Solution Manual for Fundamentals of Hydraulic Engineering Systems 4th Edition by Houghtalen

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Chapter 1 – Problem Solutions

1.2.1 1.2.1

E

E11= energy req’d to bring ice temperature to 0= energy req’d to bring ice temperature to 0

°°

CC

E

E11 = (250 L)(1000 g/L)(20 = (250 L)(1000 g/L)(20

°°

C)(0.465 cal/g·C)(0.465 cal/g·

°°

C)C)

E

E11 = 2.33x10 = 2.33x1066 cal cal

E

E22 = energy required to melt ice = energy required to melt ice

E

E22 = (250 = (250 L)(1000 g/L)(79.7 cal/g·L)(1000 g/L)(79.7 cal/g·

°°

C)C)

E

E22 = 1.99x10 = 1.99x1077 cal cal

E

E33 = energy = energy required to raise the water temperature torequired to raise the water temperature to

20 20

°°

CC E E33 = (250 L)(1000 g/L)(20 = (250 L)(1000 g/L)(20

°°

C)(1 cal/g·C)(1 cal/g·

°°

C)C) E E33 = 5.00x10 = 5.00x1066 cal cal E

Etotaltotal = E = E11 + E + E22+ E+ E33 = = 2.72x102.72x1077calcal

______________________________________ ____________________________________________

1.2.2 1.2.2

At 0.9 bar (ambient pressure), the boiling temperature At 0.9 bar (ambient pressure), the boiling temperature of water is 97

of water is 97

°°

C (see Table 1.1).C (see Table 1.1). E

E11= energy required to bring the = energy required to bring the water temperature towater temperature to

97 97

°°

CC E E11= (1200 g)(97= (1200 g)(97

°°

C - 45C - 45

°°

C)C) E E11= 6.24x10= 6.24x1044 cal cal E

E22 = energy = energy required to vaporize the waterrequired to vaporize the water

E

E22 = (1200 g)(597 cal/g) = (1200 g)(597 cal/g)

E

E22= 7.16x10= 7.16x1055 cal cal

E

Etotaltotal = E = E11 + E + E22== 7.79x107.79x1055calcal

1.2.3 1.2.3

E

E11 = energy required to change water to ice = energy required to change water to ice

E

E11= (100 g)(79.7 cal/g)= (100 g)(79.7 cal/g)

E

E11= 7.97x10= 7.97x1033 cal cal

E

E22 = energy required to change vapor to ice = energy required to change vapor to ice

E

E22 = (100 g)(597 cal/g) + (100 g)(79.7 cal/g) = (100 g)(597 cal/g) + (100 g)(79.7 cal/g)

E

E22 = 6.77x10 = 6.77x1044 cal cal

Total energy removed to freeze water and vapor. Total energy removed to freeze water and vapor.

E

Etotaltotal = E = E11 + E + E22 = 7.57x10 = 7.57x1044 cal cal

____________________________________ ______________________________________________

1.2.4 1.2.4

E

E11 = energy needed to vaporize the water = energy needed to vaporize the water

E

E11 = (100 = (100 L)(1000 g/L)(597 cal/g)L)(1000 g/L)(597 cal/g)

E

E11 = 5.97x10 = 5.97x1077 cal cal

The energy remaining (E The energy remaining (E22) is:) is:

E

E22 = E – E = E – E11

E

E22 = 6.80x10 = 6.80x1077 cal – 5.97x10 cal – 5.97x1077 cal cal

E

E22 = 8.30x10 = 8.30x1066 cal cal

The temperature change possible with the remaining The temperature change possible with the remaining energy is:

energy is:

8.30x10

8.30x1066 cal = cal = (100 L)(1000 g/L)(1 cal/g·(100 L)(1000 g/L)(1 cal/g·

°°

C)(C)(

Δ

T)T)

Δ

T = 83T = 83

°°

C, making the C, making the temperaturetemperature T = 93

T = 93

°°

C when it evaporates.C when it evaporates. Therefore, based on Table 1.1, Therefore, based on Table 1.1,

