ch4

4-1 If atmospheric pressure is 14.7 lbf/in.2, what is the pressure at a depth of 10 ft of water? Approach:

Use the equation for pressure as a function of depth in an incompressible fluid: + P = atmP .ghρ

Assumptions:

1. The density of the water is constant. 2. The water is at room temperature (70ºF).

Solution:

At an assumed temperature of 70ºF, the density of water (from Table B-6) is 62.2

2H Oρ = 3lbm/ftThe pressure in an incompressible fluid as a function of depth is

+ P = atmP ghρ

= 14.7 2lbf in. + ( )( )( )2

3 22 2

1lbf 1ft62.2 lbm/ft 32.2ft/s 10ft32.2lbm ft/s 144in.

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟⋅⎝ ⎠⎝ ⎠⎣ ⎦

P = 19.0 2lbf in. Answer 4-2 Hoover dam stands at a height of 725 ft above the Colorado river. Assuming atmospheric pressure is 14.7

lbf/in.2, calculate the pressure in the reservoir at the base of the dam. Approach:

Use the equation for pressure as a function of depth in an incompressible fluid: + P = atmP .ghρ

Assumptions:

1. The density of the water is constant. 2. The water is at room temperature

(70ºF).

Solution:

At an assumed temperature of 70ºF, the density of water (from Table B-6) is 62.2

2H Oρ = 3lbm/ftThe pressure in an incompressible fluid as a function of depth is

+ P = atmP ghρ

= ( )2

2 3 2

2

lbf lbm ft 1 lbf 1ft14.7 62.2 32.2 725ftlbm ftin. ft s 144in.32.2

s

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠ ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

2

= 2

lbf328in.

Answer

4 - 1

4-3 In a manometer containing liquid mercury, the height is read as 6 in. If atmospheric pressure is 100 kPa, what pressure is the manometer reading (in kPa)?

Approach:

Use the equation for pressure as a function of depth in an incompressible fluid: +P = atmP .ghρ

Assumptions:

1. The density of the mercury is constant. 2. The mercury is at room temperature (20ºC).

Solution:

At an assumed temperature of 20ºC, the density of mercury (from Table B-6) is Hgρ = 13,579 3kg/m

The pressure in an incompressible fluid as a function of depth is P = atmP + Hg ghρ

P = 100 + kPa ( )( )( )3 2 0.3048 m 1 kPa13,579kg/m 9.8m/s 6in.12 in. 1000 Pa

⎡ ⎤⎛ ⎞⎛⎢ ⎥⎜ ⎟⎜

⎝ ⎠⎝⎣ ⎦

⎞⎟⎠

120kPaP = Answer

4-4 A vat in a chemical processing plant contains liquid ethylene glycol at 20oC. The air space at the top of the closed vat is maintained at 110 kPa. If the depth of the liquid is 0.8 m, what is the pressure at the bottom of the tank?

Approach:

Use the equation for pressure as a function of depth in an incompressible fluid: + P = P∞ .ghρ

Assumptions:

1. The density of the ethylene glycol is constant.

Solution:

At a temperature of 20ºC, the density of water (from Table B-6) is 31117 kg mρ =

The pressure in an incompressible fluid as a function of depth is P = + P∞ ghρ

( )3 2

1000 Pa kg m110 kPa 1117 9.81 0.8 m1 kPa m s

P ⎛ ⎞ ⎛ ⎞⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

118,766 Pa 119 kPaP = = Answer

4 - 2

4-5 At great ocean depths, the hydrostatic pressure is very high. Suppose that seawater density varies with pressure according to

1 2lno

P C Cρρ

⎛ ⎞= +⎜ ⎟

⎝ ⎠

where C1 = 2.24 x 109 Pa, C2 = 1 x 105 Pa, and oρ =1024 kg/m3. Assume this relation holds at any depth, z, and use it to find the pressure at a depth of 3000 m. What would the pressure be if seawater density were assumed to be constant at 1024 kg/m3? Assume atmospheric pressure is 1 x 105 Pa.

Approach: Solve the given equation for density as a function of pressure. Substitute into Eq. 4-7, separate variables, and integrate.

Assumptions:

1. Water density is only a function of pressure.

Solution:

In a compressible fluid, static pressure is related to depth by

ddzP gρ= −

Rearrange the function given in the problem statement to get

2

1

exp P CC

ρ ρ⎧ ⎫−

= ⎨ ⎬⎩ ⎭

Substituting yields

2

1

d expdz

P CP gC

ρ⎧ ⎫−

= − ⎨ ⎬⎩ ⎭

Integrate between the two depths shown in the figure:

( )2 2

1 12

1

d

exp

P z

P z

P g dzP C

C

ρ= −⎧ ⎫−⎨ ⎬⎩ ⎭

∫ ∫

Integrating the right hand side and rearranging the left produces:

1

2 22 1

1

exp ( )P

P

C PdP g z z

Cρ

⎛ ⎞−= − −⎜ ⎟

⎝ ⎠∫

To make further progress, define 2

1

C PuC−

= , so that 1

d dPuC

= − , and

2

11 2exp( ) ( )

u

uC u du g z zρ− = −∫ 1−

1zIntegrating the left-hand side 2 1

1 2(e e ) ( )u uC g zρ− = −Use the known boundary condition to evaluate u2

at 2z z= 52 1 10 PaP P= = ×

∴ 5 5

2 22

1 1

1 10 Pa 1 10 Pa 0C PuC C− × − ×

= = =

1z

Thus 1

1 2(1 e ) ( )uC g zρ− = −

4 - 3

Solving for u1 produces

2 11

1

( )ln 1 g z zuC

ρ⎡ ⎤−= −⎢ ⎥

⎣ ⎦

3 2

9

kg m1024 9.81 (3000m)m sln 1

2.24 10 Pa

⎡ ⎤⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎢ ⎥= −

×⎢ ⎥⎢ ⎥⎣ ⎦

0.0136= −

We may now calculate the unknown pressure at station 1 through

2 11

1

C P uC−

=

Solving for P1

9 51 1 2 (2.24 10 )( 0.0135) 1 10P Cu C= − + = − × − + ×

71 3.044 10 PaP = × Answer

If density is constant at 2

kg1024m

*1 2P ghρ= + P

53 2

kg m1024 (9.81) (3000m) 1 10 Pam s

⎛ ⎞= +⎜ ⎟⎝ ⎠

×

73.024 10 Pa= × Answer

Comments: There is very little difference between the two calculations. The compressibility of water due to pressure change is not an important factor in determining hydrostatic pressure in the ocean. Note that in reality, density varies with depth due to changes in temperature and salinity. These effects are generally more significant than the change of density with pressure.

4 - 4

4-6 A large tank contains a liquid solution whose density varies with depth as shown in the table below. A gas space at the top of the tank contains air at 60 psia. Find the pressure at a depth of 30 ft using numerical integration.

depth, ft density, lbm/ft3

0 40.2 5 41.0

10 42.7 15 44.9 20 47.7 25 50.9

Approach: Integrate Eq. 4-7 from the surface of the liquid to the desired depth of 30 ft. For simplicity, use the trapezoidal rule.

Assumptions: 1. Density varies linearly with depth between given

data points.

30 54.6 Solution:

For a variable density fluid

d ( )dzP z gρ= −

First find the pressure at 5 ft by numerical integration.

