostatic formula from point B to point C:
2-8 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2-
With the oil weight known, we can now apply hydrostatics from point A to point C:
P2 If the absolute pressure at the interface
between water and mercury in Fig. P2 is 93 kPa, what, in lbf/ft 2 , is ( a ) the pressure at the
surface, and ( b ) the pressure at the bottom
of the container?
Fig. P2.
Water
75° 75°
Mercury
28 cm
8 cm
32 cm
Solution : Do the whole problem in SI units and then convert to BG at the end. The bottom width and the slanted 75-degree walls are irrelevant red herrings. Just go up and down:
2- 10 Solutions Chapter Ma 2 nu • al Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2- 10
P2 The U-tube at right has a 1-cm ID and contains mercury as shown. If 20 cm 3 of water is poured into the right-hand leg, what will be the free surface height in each leg after the sloshing has died down?
Solution: First figure the height of water added:
Then, at equilibrium, the new system must have 25 cm of water on the right, and a 30-cm length of mercury is somewhat displaced so that ―L‖ is on the right, 0 m on the bottom, and ―0 − L‖ on the left side, as shown at right. The bottom pressure is constant:
Thus right-leg-height = 9 + 25 = 34 cm Ans. left-leg-height = 20 − 9 = 10 cm Ans.
P2 The hydraulic jack in Fig. P2. is filled with oil at 56 lbf/ft 3. Neglecting piston weights, what force F on the handle is required to support the 2000-lbf weight shown?
Fig. P2. 20
Solution: First sum moments clockwise about the hinge A of the handle:
or: F = P/16, where P is the force in the small (1 in) piston. Meanwhile figure the pressure in the oil from the weight on the large piston:
2- 11 Solutions Chapter Ma 2 nu • al Pressure Distribution in a Fluid• Fluid Mechanics, Eighth Edition 2- 11
Therefore the handle force required is F = P/16 = 222/16 ≈ 14 lbf Ans.
P2 In Fig. P2 all fluids are at 20°C. Gage A reads 350 kPa absolute. Determine (a) the height h in cm; and (b) the reading of gage B in kPa absolute.
Solution: Apply the hydrostatic formula from the air to gage A:
Fig. P2. 21
Then, with h known, we can evaluate the pressure at gage B:
P2 The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. P2. If the tank accidentally contains 2 cm of water plus gasoline, how many centimeters ―h‖ of air remain when the gage reads ―full‖ in error?
Fig. P2. 22
Solution: Given γgasoline = 0(9790) = 6657 N/m 3 , compute the gage pressure when ―full‖:
Set this pressure equal to 2 cm of water plus ―Y‖ centimeters of gasoline:
Therefore the air gap h = 30 cm − 2 cm(water) − 27 cm(gasoline) ≈ 0 cm Ans.
2- 13 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 13
*P2 As measured by NASA‘s Viking landers, the atmosphere of Mars, where g = 3 m/s 2 , is almost entirely carbon dioxide, and the surface pressure averages 700 Pa. The temperature is cold and drops off exponentially: T ≈ T – Cz – o e , where C ≈ 1-5 m and T o ≈ 250 K. For example, at 20,000 m altitude, T ≈ 193 K. ( a ) Find an analytic formula for the variation of pressure with altitude. ( b ) Find the altitude where pressure on Mars has dropped to 1 pascal.
Solution : ( a ) The analytic formula is found by integrating Eq. (2) of the text:
( b ) From Table A for CO 2 2 , R = 189 m /(s -K). Substitute p = 1 Pa to find the altitude:
P2 For gases over large changes in height, the linear approximation, Eq. (2), is inaccurate. Expand the troposphere power-law, Eq. (2), into a power series and show that the linear approximation p ≈ p a – ρa g z is adequate when
Solution : The power-law term in Eq. (2) can be expanded into a series:
Multiply by p a, as in Eq. (2), and note that p a nB / T o = ( p a/ RT o) gz = ρa gz. Then the series may be rewritten as follows:
2- 14 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 14
For the linear law to be accurate, the 2nd term in parentheses must be much less than unity. If the starting point is not at z = 0, then replace z by δ z :
P2 This is an experimental problem: Put a card or thick sheet over a glass of water, hold it tight, and turn it over without leaking (a glossy postcard works best). Let go of the card. Will the card stay attached when the glass is upside down? Yes : This is essentially a water barometer and, in principle, could hold a column of water up to 10 ft high!