P = 0.777 atm P = 0.777 atm

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1.2.5 1.2.5

E

E11 = energy required to raise the temperature to 100 = energy required to raise the temperature to 100

°°

CC

E

E11 = (5000 g)(100 = (5000 g)(100

°°

C – 25C – 25

°°

C)C)

E

E22 = 3.75x10 = 3.75x1055 cal cal

E

E22 = energy required to vaporize 2.5 kg of water = energy required to vaporize 2.5 kg of water

E

E22= (2500 g)(597 cal/g)= (2500 g)(597 cal/g)

E

E22 = 1.49x10 = 1.49x1066 cal cal

E

Etotaltotal = E = E11 + E + E22 = 1.87x10 = 1.87x1066 cal cal

Time required = (1.87x10

Time required = (1.87x1066 cal)/(500 cal/s) cal)/(500 cal/s) == 3740 sec = 62.3 min 3740 sec = 62.3 min ______________________________________ ____________________________________________ 1.2.6 1.2.6 E

E11 = energy required to melt ice = energy required to melt ice

E

E11 = (5 = (5 slugs)(32.2 lbm/slug)slugs)(32.2 lbm/slug)(32(32

°°

F - 20F - 20

°°

F)(0.46F)(0.46

BTU/lbm·

°°

F) + (5 F) + (5 slugs)(32.2 lbm/slug)(144slugs)(32.2 lbm/slug)(144 BTU/lbm)

BTU/lbm)

E

E11= 2.41 x 10= 2.41 x 1044 BTU BTU

To melt the ice, the

To melt the ice, the temperature of the water willtemperature of the water will decrease to:

decrease to:

2.41 x 10

2.41 x 1044 BTU = (10 BTU = (10 slugs)(32.2 lbm/slug)slugs)(32.2 lbm/slug)(120(120

°°

F –F – T

T11)(1 BTU/lbm·)(1 BTU/lbm·

°°

F)F)

T

T11 = 45.2 = 45.2

°°

FF

The energy lost by the water (to lower its temp. to The energy lost by the water (to lower its temp. to 45.2

45.2

°°

F) is that required to mF) is that required to melt the ice. elt the ice. Now you haveNow you have 5 slugs of water at 32

5 slugs of water at 32

°°

F and 10 slugs at 45.2F and 10 slugs at 45.2

°°

F.F. Therefore, the final temperature of the water is: Therefore, the final temperature of the water is:

[(10 slugs)(32.2

[(10 slugs)(32.2 lbm/slug)(45.2lbm/slug)(45.2

°°

F – TF – T22)(1)(1

BTU/lbm· BTU/lbm·

°°

F)]F)] = [(5

= [(5 slugs)(32.2 lbm/slug)(Tslugs)(32.2 lbm/slug)(T22 - 32 - 32

°°

F)(1 BTU/lbm·F)(1 BTU/lbm·

°°

F)]F)]

T

T22 = 40.8 = 40.8

1.3.1 1.3.1

The weight of water in the

The weight of water in the container is 814 N.container is 814 N.

m = W/g = (814 N)/(9.81 m/sec m = W/g = (814 N)/(9.81 m/sec22) = 83.0 kg) = 83.0 kg At 20 At 20˚˚C, 998 kg = 1 mC, 998 kg = 1 m 3 3

Therefore, the volume can be determined by Therefore, the volume can be determined by

Vol = (83.0 kg)(1 m Vol = (83.0 kg)(1 m33/998 kg)/998 kg) Vol = 8.32 x 10 Vol = 8.32 x 10-2-2 m m33 ____________________________________ ______________________________________________ 1.3.2 1.3.2 F =

F = m·a m·a Letting Letting a = a = g resulg results in ts in Equation Equation 1.11.1

W = m·g, dividing both sides of the equation by volume W = m·g, dividing both sides of the equation by volume

yields yields γ γ = =ρρ·g·g ____________________________________ ______________________________________________ 1.3.3 1.3.3