1 1

2( )

P z

P zdP z gdzρ= −∫ ∫

Use: 2

lbf60in

ρ = 0z = 1 5ftz = −

Approximating the integral with the trapezoid rule,

1 1( )2

P P g z zρ ρ+⎛ ⎞− = − −⎜ ⎟⎝ ⎠

Solving for P1 and substituting values

11 1( )

2P g z z Pρ ρ+⎛ ⎞= − − +⎜ ⎟

⎝ ⎠

2

3 2 2

2

40.2 41.0 lbm ft 1lbf 1ft lbf32.2 ( 5 0)ft 60lbm ft2 ft s 144in i32.2

s

⎛ ⎞⎜ ⎟⎛ ⎞+⎛ ⎞ ⎛ ⎞− − − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

2n+

Now find the pressure at depth 2

1 22 2 1 1z P( )

2P g zρ ρ+⎛ ⎞= − − +⎜ ⎟

⎝ ⎠[ ]2

41.0 42.7 ft 1 132.2 10 ( 5) 61.412 s 32.2 144+⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − − − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠62.86=

Continue this process using

11 1( )

2i i

i iP g zρ ρ +

+ +

+⎛ ⎞= − − +⎜ ⎟⎝ ⎠

i iz P

until you find P6. Intermediate values of pressure are given in the table below. A wide variety of software tools can be used to automate this calculation.

depth, ft density 3lbm/ft pressure, psia 0 40.2 60 5 41.0 61.4

10 42.7 62.9 15 44.9 64.4 20 47.7 66.0 25 50.9 67.7 30 54.5 69.5

The final result is 6 2

lbf69.5in.

P = Answer

4 - 5

4-7 A manometer is attached to a rigid tank containing gas at pressure P. The manometer fluid is mercury at 20°C. Using data on the figure below, find the pressure in the tank.

Approach:

Use the equation for pressure as a function of depth in an incompressible fluid:

P = atmP + .ghρ Assumptions:

1. The density of the mercury is constant.

Patm = 101 kPa Solution:

At an assumed temperature of 20ºC, the density of mercury (from Table B-6) is Hgρ = 13,579 3kg/m

The pressure in an incompressible fluid as a function of depth is P = atmP + Hg ghρ

( )3 2

kg m 1 m 1 kPa101kPa 13,579 9.8 16 3 cmm s 100cm 1000Pa

⎛ ⎞⎛⎛ ⎞⎛ ⎞= + − ⎜ ⎟⎜⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝

⎞⎟⎠

P = 118 kPa Answer

4-8 Write an equation for the mass of the piston, pm , in terms of

2H Oρ , l , θ , and pA . See the figure below. Approach:

Perform a force balance on the piston. Use the equation for pressure as a function of depth in an incompressible fluid to find the gas pressure.

Assumptions:

1. The density of the water is constant. Solution:

A force balance on the piston produces atm p p gas pP A m g P A+ =

The gas pressure is given by singas H O atmP g lρ θ

2= + P

Substituting the second equation into the first ( )

2sinatm p p H O atm pP A m g g l P Aρ θ+ = +

Solving for piston mass 2

sinp H O pm l A Answer ρ θ=

4 - 6

4-9 Liquid water is contained in a piston-cylinder assembly as shown in the figure below. An inclined manometer filled with water is attached to the bottom of the cylinder. Using data given on the figure, calculate the force exerted by the spring on the piston.

Approach:

Calculate the pressure at the bottom of the tank starting at the spring and working your way downward. Derive a second equation for pressure at the bottom of the tank by considering the pressure in the inclined section. Equate the two expressions for pressure and solve for spring force.

Assumptions:

1. The density of the water is constant. 2. The water is at 70ºF.

Solution:

The pressure at the bottom of the tank is: pistons

b atmpiston piston

m gFP P gh

A Aρ= + + +

where Fs is the force exerted by the spring. Considering the inclined section, the pressure at the bottom may also be written

( )sinb aP gs Pρ θ= + tm Equating these two expressions

( )sin pistons

piston piston

m gFgs g

A Ahρ θ ρ= + +

Solving for spring force,

( )sin pistons piston

piston

m gF gs gh A

Aρ θ ρ

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠

Using the density of water at 70ºF from Table B-6,

( )( )( )( ) ( )

( )( ) ( )( )( ) ( )

23 2 2

2

22 3 2 2

2

1ft 1ft62.2 lbm ft 32.17 ft s 5.5 in. sin45° 30in.12in. 144in.

1ft 1ft0.5 lbm 32.17 ft s 62.2 lbm ft 32.17 ft s 2in. 30in.12in. 144in.

sF⎡ ⎤⎛ ⎞⎛ ⎞

= ⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞− − ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

249.5lbm ft s 1.54 lbfsF = ⋅ = Answer

4 - 7

4-10 A tank contains air at 80oF. A manometer connected to the tank contains liquid mercury, also at 80oF. Assuming atmospheric pressure is 14.2 psia, and using data on the figure below, calculate the density of the air in the tank.

Approach: Calculate the pressure in the tank using the manometer relations. Then use the ideal gas law to find the air density.

Assumptions:

1. The density of the mercury is constant. 2. Air may be considered as an ideal gas.

Solution:

First find the pressure of the air using

atm HgP P ghρ= + Using data for the density of mercury in Table B-6:

( )3 2

32 3

2 3

lbm ft845 32.17 15.4 2.1 in.lbf ft s14.2 20.7 psiain. lbm ft 12 in.32.17

lbf s 1 ft

P

⎛ ⎞⎛ ⎞ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= + =

⋅⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⋅⎝ ⎠⎝ ⎠

From the ideal gas law:

( )

( )3

lbm20.7 psia 28.97lbmol

psia ft10.73 80 460 Rlbmol R

PMRT

ρ

⎛ ⎞⎜ ⎟⎝ ⎠= =

⎛ ⎞⋅+⎜ ⎟⋅⎝ ⎠

30.104 lbm/ft= Answer

4-11 Two piston-cylinder assemblies are connected by a tube, as shown below. The diameter of each cylinder is 8 cm and the mass of each piston is 0.4 kg. A mass rests on top of each piston. The fluid in the tube is mercury, at 20°C. Using data on the figure below, calculate the unknown mass m2.

Approach: Calculate the pressure at the lowest point in the connecting tube by two paths: one through the right cylinder and the other through the left cylinder. Equate these pressures and solve for the unknown mass.

Assumptions:

1. The density of the mercury is constant. Solution:

The pressure at the lowest point in the tube may be calculated using either the right side or the left side. This produces:

3 31 21 22 2 2 2atm atm

m g m gm g m gP gh PR R R R

ρ ρπ π π π

+ + + = + + + gh

m1 = 5 kg

h1 = 1.5 cm h2 = 4.5 cm

where m3 = piston mass, and R = cylinder radius. Simplifying

1 21 22 2

m mh hR R

ρ ρπ π

+ = + → ( ) 22 1 1 2m m h h Rρ π= + −

Using the density of mercury from Table A-6 at 20°C,

( ) ( )2

22 3 2

kg 1 m 1 m5 kg 13,579 1.5 4.5 cm 4 cmm 100 cm 10,000 cm

m π⎛ ⎞⎛ ⎞

= + − ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2.95 kgm = Answer

4 - 8

4-12 In the device shown below, calculate the gage pressure of the gas in the tank.

Approach: Use the expression for variation of pressure with depth in an incompressible fluid.

Assumptions:

1. The densities of the oil and water are constant. Solution:

Pressure in a static fluid is only a function of horizontal location, therefore

A BP P= 1 1 2 2B atmP P h g h gρ ρ= + +

The gage pressure is given by 1 1 2 2g B atmP P P h g h gρ ρ= − = +

( )( ) ( )( )2

2

3 33

32.2 ft/s1ft 32.2 132.2 lbm ft/s3in 49 lbm/ft 8.6 2.2 62.2

1lbf 12in. 32.2 12gP

⎛ ⎞⎜ ⎟⎛ ⎞⋅ ⎛ ⎞⎛ ⎞⎜ ⎟= + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

0.316psiggP = Answer

4 - 9

4-13 In the manometer shown below, 2 g of oil and 11 g of water are introduced. The oil has a density of 620 kg/m3. Find the length l, to which the water rises in the inclined section.