P2 A correlation of computational fluid dynamics results indicates that, all other things being equal, the distance traveled by a well-hit baseball varies inversely as the 0. power of the air density. If a home-run ball hit in NY Mets Citi Field Stadium travels 400 ft, estimate the distance it would travel in ( a ) Quito, Ecuador, and ( b ) Colorado Springs, CO.
Solution : Citi Field is in the Borough of Queens, NY, essentially at sea level. Hence the standard pressure is p o ≈ 101,350 Pa. Look up the altitude of the other two cities and calculate the pressure:
( a ) Quito : z ≈ 2850 m , pQ = po [1−
(0)(2850)
]5.
288.
= (101350)(0) = 71,500 Pa
( b ) Colorado Springs : z ≈ 1835 m , pCC = po [1−
(0)(1835)
]5.
288.
= (101350)(0) = 81,100 Pa
Then the estimated home-run distances are:
( a ) Quito : X = 40 0(
101350
)0 = 400(1) ≈ 454 ft Ans .( a ) 71500
( b ) Colorado Spri ngs : X = 40 0(
101350
)0 = 400(1) ≈ 433 ft Ans .( b ) 81100
The Colorado result is often confirmed by people who attend Rockies baseball games.
2- 16 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 16
P2 In Fig. P2 determine Δp between points A and B. All fluids are at 20°C.
Fig. P2. 31 Solution: Take the specific weights to be Benzene: 8640 N/m 3 Mercury: 133100 N/m 3 Kerosene: 7885 N/m 3 Water: 9790 N/m 3 and γair will be small, probably around 12 N/m 3. Work your way around from A to B:
P2 For the manometer of Fig. P2, all fluids are at 20°C. If pB − pA = 97 kPa, determine the height H in centimeters. Solution: Gamma = 9790 N/m 3 for water and 133100 N/m 3 for mercury and (0)(9790) = 8096 N/m 3 for Meriam red oil. Work your way around from point A to point B:
Fig. P2. 32
2- 17 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 17
P2 In Fig. P2 the pressure at point A is 25 psi. All fluids are at 20°C. What is the air pressure in the closed chamber B?
Solution: Take γ = 9790 N/m 3 for water, 8720 N/m 3 for SAE 30 oil, and (1)(9790) = 14196 N/m 3 for the third fluid. Convert the pressure at A from 25 lbf/in 2 to 172400 Pa. Compute hydrostatically from point A to point B:
Fig. P2. 33
P2 To show the effect of manometer dimensions, consider Fig. P2. The containers (a) and (b) are cylindrical and are such that pa = pb as shown. Suppose the oil-water interface on the right moves up a distance Δh < h. Derive a formula for the difference pa − pb when (a) and (b) d = 0. What is the % difference? Fig. P2. 34 Solution: Take γ = 9790 N/m 3 for water and 8720 N/m 3 for SAE 30 oil. Let ―H‖ be the height of the oil in reservoir (b). For the condition shown, pa = pb, therefore (1) Case (a), When the meniscus rises Δh, there will be no significant change in reservoir levels. Therefore we can write a simple hydrostatic relation from (a) to (b): 2- 19 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 19 P2 In Fig. P2 both the tank and the slanted tube are open to the atmosphere. If L = 2 m, what is the angle of tilt φ of the tube? Fig. P2. 36 Solution: Proceed hydrostatically from the oil surface to the slanted tube surface: P2 The inclined manometer in Fig. P2 contains Meriam red oil, SG = 0. Assume the reservoir is very large. If the inclined arm has graduations 1 inch apart, what should θ be if each graduation repre- sents 1 psf of the pressure pA? Fig. P2. 37 Solution: The specific weight of the oil is (0)(62) = 51 lbf/ft 3. If the reservoir level does not change and ΔL = 1 inch is the scale marking, then 2- 20 ChSolutions apter 2 Ma • nu Pressure Distribution in a Fluid al • Fluid Mechanics, Eighth Edition 2- 20 oil water mercury P2 If the pressure in container A is 200 kPa, compute the pressure in Fig. P2 B container B. A Water 18 cm Solution : The specific weights are γ = (0)(9790) = 7832 N/m 3 , γ = 133,100 N/m 3 , and γ = 9790 N/m 3. Solution : Begin at B and proceed around to A. 16 cm 8 cm Oil , SG = 0. Mercury 200,000+ (9790)(0 m )+ (133100)(0 − 0 m ) − ( 0 x 9790)(0 m ) = pA Solve for pA = 219,000 Pa = 219 kPa Ans.
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