γγ

= =

ρρ

·g = (13,600 kg/m·g = (13,600 kg/m33)(9.81 m/s)(9.81 m/s22)) S.G. =

S.G. =

γγ

liguidliguid//

γγ

water at 4water at 4°°CC

S.G. = (133,000 N/m S.G. = (133,000 N/m33)/(9810 N/m)/(9810 N/m33)) S.G. = 13.6 (mercury) S.G. = 13.6 (mercury) ____________________________________ ______________________________________________ 1.3.4 1.3.4

The force exerted on the tank bottom is equal to the The force exerted on the tank bottom is equal to the weight of the water body.

weight of the water body.

F = W = m

F = W = m .. g = [ρ g = [ρ..(Vol)] (g)(Vol)] (g)

F = [1.94 slugs/ft

F = [1.94 slugs/ft33 ( (ππ · (5 ft) · (5 ft)22 · 3 ft)] (32.2 ft/sec · 3 ft)] (32.2 ft/sec22))

F = 1.47 x 10 F = 1.47 x 1044 lbs lbs

(Note: 1 slug = 1 lb·sec (Note: 1 slug = 1 lb·sec22/ft)/ft) F

F

= 133 kN/m = 133 kN/m33

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1.3.5 1.3.5

Weight of water on earth = 7.85 kN Weight of water on earth = 7.85 kN

m = W/g

m = W/g = (7,850 N)/(9.81 m/s= (7,850 N)/(9.81 m/s22))

m = 800 kg m = 800 kg

Note: mass on

Note: mass on moon is the samoon is the same as mass on eame as mass on earthrth

W (moon) = mg = (800 kg)[(9.81 m/s W (moon) = mg = (800 kg)[(9.81 m/s22)/(6)])/(6)] W(moon) = 1310 N W(moon) = 1310 N ______________________________________ ____________________________________________ 1.3.6 1.3.6 W = mg = (0.258 slug)(32.2 ft/s W = mg = (0.258 slug)(32.2 ft/s22)) W = 8.31 lb W = 8.31 lb Note: a slug ha

Note: a slug has units of (lb·s units of (lb·ss22)/(ft))/(ft)

Volume of 1 gal = 0.134 ft Volume of 1 gal = 0.134 ft33

S.G. =

γγ

liguidliguid//

γγ

water at 4water at 4°°CC

γγ

= (8.31 lb)/(0.134 ft = (8.31 lb)/(0.134 ft33) = 62.0 lb/ft) = 62.0 lb/ft33 S.G. = (62.0 lb/ft S.G. = (62.0 lb/ft33)/(62.4 lb/ft)/(62.4 lb/ft33)) S.G. = 0.994 S.G. = 0.994 ______________________________________ ____________________________________________ 1.3.7 1.3.7

Density can be expressed as: Density can be expressed as:

ρ

ρ = m/Vol = m/Vol

and even though volume changes

and even though volume changes with temperature,with temperature, mass doe

mass does not. s not. Thus,Thus,

((ρρ11)(Vol)(Vol11) = (ρ) = (ρ22)(Vol)(Vol22) = constant; or) = constant; or

Vol

Vol22 = ( = (ρρ11)(Vol)(Vol11)/()/(ρρ22))

Vol

Vol22 = (1000 kg/m = (1000 kg/m33)(100 m)(100 m33)/(958 kg/m)/(958 kg/m33))

Vol

Vol22 = 104.4 m = 104.4 m33 (or a (or a 4.4% change 4.4% change in volume)in volume)

1.3.8 1.3.8 (1 N)[(1 lb)/(4.448 N)] = (1 N)[(1 lb)/(4.448 N)] = 0.2248 lb0.2248 lb ____________________________________ ______________________________________________ 1.3.9 1.3.9 (1 N (1 N