Approach: Use the given mass and density of the oil and the diameter of the tube to determine the length l2. Similarly for the water, determine the sum of l1 and l. Calculate the pressure at the lowest point in the tube by two paths: one through the inclined section and the other through the straight section. Equate these pressures and solve for the unknown lengths.

Assumptions:

1. The densities of the water and oil are constant. 2. The system is at room temperature.

Solution:

The mass of the oil is given by: 2

0 0 2Dm lρ π

⎡ ⎤⎛ ⎞= ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2

Rearranging

( )0

2 2 2

0 3

1 kg2 gm1000 g

kg 1 m620 π 0.5 cm2 m 100 cm

0.041 mm

lDρ π

⎛ ⎞⎜ ⎟⎝ ⎠= =

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

The volume of water is given by

( )2

12wDV lπ ⎛ ⎞= +⎜ ⎟

⎝ ⎠l

Using the definition of density,

( )2

12

w ww

w

m mV D l l

ρπ

= =⎛ ⎞ +⎜ ⎟⎝ ⎠

Solving for the sum of the lengths and using the density of water at 20°C from Table A-6,

( )( )

( )1 2 2

3

1 kg11gm1000 g

0.141mkg 1 m998.2 π 0.5 cm2 m 100 cm

w

w

ml l

Dρ π

⎛ ⎞⎜ ⎟⎝ ⎠+ = = =

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Define a new variable 3 1 0.141ml l l= + =

The pressure at the bottom of the straight section is the same as the pressure at the bottom of the inclined section. Therefore:

0 2 1 sinatm w atm wP gl gl P glρ ρ ρ θ+ + = + 0 2 1 sinw wl l lρ ρ ρ θ+ =

Since 1 3l l l= −

( )0 2 3 sinw wl l l lρ ρ ρ+ − = θ Rearranging

[ ]0 2 3 sinw wl l l wρ ρ ρ θ ρ+ = + 0 2 3

sinw

w w

l ll

ρ ρρ θ ρ

+=

+

Substituting values

4 - 10

( ) ( )3 3

3 3

kg kg620 0.041m + 998.2 0.141mm m

kg kg998.2 sin40° 998.2m m

l

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

⎛ ⎞ +⎜ ⎟⎝ ⎠

0.101 ml = 10.1cm= Answer

4 - 11

4-14 A manometer connects a large water tank open to the atmosphere to a closed spherical tank of air. The manometer contains both oil and water. Using data on the figure, find the gage pressure of the air in tank A.

Approach: Use the equation for pressure as a function of depth in an incompressible fluid to trace pressure through the system.

Assumptions:

1. The densities of the water and oil are constant. 2. The system is at room temperature.

Solution:

The gage pressure at point 1 is ( )1 1 1atm atmP P gh Pρ= + −

( )3 2

kg m 1 m999.5 9.8 28 11 cmm s 100 cm

⎛ ⎞⎛ ⎞ ⎛ ⎞⎟= −⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠⎝ ⎠ ⎝

1P

⎠

1665 Pa=At point 2,

2 2 2P ghρ+ =

2 1 2P P gh2ρ= −

( ) ( )

2

2 2

1665Pakg m 1 m999.5 0.7 9.8 14 8 cmm s 100 cm

P =

⎛ ⎞ ⎛ ⎞ ⎛− −⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝

⎞⎟⎠

1254 Pa= Finally, relate P2 to P3 using

3 2 1 3P P ghρ= +

( )( )( )311254 Pa 999.5 9.8 24 9

100P ⎛ ⎞= + − ⎜ ⎟

⎝ ⎠

3 2.72 kPaP = Answer

4 - 12

4-15 The manometer in the figure is designed to measure small changes in pressure. Using the data in the table, a. Determine the initial gage pressure of the gas in the sphere. b. The pressure is increased so that h2 becomes 2.0 cm. During this process, none of the liquid

interfaces change in diameter. Find the final gage pressure of the gas.

Unchanged quantities

Initial value Final value

d 1 cm h1 9.1 cm D 9 cm h2 2.2 cm 2.0 cm

SG1 0.86 h3 4.6 cm SG2 11.3 h4 10.7 cm

Approach: Use the equation for pressure as a function of depth in an incompressible fluid to find the initial gage pressure. For the second part, note that the volume of each liquid must remain constant.

Assumptions:

1. The densities of the two liquids are constant.

Solution:

a) From the left side of the manometer, the pressure at point 2 is

( )2 1 1gasP P g h hρ= + − 2

From the right side of the manometer, the pressure at point 2 is ( ) (2 4 1 4 3 2 3 2P P g h h g h hρ ρ= + − + − )

Equating the last two equations ( ) ( ) ( )1 1 2 4 1 4 3 2 3 2gasP g h h P g h h g h hρ ρ ρ+ − = + − + −

( ) ( ) ( )4 1 4 3 2 3 2 1 1 2gasP P g h h g h h g h hρ ρ ρ− = − + − − −

( ) ( )

( )

3 2 3 2

3 2

kg m 1m kg m 1m0.86 1000 9.81 10.7 4.6 cm 11.3 1000 9.81 4.6 2.2 cmm s 100cm m s 100c

kg m 1m0.86 1000 9.81 9.1 2.2 cmm s 100cm

gageP⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞− − ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠2590Pa 2.59kPagageP = =

m

h

b) The change in height h2 is The volume of each fluid must remain constant. Therefore

2 2.0 2.2 0.2cm.h∆ = − = −

( ) ( )2 22 1/ 2 / 2d h Dπ π∆ = ∆

( ) ( )( )

222

1 22

1cm 0.2cm0.00247cm

9cmd hh

D−∆

∆ = = = −

The height h3 must increase by the same amount as h2 decreases, so 3 0.2cm.h∆ = Similarly, With these considerations

4 0.00247cm.h∆ =

1 2 3 49.1 0.00247 9.098 2.0 4.6 0.2 4.8 10.7 0.00247 10.7025h h h h= − = = = + = = + = Recomputing the gage pressure with the above values gives

3003 Pa 3.00kPagageP = = Answer

4 - 13

4-16 A glass tube containing oil is inserted into a tank of water, as shown in the figure below. Using data on the figure, calculate the oil density. Assume the temperature is 20oC.

Approach: Use the equation for pressure as a function of depth in an incompressible fluid to trace pressure through the system.

Assumptions:

1. The densities of the water and oil are constant. 2. The system is at room temperature.

Solution:

The pressure at the bottom of the oil must equal the pressure at the same depth of water; that is, at L4.

23 4atm oil atm H OP gL Pρ ρ+ = + gL

L L L L− + =

4 2 3 1L L L L∴ = + − 2 4 3 1

Substituting L4 and rearranging

( )

2 2 3 1

3

H Ooil

L L LL

ρρ

+ −=

Using the density of water from Table A-6 at 20ºC

L1 = 11.5 cm L2 = 9.6 cm L3 = 4.45 cm

( )3

3

kg998.2 9.6 4.45 11.5 cmkgm 572

4.45cm moilρ

⎛ ⎞ + −⎜ ⎟⎝ ⎠= = Answer

4 - 14

4-17 A hydraulic lift is used to raise a crate, as shown in the figure below. The large tube has a diameter of 3 ft and the small tube has a 1 in. diameter. The fluid is oil with a specific gravity of 0.86. The plate under the crate has negligible mass. Initially the system is at rest with the oil heights of 1.4 ft and 6 in., as shown.

a. Find the mass of the crate, in lbm. b. Oil is poured into the small tube until the crate rises 1 in. Calculate the volume of oil added.