⋅⋅

m)[(3.281 ft)/(1 m)][(0.2248 lb)/(1 m)[(3.281 ft)/(1 m)][(0.2248 lb)/(1 N))]N))] = = 7.376 x 107.376 x 10-1-1 ft ft ____________________________________ ______________________________________________ 1.4.1 1.4.1

[[

μμ

(air)/(air)/

μμ

(H(H22O)]O)]2020°°CC= (1.817x10= (1.817x10-5-5)/(1.002x10)/(1.002x10-3-3))

[[

μμ

(air)/(air)/

μμ

(H(H22O)]O)]8080°°CC= (2.088x10= (2.088x10-5-5)/(0.354x10)/(0.354x10-3-3))

[[

νν

(air)/(air)/

νν

(H(H22O)]O)]2020°°CC= (1.509x10= (1.509x10-5-5)/(1.003x10)/(1.003x10-6-6))

[[

νν

(air)/(air)/

νν

(H(H22O)]O)]8080°°CC= (2.087x10= (2.087x10-5-5)/(0.364x10)/(0.364x10-6-6))

[[

Note: The rat

Note: The ratio of absolute and io of absolute and kinematic visckinematic viscosities ofosities of air and water increases with temperature because the air and water increases with temperature because the viscosity of air increases with temperature, but that of viscosity of air increases with temperature, but that of water decreases w

water decreases with temperature. ith temperature. Also, the values ofAlso, the values of kinematic viscosity (

kinematic viscosity (

νν

) for air and water are much) for air and water are much closer than those

closer than those of absolute viscof absolute viscosity. osity. Why?Why? ____________________________________ ______________________________________________

1.4.2 1.4.2

μμ

(water)(water)2020°°CC= 1.002x10= 1.002x10-3-3 N N

⋅⋅

sec/msec/m22

νν

(water)(water)2020°°CC= 1.003x10= 1.003x10-6-6 m m22/s/s (1.002x10 (1.002x10-3-3 N N

⋅⋅

sec/msec/m22)·[(0.2248 lb)/(1 N)]·)·[(0.2248 lb)/(1 N)]· [(1 m) [(1 m)22/(3.281 ft)/(3.281 ft)22] =] = 2.092x102.092x10-5-5lblb (1.003x10 (1.003x10-6-6mm22/s)[(3.281 ft)/s)[(3.281 ft)22/(1 m)/(1 m)22] =] = 1.080x10 1.080x10-5-5 ft ft22/s/s lb lb (air)/

(air)/ (H(H22O)]O)]2020 CC= 1.813x10= 1.813x10-2-2

(air)/

(air)/ (H(H22O)]O)]8080 CC= 5.90x10= 5.90x10-2-2

(air)/

(air)/ (H(H22O)]O)]2020 CC= 15.04= 15.04

(air)/

(air)/ (H(H22O)]O)]8080 CC= 57.3= 57.3

sec/ft sec/ft22

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1.4.3 1.4.3

(a) 1 poise = 0.1 N

⋅⋅

sec/msec/m22 (0.1 N (0.1 N

⋅⋅

sec/msec/m22)[(0.2248 lb)/(1 N)][(1 m))[(0.2248 lb)/(1 N)][(1 m)22/(3.281 ft)/(3.281 ft)22] =] = 2.088x10 2.088x10-3-3 lb lb alternatively, alternatively, 1 lb

1 lb

⋅⋅

sec/ftsec/ft22= 478.9 poise= 478.9 poise (b) 1 stoke = 1 cm (b) 1 stoke = 1 cm22/sec/sec

(1 cm

(1 cm22/s)[(0.3937 in)/s)[(0.3937 in)22/(1 cm)/(1 cm)22][(1 ft)][(1 ft)22/(12 in)/(12 in)22] =] = 1.076x10

1.076x10-3-3 ft ft22/sec/sec

alternatively, alternatively,

1 ft

1 ft22/sec = 929.4 stoke/sec = 929.4 stoke

______________________________________ ____________________________________________

1.4.4 1.4.4

Assuming a

Assuming a Newtonian relationship:Newtonian relationship:

ττ

= =

μμ

(dv/dy) =(dv/dy) =

μμ

((ΔΔv/v/ΔΔy)y)

ττ

= (2.09x10 = (2.09x10-5-5 lb lb

⋅⋅

sec/ftsec/ft22)[(5 ft/sec)/(0.25 ft)])[(5 ft/sec)/(0.25 ft)]

ττ

= (4.18x10 = (4.18x10-4-4 lb/ft lb/ft22)) F = F =

ττ

·A = (4.18x10·A = (4.18x10-4-4 lb/ft lb/ft22)(10 ft)(30 ft))(10 ft)(30 ft) F = 0.125 lbs F = 0.125 lbs ______________________________________ ____________________________________________ 1.4.5 1.4.5 v = y

v = y22 – 2y, where y is in inches and v – 2y, where y is in inches and v is in ft/sis in ft/s

Making units consistent yields Making units consistent yields

v = 144y

v = 144y22 – 24y, where y is in ft and – 24y, where y is in ft and

νν

is in ft/s is in ft/s Taking the first derivative w/respect to y: Taking the first derivative w/respect to y:

dv/dy = 288y – 24 sec dv/dy = 288y – 24 sec-1-1

ττ

= =

μμ

(dv/dy)(dv/dy)

ττ

= (0.375 N = (0.375 N

⋅⋅

sec/msec/m22)( 288y – 24 sec)( 288y – 24 sec-1-1))

1.4.5 (cont.) 1.4.5 (cont.) Solutions: Solutions: y = 0 ft, y = 0 ft,

ττ

= - 9.00 N/m = - 9.00 N/m22 y = 1/12 ft, y = 1/12 ft,

ττ

= 0 N/m = 0 N/m22 y = 1/6 ft, y = 1/6 ft,

ττ

= 9.00 N/m = 9.00 N/m22 y = 1/4 ft, y = 1/4 ft,

ττ

= 18.0 N/m = 18.0 N/m22 y = 1/3 ft, y = 1/3 ft,

ττ

= 27.0 N/m = 27.0 N/m22 ____________________________________ ______________________________________________ 1.4.6 1.4.6

Based on the geometry of

Based on the geometry of the inclinethe incline

T

Tshear forceshear force = W(sin15 = W(sin15

°°

) =) =

τ⋅

A =A =

μμ

(dv/dy)A(dv/dy)A

Δ

y = [(y = [(

μμ

)()(

Δ

v)(A)] / [(W)(sin15v)(A)] / [(W)(sin15

°°

)])]

Δ

y = [(1.29 Ny = [(1.29 N

⋅⋅

sec/msec/m22)(0.025 m/sec))(0.025 m/sec) (0.50m)(0.75m) (0.50m)(0.75m)]/[(220 ]/[(220 N)(sin15N)(sin15

°°

)])] y = 2.12 x 10 y = 2.12 x 10-4-4 m = 2.12 x 10 m = 2.12 x 10-2-2 cm cm ____________________________________ ______________________________________________ 1.4.7 1.4.7 ∑

∑FFyy = 0 = 0 (constant velocity motion)(constant velocity motion)

W = T

W = Tshear forceshear force = =

τ⋅

A; A; where where A is A is the surthe surface areface areaa

(of the cylinder) in contact with the

(of the cylinder) in contact with the oil film:oil film:

A = (

ππ

)[(5.48/12)ft])[(5.48/12)ft][(9.5/12)ft] = 1.14 [(9.5/12)ft] = 1.14 ftft22 Now,

Now,

ττ

= W/A = (0.5 lb)/(1.14 ft = W/A = (0.5 lb)/(1.14 ft22) = 0.439 lb/ft) = 0.439 lb/ft22