Approach:

Use the equation for pressure as a function of depth in an incompressible fluid on the left-hand tube to find the pressure just under the plate. Also calculate this pressure based on the crate mass and the pressure above the crate. For the second part, recognize that the total volume of the oil remains constant.

Assumptions: 1. The density of the oil is constant. 2. The system is at room temperature.

Solution:

a) Consider the pressure just under the plate. The following relation holds:

2 11

( ) atm atmmgg h h P PA

ρ − + = +

where h1 is the height of oil in the large tube, h2 is the height of oil in the small tube, and A1 is the area of the plate. Solving for m, the mass of the crate, gives

2

21 2 1 1 2 1 3

lbm( ) 0.86 ( ) 0.86 62.4 (1.5ft) (1.4 0.5)ft = 34l lbmftH Om A h h A h hρ ρ π⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠

b) Let be the new height of oil in the large tube and be the new height in the small tube. From part a, 1 *h 2 *h

2 11

( * *)mg g h hA

ρ= −

Solving for 2 *h

2 12 21

3

12in.341 lbm1ft

* 7 in. 17.8 in.lbm62.4 (0.86) (1.5) ftft

mh hAρ π

⎛ ⎞⎜ ⎟⎝ ⎠= + = + =

⎛ ⎞⎜ ⎟⎝ ⎠

The initial height h2 was 1.4 ft. or 16.8 in. So, the height in each tube has risen by 1 in. Recognizing this, the volume of oil added must be

2 2 3(0.5in.) (1 in.) (18in.) (1 in.)=1019 in.V π π= + Answer

4 - 15

4-18 An underwater gate 8 m wide is held closed by a force, F, as shown in the figure. If the force is applied at the middle of the gate, what is the minimum value required to keep the gate closed?

Approach: Find the resultant hydrostatic force due to the water and its point of application. Take moments about the hinge to determine the unknown force.

Assumptions:

1. The water density is constant.

Solution:

The resultant force, RF , is given by sinR cF g y Aρ θ=

The location of the centroid of the gate is

21 2c

ay a= +

23 3.46m

sin 60a = =

2m 3.46msin(60) 2cy = + 4.04m=

Using the density of water at room temperature (see Table A-6)

3 2

kg m996 9.81 sin(60)(4.04m)(3.46 m)(8 m)m sRF ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

59.48 10 NRF = ×The point of application, ,py is

,xx cp c

c

Iy y

y A= +

The moment of inertia for a rectangular plate

3

4 2, 12xx c

a aI =

3 2

4 2 2

4 212 12p c cc c

a a ay y yy a a y

= + = +

2(3.46m)4.04m

12(4.04)m= + 4.29m=

To find F, take moments about the hinge 3 1 2 pa a a y= + −

32 3.46 4.29

sin(60)a = + − 1.48m=

232 R

aF F a⎛ ⎞ =⎜ ⎟⎝ ⎠

3

2

2 RF aF

a=

52(9.48 10 )N(1.48m)3.46m

×=

Front view of gate:

Moment diagram:

58.12 10 NF = × Answer

4 - 16

4-19 A horizontal pipe 1.4 m in diameter is half filled with liquid oxygen (SG=1.18). The gas above the liquid is at a pressure of 250 kPa. The pipe is closed on both ends by vertical, flat surfaces. Find the magnitude of the resultant force of the fluid and gas acting on one of the end surfaces.

Approach:

The total force on the end plate is the sum of the force due to the gas and the liquid. The gas force is just pressure times area while the liquid force is found using hydrostatic relations.

Assumptions:

1. The oxygen density is constant.

Solution:

The area, A, of half of the end surface is

2

2RA π

=2

2(1.4m) 0.770m2

π=

The force exerted by the gas is

( ) 21

1000Pa250kPa (0.77 m )1 kPa

F PA ⎛ ⎞= = ⎜ ⎟⎝ ⎠

51.92 10 N= ×The force exerted by the liquid is 2 sin cF PA g y Aρ θ= + where

4 (4)(0.7 m) 0.297 m3 3c

Ryπ π

= = =

Substituting: 2 2

2 3 2

1000 Pa kg m(250kPa) (0.77m ) 1000 (1.18) 9.81 (sin90 )(0.297 m)(0.77m )1 kPa m s

F ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

52 1.95 10 NF = ×

The total force is 5

1 2 3.87 10 NF F F= + = × Answer

4 - 17

4-20 A gravity dam made of concrete ( ρ = 2200 kg/m3) holds back water which is 5.5 m deep, as shown. The bottom of the dam rests on the soil and is held in place by friction. Calculate the minimum coefficient of friction between the dam and the soil so that the dam does not slide.

Approach:

Find the resultant hydrostatic force on the dam. Use a free-body diagram of the dam to do a force balance including the dam’s weight, the frictional force and the hydrostatic force. The hydrostatic force in the horizontal direction equals the frictional force at the minimum value of friction coefficient.

Assumptions:

1. The water density is constant.

Solution:

The resultant force, RF , is given by sinR cF g y Aρ θ=

2

3 1

7.7tan9.1 3.1

aa a

θ = =− −

51.3θ =

( )

1o

5.5 6.97 msin sin 51.3

hbθ

= = =

Using the density of water at an assumed temperature of o25 C

3 2

kg m 6.97 m997 9.81 sin(51.3) (6.97 m)(1m)m s 2RF ⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

where the force on a 1m deep section of dam has been considered. Evaluating: 51.88 10 N=188 kNRF = ×The resultant force may be resolved into horizontal and vertical components , sinR x RF F θ= (188kN)sin(51.3 )= 148kN=

, cosR z RF F θ= (188kN)cos(51.3 )= 115kN= The volume of the dam is

1 2 2 3 11 ( ) (12

V a a a a a⎡ ⎤= + −⎢ ⎥⎣ ⎦m)

(3.1)(7.7) (0.5)(7.7)(9.1 3.1)= + − 347.0m=A free body diagram of the dam shows: ,( )f R zF F mgµ= + The dam will not slide if the frictional force equals the horizontal force:

,f R xF F= ,

,

R x

R z c

FF Vg

µρ

=+

( )33 2

1000 N(148kN)1kN

1000 N kg m(115kN) 2200 47 m 9.811 kN m s

µ

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎟⎠

0.131+⎜ ⎟ ⎜ ⎟ ⎜

⎝ ⎠ ⎝ ⎠ ⎝

=

Free body diagram:

Answer

4 - 18

4-21 A conduit leading from a reservoir is closed by a square gate pivoted along its midline, as shown. Calculate the force of the gate on the stop that holds it closed.

Approach:

Find the resultant hydrostatic force due to the water and its point of application. Take moments on the gate about the stop to determine the unknown force.

Assumptions:

1. The water density is constant. 2. the system is at room temperature.

Solution:

For a rectangular plate, the centroid is at the center, so 18ftcy =The resultant force is sinR cF g y Aρ θ=

23 2

2

lbm ft 1lbf62.4 32.2 sin(90 )(18ft)(3)(3)ftlbm ftft s 32.2

s

RF

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

41.01 10 lbf= ×

This force is applied at

1xx cP c

c

Iy y

y A= +

4

212cc

ayy a

= +

2318

(12)(18)= +

18.0417 ft=We have used a large number of significant digits because we will need them in the next step. Taking moments about the pivot, (1.5) ( )R P cF F y y= − Solving for F,

( )

1.5R P cF y y

F−

=

4(1.01 10 lbf)(18.0417 18)ft

1.5ft-F ×

=

281 lbfF = Answer

4 - 19

4-22 A square plate, called a paddle, covers a passage in a canal lock, as shown. The angle, α, is 15o. Find the vertical force, F, needed to open the paddle.