ττ

= =

μμ

(dv/dy) =(dv/dy) =

μμ

((ΔΔv/v/ΔΔy), wherey), where Δ

Δv = v (the velociv = v (the velocity of the cylity of the cylinder). nder). Thus,Thus,

v = ( v = (

ττ

)()(ΔΔy)/y)/

μμ

v = [(0.439 lb/ft v = [(0.439 lb/ft22){(0.002/12)ft}){(0.002/12)ft}] / ] / (0.016 lb·s/ft(0.016 lb·s/ft22)) v = 4.57 x 10 v = 4.57 x 10-3-3 ft/sec ft/sec sec/ft sec/ft22

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1.4.8 1.4.8

ττ

= =

μμ

((

Δ

v/v/

Δ

y)y)

ττ

= (0.0065 lb = (0.0065 lb

⋅⋅

sec/ftsec/ft22)[(1 ft/s)/(0.5/12 ft))[(1 ft/s)/(0.5/12 ft)

ττ

= 0.156 lb/ft = 0.156 lb/ft22 F = (

F = (

ττ

)(A) = (2 sides)(0.156 lb/ft)(A) = (2 sides)(0.156 lb/ft22)(2 ft)(2 ft22)) F = 0.624 lb F = 0.624 lb ______________________________________ ____________________________________________ 1.4.9 1.4.9

μμ

= =

ττ

/(dv/dy) = (F/A)/(/(dv/dy) = (F/A)/(

Δ

v/v/

Δ

y);y);

Torque = Force·distance = F·R; R = radius Torque = Force·distance = F·R; R = radius

Thus;

μμ

= = (Torque/R)/[((Torque/R)/[(A)(A)(

Δ

v/v/

Δ

y)]y)]

μμ

= = )) )( )( )( )( )( )( 2 2 (( )) // )( )( )( )( )( )( 2 2 (( // 3 3 ω ω π π ω ω π π R R hh y y Torque Torque y y R R h h R R R R Torque Torque

₌₌

⋅⋅

Δ

⋅⋅

μμ

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⋅⋅

rpm rpm rad rad rpm rpm m m m m m m m m N N 60 60 sec sec // 2 2 )) 2000 2000 )( )( 04 04 .. 0 0 (( )) 025 025 .. 0 0 )( )( 2 2 (( )) 0002 0002 .. 0 0 )( )( 50 50 .. 1 1 (( 3 3 π π π π ______________________________________ ____________________________________________ 1.4.10 1.4.10

μμ

= (16)(1.002x10 = (16)(1.002x10-3-3 N N

⋅⋅

sec/msec/m22))

μμ

= 1.603x10 = 1.603x10-2-2 N N

⋅⋅

sec/msec/m22 Torque = Torque = R R

_{∫ ∫}

_r_r_dF_dF

₌₌

_∫∫

RR_r_r

_⋅⋅

_dA_dA τ τ 0 0 00 )) (( Torque = Torque =

_∫∫

Δ

R R dA dA y y vv r r 0 0 )) )( )( )( )( (( μ μ Torque = Torque =

_∫∫

Δ

−−

R R dr dr r r y y r r r r 0 0 )) 2 2 )( )( 0 0 )) )( )( (( )( )( )( )( (( μ μ ω ω π π Torque = Torque =

_∫∫

Δ

R R dr dr r r y y ₀₀ 3 3 )) (( )) )( )( )( )( 2 2 (( π π μ μ ω ω Torque = Torque = ⎥⎥ ⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎢⎢ ⎣⎣ ⎡⎡ ⋅⋅ ⋅⋅ −− 4 4 )) 1 1 (( 0005 0005 .. 0 0 sec) sec) // 65 65 .. 0 0 )( )( sec/ sec/ 10 10 603 603 .. 1 1 )( )( 2 2 (( 22 22 mm44 m m rad rad m m N N π π Torque = 32.7 N Torque = 32.7 N 1.5.1 1.5.1 h = [(4)( h = [(4)(

σ

)(sin θ)(sinθ)] / [()] / [(

γγ

)(D)])(D)] But sin 90 But sin 90˚˚ = 0, σ = 0,σ = 7.132x10 = 7.132x10 -2 -2 N/m N/m and

and

γγ

= 9790 N/m = 9790 N/m33 (at 20 (at 20˚˚C)C)

thus, D = [(4)(

σ

)] / [()] / [(

γγ

)(h)]; for h = 3.0 cm)(h)]; for h = 3.0 cm D = [(4)( 7.132x10 D = [(4)( 7.132x10-2-2 N/m)] / [(9790 N/m N/m)] / [(9790 N/m33)(0.03m)])(0.03m)] D = 9.71 x 10 D = 9.71 x 10-4-4 m = 9.71 x 10 m = 9.71 x 10-2-2cm; cm; thus,thus, for h = 3.0 cm, D = 0.0971 cm for h = 3.0 cm, D = 0.0971 cm for h = 2.0 cm, D = 0.146 cm for h = 2.0 cm, D = 0.146 cm for h = 1.0 cm, D = 0.291 cm for h = 1.0 cm, D = 0.291 cm ____________________________________ ______________________________________________ 1.5.2 1.5.2

The concept of a line

The concept of a line force is logical for two reasons:force is logical for two reasons: 1)

1) The surface tension acts along the The surface tension acts along the perimeter ofperimeter of the tube pulling the

the tube pulling the column of water upwardscolumn of water upwards due to adhesion between the water and the due to adhesion between the water and the tube.

tube. 2)

2) The surface tension is be The surface tension is be multiplied by themultiplied by the tube perimeter, a length, to obtain the

tube perimeter, a length, to obtain the upwardupward force used in force balance

force used in force balance development of thedevelopment of the equation for capillary rise.

equation for capillary rise.

____________________________________ ______________________________________________ 1.5.3 1.5.3

σ

= [(h)( = [(h)(

γγ

)(D)] / [(4)(sin)(D)] / [(4)(sin

θθ

)])]

σ

= = [(0.6/12)ft(1.[(0.6/12)ft(1.94slug/ft94slug/ft33)(32.2 lb/ft)(32.2 lb/ft33)(0.02/12)ft]/)(0.02/12)ft]/ [(4)(sin 54 [(4)(sin 54

°°

)])] = 3.65x10 = 3.65x10-1-1NN sec/msec/m22 m m = 1.61 x 10 = 1.61 x 10-3-3 lb/ft lb/ft

(6)

1.6.1 1.6.1 1.5.4 1.5.4 E Ebb = - = -

Δ

P/(P/(

Δ

Vol/Vol) = 9.09 x 10Vol/Vol)= 9.09 x 10 9 9 N/m N/m22 Capillary rise in the 0.25 cm. tube

Capillary rise in the 0.25 cm. tube is found using:is found using:

____________________________________ ______________________________________________ h = [(4)( h = [(4)(

σ

)(sin)(sin

θθ

)] / [()] / [(

γγ

)(D)])(D)] where where

σ

= (6.90 x 10 = (6.90 x 10-2-2)(1.2) = 8.28 x 10)(1.2) = 8.28 x 10-2-2 N/m N/m 1.6.21.6.2 P P11 = 25 bar = 25 x 10 = 25 bar = 25 x 1055 N/m N/m22 = 2.50 x 10 = 2.50 x 1066 N/m N/m22 and and

γγ

= (9752)(1.03) = 1.00 x 10 = (9752)(1.03) = 1.00 x 1044 N/m N/m33

Δ

Vol/Vol = -Vol/Vol = -

Δ

P/EP/E b b

h = h = )) 0025 0025 .. 0 0 )( )( // 10 10 00 00 .. 1 1 (( )) 30 30 )(sin )(sin // 10 10 28 28 .. 8 8 (( 4 4 3 3 4 4 2 2 m m m m N N m m N N

⋅⋅

−−

Δ

Vol/Vol = -(4.5 x 10Vol/Vol = -(4.5 x 1055 N/m N/m22 – 2.5 x 10 – 2.5 x 1066 N/m N/m22)/)/ (2.2x10 (2.2x1099 N/m N/m22)) h = 6.62x10 h = 6.62x10-3-3 m = 0.662 cm m = 0.662 cm ____________________________________ ______________________________________________