Approach:

Find the resultant hydrostatic force due to the water and its point of application on both sides of the paddle. Take moments about the bottom of the paddle to determine the unknown force.

Assumptions:

1. The water density is constant. 2. The system is at room temperature.

Solution: We need the resultant force on the paddle. Referring to the figure below,

140 ftcos

s bα

+ =

140 ftcos

s bα

= −( )40 2 39.4 ft

cos 15= − =

°

Atmospheric pressure acts on both side of the lock, and its effects on the paddle will cancel. For simplicity, we set Patm = 0. The centroid is at the middle of the paddle; therefore, the resultant force on the upstream side is

1 1 sin2R atmb

1F P g s aρ⎡ ⎛ ⎞= + +⎜ ⎟⎢ ⎝ ⎠⎣ ⎦bθ ⎤

⎥ where o o1 90 75θ α= − =

( )( )( ) 21 3 2

2

lbm ft 2 1 lbf0 62.4 32.2 39.4 ft sin 75 2 2 ftlbm ftft s 2 32.2

s

RF

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞⎛ ⎞= + + ° ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠⎝ ⎠ ⎜ ⎟⎜ ⎟⎝ ⎠

9740 lbf=

Now find the resultant force on the downriver side of the paddle. In the figure to the right,

o2 1180θ θ= − o o180 75= − 115= °

( )( )2 o

40 34 ft2 ft 4.21 ft

cos 15s

−= − =

2 2 sin2R atmb

2F P g s aρ θ⎡ ⎤⎛ ⎞= + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( )( )

b

( ) ( ) ( )22 162.4 32.2 4.21 sin 115 2 22 3RF ⎛ ⎞ ⎛= + °⎜ ⎟ ⎜

⎝ ⎠ ⎝ 2.2⎞⎟⎠

note 2 1179 lbfRF = 1 2sin sinθ θ= Now find the point of application of the force on the upriver side (FR1)

2

1 1

11

212

2 sin

patm

b by sPbs

gρ θ

= + +⎡ ⎤

+ +⎢ ⎥⎣ ⎦

( )2 2

1

2 ft239.4 ft ft22 12 39.4 0 ft2

py = + +⎡ ⎤+ +⎢ ⎥⎣ ⎦

40.42 ft=

4 - 20

Next find the point of application of the force on the downriver side (FR2)

2

2 2

22

212

2 sin

patm

b by sPbs

gρ θ

= + +⎡ ⎤

+ +⎢ ⎥⎣ ⎦

( )2 22 ft24.21 ft ft22 12 4.21 0 ft2

= + +⎡ ⎤+ +⎢ ⎥⎣ ⎦

5.28 ft=

Define the distance x1 as 1 1 1px y s= −

p

40.42 39.4= − 1.004 ft= Similarly define x2 as

2 2 2x y s= − 5.28 4.21= − 1.064 ft= Take moments about point A ( ) (2 2 1n R R )1F b F b x F b x+ − = −

( ) ( ) ( ) ( )1 1 2 2 9740 2 1.004 1179 2 1.0642

R Rn

F b x F b xF

b− − − − − −

= =

4278 lbf=To get the force in the vertical direction

1cos nFF

θ =

( )1

4278 16,529 lbfcos cos 75

nFF

θ= = = Answer

4 - 21

4-23 A cylinder 5 m long and 4 m in diameter is wedged into a rectangular opening in the bottom of a tank of water. The cylinder seals the opening, which is also 5 m long. The center of the cylinder is 1 m above the floor of the tank and the water depth is 8 m. Find the net force of the water on the cylinder.

Approach: Because of symmetry, the only net force on the cylinder is in the vertical direction. The weight of the water above the cylinder provides a downward force, while the water under the cylinder provides an upward force. Atmospheric pressure cancels out and need not be considered.

Assumptions:

1. The water density is constant.

Solution: By symmetry, there are no horizontal forces. Forces due to atmosphere pressure will cancel and are not included. The force vF is

1vF ghAρ=

1( )(2 )g h R RLρ= − 2 53 2

kg m997 9.8 (7 2)m (2)(2)(5)m 9.78 10 Nm s

⎛ ⎞⎛ ⎞= − = ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

where , the length of the cylinder. The mass of water, , in the region above one half of the cylinder is

5mL =

1m

2 21

14

m R Rρ π⎛ ⎞= −⎜ ⎟⎝ ⎠

L

( )2

2 2 2 4279kg1 3

kg (2)997 2 m 5mm 4

m π⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠ ⎣ ⎦

=

The total downward force is 1 1 12m gVC VF F= + 59.78 10 2(4279)(997)= × +

61.06 10 N= ×The force 2VF due to the water pressure at the bottom of the tank is

2VF ghAρ= 1 2 4( )(2 )g h h hρ= + To find , note 4h

1 12 1sin sin 302

hR

θ − −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( )3 cos 2cos 30 1.73mh R θ= = =

4 3h R h= − 2 1.73= − 0.268m= 2 1 2( )2V 4F g h h hρ= + 5(997)(9.81)(7 1)2(0.268) 2.10 10 N= + = ×

The mass of the water in the region below one half of the cylinder is 2m

22 2 2 3

12 360

m h R h h R Lθρ π⎡ ⎤= − −⎢ ⎥⎣ ⎦

21 30997 (1)(2) (1)(1.732) (2) 52 360

π⎡ ⎤= − −⎢ ⎥⎣ ⎦432.6 kg=

The net upward force is 2 2 22VC VF F m g 52.1 10 2(433)(997)= × −= − 52.01 10 N= ×

The net force which pushes the cylinder downward into the hole is 1 2VC VCF F F= − 6 51.06 10 2.01 10= × − × 58.61 10 N= × 861 kN= Answer

4 - 22

4-24 A semi-circular gate hinged at the bottom holds back a tank of water 4 ft deep. If the gate is 15 ft wide, what force, F, is required to keep the gate closed?

Approach: Find the resultant hydrostatic force due to the water and its point of application. Take moments about the hinge to determine the unknown force.

Assumptions:

1. The water density is constant.

Solution:

Atmosphere forces cancel and will not be considered.

The horizontal force is sinH cF g y Aρ θ=

2

sin(90 )2 2HR RF g RL g Lρ ρ⎛ ⎞= =⎜ ⎟

⎝ ⎠

23

3 2

2

lbm ft 1lbf 462.4 32.2 (15)ftlbm ftft s 232.2

s

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠ ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

7488lbf=The point of application of the force is:

,xx CP C

C

Iy y

y A= +

3

2 2 6R

12 ( )2

R LR RR RL

= + = +⎛ ⎞⎜ ⎟⎝ ⎠

4 4 2.667 ft2 6

= + =

The mass of water in the quarter circle is

2 2

23

lbm (4)62.4 ft (15ft)4 ft 4Rm Lπ πρ

⎛ ⎞ ⎡ ⎤⎛ ⎞= =⎜ ⎟ ⎢ ⎥⎜ ⎟⎝ ⎠⎝ ⎠ ⎣ ⎦

11762lbm=

Taking moments about the hinge

( )43 H P

RFR mg F R yπ

⎛ ⎞= + −⎜ ⎟⎝ ⎠

43

PH

R yF mg FRπ−⎛ ⎞= + ⎜ ⎟

⎝ ⎠

( ) 2

2

ft 1lbf 4 4 2.66711762lbm 32.2 (7488lbf)lbm fts 332.2

s

Fπ

⎛ ⎞⎜ ⎟ −⎛ ⎞ ⎛= +⎜ ⎟⎜ ⎟ ⎜⋅⎝ ⎠ ⎝⎜ ⎟⎜ ⎟⎝ ⎠

4⎞⎟⎠

7488lbf= Answer

4 - 23

4-25 A hemisphere filled with oil (SG = 0.72) is inverted on a flat surface. A narrow tube partially filled with oil protrudes from the top of the hemisphere, which has a mass of 5 kg. At what height, h, will the hemisphere lift off the surface? Neglect the weight of the tube and the oil it contains.

Approach:

Balance the upward hydrostatic force against the weight of the oil and the container. When the hydrostatic force is greater, the hemisphere will lift.

Assumptions:

1. The oil density is constant 2. The weight of the oil in the tube is small. 3. the weight of the tube is small.

Solution: Atmospheric pressure forces cancel out and will not be considered. There is an upward force due to static pressure given by 2( )VF g h R Rρ π= + This force is balanced by the weight of the oil in the container and the weight of container itself. The mass of oil is

31 42 3

m Rρ π⎛ ⎞= ⎜ ⎟⎝ ⎠

So the force balance is V cF m g m g= +

where is the mass of the container. Substituting: cm

2 32( )3 cg h R R g R m gρ π ρ π⎛ ⎞+ = +⎜ ⎟

⎝ ⎠

Solving for h

3

2

23

cmRh R

R

πρ

π

⎛ ⎞ +⎜ ⎟⎝ ⎠= −

3

3

2

2 5kg(0.1m)1000 kg3 (0.72)

m 0.1m(0.1 m)

h

π

π

⎛ ⎞ +⎜ ⎟ ⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠= −

0.188 m= Answer

4 - 24

4-26 If the “tip of the iceberg” (that is, the volume of the iceberg above the water surface) is 79 m3, what is the volume of the submerged iceberg? For seawater density use 31.027g/cm .seawaterρ =

Approach:

Use the formula for submerged volume of a floating object. Assumptions:

none

Solution:

For a floating body, the submerged volume is

( )ice icesub total sub tip

sea sea

V V V Vρ ρρ ρ

= = +

Gathering terms

1 ice icesub tip

sea sea

V Vρ ρρ ρ

⎛ ⎞ ⎛− =⎜ ⎟ ⎜

⎝ ⎠ ⎝

⎞⎟⎠

Solving for submerged volume (the density of ice is in Table A-3)

3

3

3

3

92079m1027

kg 1 m1 920m 100 cm1

g 1 kg1.027cm 1000 g

icetip

seasub

ice

sea

VV

ρρρρ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =

⎛ ⎞− ⎜ ⎟⎝ ⎠−

⎛ ⎞⎜ ⎟⎝ ⎠

3679 m= Answer

4-27 A small boat has a mass of 650 lbm when empty. If the volume of the hull is 166 ft3, determine the

maximum load the boat can carry in fresh water. Approach:

Draw a free body diagram of the boat. Include the buoyant forces and the weight of the boat and the load.

Assumptions:

1. The center of mass of the load is on the same vertical line as the center of gravity of the hull.

2. Water temperature is near room temperature.

Solution:

At the maximum load, the weight of the boat plus the weight of the load equals the buoyant force when the hull is submerged as much as possible without flooding. The buoyant force is directed along the center of mass of the boat but is offset in the figure for clarity. Designating as the mass of the boat and as the mass of the load: bm m

b Bm g m g F gVρ+ = = bm m Vρ+ =

bm V mρ= −

33

lbm62.3 (166ft ) 650lbmft

⎛ ⎞= −⎜ ⎟⎝ ⎠

9692lbm= Answer

Comments:

In reality, one would need a considerable margin of safety and the boat should carry a much lower load than the maximum. A boat submerged to the gunwales is in danger of flooding with any passing wave.

4 - 25

4-28 A layer of oil 6 cm thick covers a layer of water. A cylinder made of soft pine floats in this two-layer fluid, as shown. Using data on the figure, find the height, x, by which the cylinder protrudes from the fluid.

Approach:

Perform a force balance on the cylinder, matching weight force with buoyancy.

Assumptions: 1. The density of the oil is constant. 2. The density of the water is constant.

Solution: The weight force on the pine cylinder, mpg, must be balanced by the buoyancy forces due to the oil and water. The displaced masses of the oil, mo, and water, mw, can be used to determine their contributions to buoyancy.

2

2

2

(6)

( 6 )

p o w

p p

o o

w w

m g m g m g

m r h

m r

m r h x

π ρ

π ρ

π ρ

= +

=

=

= − −

where h is the height of the cylinder. Substituting masses into the force balance 2 2 2(6) ( 6 )

6 ( 6 )

(0.56) 6(0.75) ( 6 )(0.56) 6(0.75) 6

p o

p o w

w w w

r h g r g r h x g

h h x

h hh h x

w

x

π ρ π ρ π ρ

ρ ρ ρ

ρ ρ ρ

= + − −

= + − −

= + − −= + − −

Solving for x ( ) ( ) ( ) ( )6 0.75 6 1 0.56 6 0.75 6 14 1 0.56x h= − + − = − + −

4.66 cmx = Answer

4 - 26

4-29 A hot air balloon has a mass of 250 kg and carries 2 passengers whose average weight is 185 lbf. The balloon, which has a diameter of 12 m, rises through atmospheric air which is at 20oC. Find the minimum possible average temperature of the air inside the balloon. Atmospheric pressure is 100 kPa.

Approach:

Perform a force balance on the balloon, using the ideal gas law to find air masses.

Assumptions: 1. The balloon is spherical. 2. Air behaves like an ideal gas under these conditions. 3. Pressure is the same inside and outside of the balloon.

Solution:

The volume of the balloon is 34

3V Rπ= ( )3 34 6 905m

3π= =

The mass of atmospheric air at 20oC displaced by this volume is

( )( )

( )

5 3 kg10 Pa 905m 28.97kmol

kJ 1000J8.314 20+273 Kkmol K 1kJ

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⋅⎝ ⎠⎝ ⎠

11

PVMmRT

= 1076kg=

The passengers have an average weight of 185 lbf. To find their average mass, use 2 185 lbfF m g= =

where m2 is the average mass of a passenger. Solving for m2 gives

2

2

2

lbm ft32.2185lbf 185lbf s=ftg 1lbf32.2s

m

⋅⎛ ⎞⎛⎜ ⎟⎜

= ⎜ ⎟⎜⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝

⎞⎟⎟⎟⎟⎠

185 lbm=

The buoyancy force is equal to the weight of the air displaced, therefore 1 .BF m g= The buoyancy force on the balloon must be balanced by the total weight of the balloon, hot air, and passengers:

1 2 32m g m g m g m g= + + 4

2m m m= − −

where m2 is the average mass of a passenger, m3 is the mass of the empty balloon, m4 is the mass of hot air in the balloon, and m1 is the mass of the 20oC air displaced by the balloon. Solving for m4

4 1 2 3m ( ) 1kg1076kg 2 185 lbm 250kg 658kg2.2 lbm

⎛ ⎞= − − =⎜ ⎟

⎝ ⎠

To find the temperature, use the ideal gas law. For the displaced air,

11

PVMmRT

=

For the hot air in the balloon

44

PVMmRT

=

The air inside the balloon is at atmospheric pressure, as is the air outside the balloon, so we may divide the last two equations to get

1 4

4 1

m Tm T

=

Solving for the temperature of the hot air gives

( )1 14

4

1076kg 20+273 K658kg

m TTm

= =

o4 479K 206 CT = = Answer

4 - 27

4-30 A rectangular gate 12 ft high and 3 ft wide is held closed by water pressure, as shown in the figure. A counterweight of mass, m, is connected to the gate by a cable which runs over a pulley and attaches to the top of the gate. The counterweight, which is partially immersed in the water, is cylindrical with a diameter of 1.5 ft and a mass of 800 lbm. Air at atmospheric pressure is above the water and on the back side of the gate. Calculate the minimum water depth, h, for which the gate will stay closed.