Δ

Vol/Vol = 9.3 x 10Vol/Vol = 9.3 x 10-4-4 = 0.093% (volume increase) = 0.093% (volume increase)

Δρ

//

ρρ

= - = -

Δ

Vol/Vol = -0.093% (density Vol/Vol = -0.093% (density decreases)decreases) 1.5.5

1.5.5

____________________________________ ______________________________________________ Condition 1: h

Condition 1: h11 = [(4)( = [(4)(

σ

11)(sin)(sin

θθ

11)] / [()] / [(

γγ

)(D)])(D)]

1.6.3 1.6.3 h

h11 = [(4)( = [(4)(

σ

11)(sin30)(sin30

°°

)] / [()] / [(

γγ

)(0.7 mm)])(0.7 mm)]

ρρ

oo = 1.94 slugs/ft = 1.94 slugs/ft33 (based on (based on temp. & pressure)temp. & pressure)

Condition 2: h

Condition 2: h22 = [(4)( = [(4)(

σ

22)(sin)(sin

θθ

22)] / [()] / [(

γγ

)(D)])(D)]

m =

ρρ

oo·Vol·Voloo = (1.94 slug/ft = (1.94 slug/ft33)(120 ft)(120 ft33) = 233 slugs) = 233 slugs

h

h22 = [(4)(0.9 = [(4)(0.9

σ

11)(sin42)(sin42

°°

)] / [()] / [(

γγ

)(0.7 mm)])(0.7 mm)]

W

W = mg = (233 slugs)(32.2 ft/sec = mg = (233 slugs)(32.2 ft/sec22) =) = 7,500 lb7,500 lb h

h22/h/h11 = [(0.9)(sin42 = [(0.9)(sin42

°°

)] / (sin30)] / (sin30

°°

) =) = 1.2041.204

ρ

ρ = =

ρρ

oo/[1+(/[1+(ΔΔVol/Vol)]; see example 1.3Vol/Vol)]; see example 1.3

alternatively, alternatively,

ρ

ρ= 1.94 slug/ft= 1.94 slug/ft33/[1+(-0.545/120)] =/[1+(-0.545/120)] = 1.95 slug/ft1.95 slug/ft33

h

h22 = 1.204(h = 1.204(h11), about a 20% increase!), about a 20% increase! _{____________________________________}_{_________________________________________}_{_____}

____________________________________ ______________________________________________ 1.6.4 1.6.4 1.5.6 1.5.6 R R P P P Pii = 30 N/cm = 30 N/cm22 = 300,000 N/m = 300,000 N/m22 = 3 bar = 3 bar

Δ

P = 30 bar – 3 bar P = 30 bar – 3 bar = 27 bar = 27x10= 27 bar = 27x1055 N/m N/m22 Amount of water that enters pipe =

Amount of water that enters pipe =

Δ

VolVol Vol

Vol pipe pipe= [(= [(

ππ

)(1.50 m))(1.50 m)22/(4)]·(2000 m) = 3530 m/(4)]·(2000 m) = 3530 m33

Δ

Vol = (-Vol = (-

Δ

P/EP/E b b)(Vol) = [(-27x10)(Vol) = [(-27x10 5 5

N/m N/m22)/)/

Δ

P = PP = Pii – P – Pee (internal pressure minus (internal pressure minus external pressure)external pressure)

(2.2x10 (2.2x1099 N/m N/m22)]*(3530 m)]*(3530 m33))

Δ

Vol = -4.33 mVol = -4.33 m33

∑

FFxx= = 0; 0; 22

ππ

(R)((R)(

σ

) –) –

Δ

P(P(

ππ

)(R )(R 22) = 0) = 0

Water in the pipe is

Water in the pipe is compressed by this amount.compressed by this amount. P = 2

P = 2

The volume of H

The volume of H22O that enters the pipe is 4.33 mO that enters the pipe is 4.33 m33

/R /R