Approach:

Draw free body diagrams of the gate, the counterweight, and the cable, and do a force balance on the counterweight and cable. Perform a moment balance on the gate. Express forces and moments in terms of the unknown height, h. Combine equations and solve for h.

Assumptions: 1. The water is incompressible. 2. The system is at room temperature.

Solution:

Begin by considering free body diagrams of the gate, the cable, and the counterweight. Atmospheric pressure acts on both sides of the gate, so its contribution will cancel and is not shown.

1

2

3

4

5

hydrostatic forcetension in thecabletension in thecablegravitybouyancy force

FFFFF

−−−

−−

From a force balance on the cable 2 3F F=

From a force balance on the counterweight 3 5 4F F F+ =

Taking moments about the gate hinge 1 1 2 2x F x F=

To find the hydrostatic force on the gate, F1, use ( )1 sinatm CF P A g y Aρ θ= +

Since Patm = 0, and o90 ,θ = / 2,Cy h=2

1 2gh aF ρ

=

where a is the depth of the gate into the page. The point of application of the hydrostatic force is

,

sin

xx CP C

atmC

Iy y P Ay A

gρ θ

= ++

Substituting the moment of inertia of a rectangular plate from Table 4-1,

3 22 212

2

Ph ah h hy

h ha= + = + =

⎛ ⎞⎜ ⎟⎝ ⎠

6 3h

Gate:

Counterweight: Cable:

4 - 28

By definition (see figure on the previous page),

113px h y h= − =

Substituting the expressions for x1and F1 into the moment balance gives

2

2 23 2h gh a x Fρ⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

The buoyancy force on the counterweight is 2

5 ( 2F g r hρ π= − ) The force balance on the counterweight becomes

22 ( 2)F g r h mgρ π+ − =

Combine this with the moment balance to obtain 2

22 ( 2)

3 2h gh a x mg g r hρ

ρ π⎛ ⎞⎛ ⎞ ⎡ ⎤= − −⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠⎝ ⎠

Since F3 = F2, the free body diagram is

Canceling g from each term and substituting values 23

22 2

lbm lbm 1.562.4 (3ft) 12ft 800 lbm 62.4 ft ( 2)ft6 ft ft 2h hπ

⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦−

Using equation solving software 5.45fth = Answer

4 - 29

4-31 Rain falling on a roof flows downward over the shingles and is collected in a gutter. The gutter, which is closed at one end and open at the other, is slightly inclined so that water runs out the open end. In a heavy downpour, rain falls steadily for several hours and the flow in the gutter reaches steady state. Due to the addition of runoff from the roof, the depth of water in the gutter increases gradually along the length of the gutter in the direction of flow. Using data in the figure below, calculate the inches per hour of rainfall that will cause the gutter to be completely filled with water at the open end. At rainfall rates higher than this, the gutter is inadequate to handle the water flow and excess water spills over the sides of the gutter before reaching the end. Assume that the exit velocity of the water from the gutter is 3 ft/s and that all the rain striking the roof is collected in the gutter. (Note that the home owner has been too cheap to install downspouts.)

Approach:

Express the velocity of rainfall in terms of the mass flow rate of rain falling on the roof. Set this mass flow rate equal to the mass flow rate of rain exiting the gutter and solve for rainfall velocity.

Assumptions:

1. All rain striking the roof is collected in the gutter.

2. The system operates in steady state. 3. The rain falls vertically.

Solution: Imagine setting a straight-sided container of bottom surface area A out in the rain. After time t, the container would fill up to a height of L. The rain is assumed to fall normal to the bottom surface. The amount of mass that accumulates in this time, t, is:

rainm V LAρ ρ= = The mass flow rate of the rain is

in.where of rainhrain

LA Lmt tρ

= =

To find the mass flow rate of rain on the slanted roof, use

A = [42 sin (60)] 85 = 3092 ft2

where A is the roof surface area projected on the bottom of the figure to the right. By conservation of mass:

rain gutter

gutter

m m

LA Atρ ρ

=

= V

where gutterm is the mass flow rate exiting the gutter wand Agutter is the cross-sectional area of the gutter. Si

gutterALt A

=V

( )( )

( )2

ft 1ft4in. 4in. 3s 12in. in.4.66

h1h3092ft3600s

Lt

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎜ ⎟⎝ ⎠

Answer

4 - 30

4-32 In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3. Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3. If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.

Approach:

Use the definition of volumetric flow rate and the relation between volumetric and mass flow rates.

Assumptions:

1. Air enters only through the cracks. 2. Air temperature is the same inside and outside.

Solution: The volumetric flow rate is related to the volume of the building by

or iV t V∆ =i

VtV

∆ =

Volumetric flow rate may be written in terms of mass flow rate using

ii

mV

ρ=

Eliminating volumetric flow rate between the last two equations

( )

( )

33

5

kg210m 1.186m

kg9.4 10 62 ms m

i

Vtm

ρ−

⎛ ⎞⎜ ⎟⎝ ⎠∆ = =

⎛ ⎞×⎜ ⎟⋅⎝ ⎠

42,735 s 11.9 h= = Answer

4-33 The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is

discovered the lower deck is already submerged to a level of 7.5 inches. At this time, a sailor turns on the bilge pump which begins to remove water at a rate of 73.8 gal/min. As an approximation, the lower deck can be modeled as a flat-bottomed container with a bottom surface area of 510 ft2 and straight vertical sides. How long will it be after the pump is turned on until the deck is clear of water?

Approach:

Apply conservation of mass. Assumptions:

1. The hull is box-shaped.

Solution: From conservation of mass:

cvi e

dmm m

dt− = which integrates to ( )i e cm m dt dm− = v∫ ∫

Since both rates are constant, ( )i e cm m t m− = v

where mcv is the mass that accumulates in the control volume in time t. In our case, mass leaves, so mcv is negative.

cv

i e i e

m Vtm m V V

ρρ ρ

−= =

− −

( )

( )

2

3

7.5510ft ft12

gal 0.0353ft57.5 73.8min 0.2642gal

i e i e

V AxV V V V

⎛ ⎞− ⎜ ⎟− − ⎝ ⎠= = =− − ⎛ ⎞

− ⎜ ⎟⎝ ⎠

where V is a positive number and x is the initial depth of water. Evaluating

146.4 min 2.44 ht = = Answer

4 - 31

4-34 On April 1, a reservoir has a water depth of 11 m. The reservoir is fed by a stream which becomes swollen with snow melt as the month progresses. The volumetric flow rate of stream water entering the reservoir during the month of April is

( )71 2.5 10 exp 0.026V t= ×

where t is the time in days and the volumetric flow rate has units of m3/day. (t = 0 at 12:01 a.m. on April 1). Water issues from the reservoir through a dam. The flow rate of the discharge at the dam is steady at a rate of 0.4 x 107 m3/day for the first 15 days of the month. At midnight on April 15, the sluice gates are adjusted to allow a higher flow rate of 6.35 x 107 m3/day. This rate remains constant until the end of the month. If the surface area of the reservoir is 2.8 x 106 m3, find the depth on April 30. Assume that the surface area remains unchanged during the month and that the effects of rainfall and evaporation are negligible.

Approach:

Apply conservation of mass, integrating over the given incoming flow rate.

Assumptions:

1. The surface area of the reservoir is constant. 2. There is no evaporation or rainfall and only one

stream entering and leaving the reservoir. 3. The water is incompressible.

Solution: By conservation of mass

cvi e

dmm m

dt= −∑ ∑

We have one stream entering with flow rate and one stream exiting through the dam with flow rate 1m 2m

1cvdm

m mdt

= − 2

1 2cvdm m dt m dt= −

Substituting m Vρ= and cv cvm Vρ= ( ) 1 2cvd V V dt V dtρ ρ ρ= −

1 2cvdV V dt V dt= −Integrating over the month

30 30 30

1 20 0 0

cvdV V dt V dt= −∫ ∫ ∫

( ) ( ) (30 15 30

7 730 days 0 day

0 0 15

2.5 10 e 0.4 10 6.35 10ctcv cvV V dt dt dt− = × − × − ×∫ ∫ ∫ )7

Defining 30 days 0 daycv cv cvV V V∆ = − and performing the integrations

( ) ( )( ) ( )( )30

7 7e 12.5 10 0.4 10 15 6.35 10 15c

cvVc

⎡ ⎤−∆ = × − × − ×⎢ ⎥

⎣ ⎦7

Noting that , 0.026c =

8 31.235 10 mcvV∆ = ×The change in reservoir depth is

8

6

1.235 10 44.12 m2.8 10

cv

cv

VL

A∆ ×

∆ = = =×

The final depth is 2 1 11 m 44.1m 55.1mL L L= + ∆ = + = Answer

4 - 32

4-35 Hydraulic fluid enters a square conduit 2 in. on a side at a velocity of 14.2 ft/s and a temperature of 60oF. The fluid leaves the conduit at 100oF. Neglecting frictional heating and kinetic energy, find the rate of heat addition to the fluid in steady state.

Approach:

Write the first law for an open system and eliminate all terms except heat and enthalpy change.

Assumptions:

1. Specific heat is constant. 2. The system operates in steady state. 3. Frictional heating in the fluid is negligible. 4. There is no change in kinetic or potential energy.

Solution: From the first law

22

ei

2 2cv

cv cv i i i e e edE

Q W m h gz m h gzdt

⎛ ⎞⎛ ⎞= − + + + − + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ VV

This simplifies to 0 (cv i eQ m h h= + − )

Assuming constant specific heat, ( )cv p e iQ mc T T= −where m Aρ= V The density of hydraulic fluid is taken at the average of the inlet and exit temperatures,

o60 100 80 F2 2

i eave

T TT

+ += = =

Using the density from Table B-6,

2

23 2

lbm ft 1ft52.5 14.2 (2)(2)in.ft s 144in.

m⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

lbm20.7s

=

Using cp from Table B-6 at o80 F

olbm Btu20.7 0.453 (100 60) Fs lbm FcvQ ⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Btu37.5s

= Answer

4 - 33

4-36 Air at 20oC and 101 kPa enters a passage between two printed circuit boards inside a desktop computer. One board contains 9 chips each dissipating 2 W and 5 chips each dissipating 1.3 W. No heat enters the passage from the other card. If the exit temperature is 28.5oC, find the volumetric flow rate of the air. Neglect kinetic and potential energy changes.

Approach:

Write the first law for an open system and eliminate all terms except heat and enthalpy change.

Assumptions:

1. Specific heat is constant. 2. The system operates in steady state. 3. Frictional heating in the fluid is negligible. 4. There is no change in kinetic or potential energy. 5. All heat dissipated in the chip enters the air by

forced convection; no heat is conducted into the card.

Solution:

From the first law

22

ei

2 2cv

cv cv i i i e e edE

Q W m h gz m h gzdt

⎛ ⎞⎛ ⎞= − + + + − + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ VV

This simplifies to 0 (cv i eQ m h h= + − )

Assuming constant specific heat, ( )cv p e iQ mc T T= −The mass flow rate is

( )

cv

p e i

Qm

c T T=

−

Using the value of cp for air from Table A-7 at 300 K

[ ]o

o

(9)(2) (5)(1.3) WkJ 1000J1.006 (28.5 20) C

kg C 1kJ

m+

=⎛ ⎞⎛ ⎞

−⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

kg0.00287s

=

To find volumetric flow rate, use

mVρ

=

The value of ρ at the average temperature of the air is needed here. The average temperature is

o20 28.5 24.25 C2aveT +

= = 297.3K=

We can either calculate ρ at 297.3K from the ideal gas law or approximate ρ by the value at 300K in Table A-7. We choose the simpler approach and use Table A-7. Then

3

3

kg0.00287 ms 0.00243kg1.18m

mVsρ

= = = Answer

4 - 34

4-37 Water flows in a pipe 1.7 cm in diameter and 480 cm long at a velocity of 8.8 m/s. The water enters at 100oC and exits at 80oC. Calculate the rate of heat removal per square centimeter of pipe wall. Neglect frictional losses and kinetic energy.

Approach:

Write the first law for an open system and eliminate all terms except heat and enthalpy change.

Assumptions:

1. Specific heat is constant. 2. The system operates in steady state. 3. Frictional heating in the fluid is negligible. 4. There is no change in kinetic or potential energy.

Solution: From the first law

22

ei

2 2cv

cv cv i i i e e edE

Q W m h gz m h gzdt

⎛ ⎞⎛ ⎞= − + + + − + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ VV

This simplifies to 0 (cv i eQ m h h= + − )

Assuming constant specific heat, ( )cv p e iQ mc T T= −To find mass flow rate, use m Aρ= V Using the density of water at the average temperature of from Table A-6 o90 C

2 2

23 4

kg m 1.7 1m965 8.8 cmm s 2 10 c 2m π

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦ m

kg1.93s

=

The heat removed is (using Table A-6 for Pc at ) oave 90 CT =

okg J1.93 4202 (80 100) Cs kg KcvQ

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠51.62 10 W 162 kW= − × = −

To get the heat flux on the wall, use

cv

w

Qq

A=

( )( )( )

5

2

1.62 10 W W63.21.7 cm 4.80cm cm

qπ

− ×= = Answer

4 - 35

4-38 In a solar collector, air is heated as it flows in a rectangular channel under the collector surface, as shown. Assume the rate of heat addition on the top surface is constant and uniform at 400 W/m2 and that all other sides of the channel are insulated. The air enters at 15oC and 101 kPa with a velocity of 5 m/s. The heat transfer coefficient inside the channel is 155 W/m2·K. Find the minimum and maximum temperatures of the collector surface.

Approach:

Write the first law for an open system and eliminate all terms except heat and enthalpy change. Use this to calculate the exit temperature of the air. Use the convection equation to find the collector temperature at both the inlet, where it is a minimum, and the outlet, where it is a maximum.

Assumptions:

1. Specific heat and density are constant. 2. The system operates in steady state. 3. Frictional heating in the fluid is negligible. 4. There is no change in kinetic or potential energy. 5. The heat transfer coefficient is constant along the

channel and independent of temperature. 6. The collector is perfectly insulated on all sides

except the top.

Solution: From the first law

22

ei

2 2cv

cv cv i i i e e edE

Q W m h gz m h gzdt

⎛ ⎞⎛ ⎞= − + + + − + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠∑ ∑ VV

With the assumptions listed above, this simplifies to 0 (cv i eQ m h h= + − )

A

Assuming constant specific heat, ( )cv p e iQ mc T T= −

To find mass flow rate, use m ρ= V with the density of air at a temperature of and 1 atm from Table A-7: o15 C

( )( )2

3 4 2

kg m 1m kg1.234 3 3cm 20cm 0.0222m s 10 cm s

m ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

kg1.93s

=

The exit temperature of the air is (with cp from Table A-7 at ) o15 C

( ) ( )2o o

W 1m400 20cm 1.6mm 100cm

15 C=20.7 Ckg kJ 1000J0.0222 1.0056s kg K 1kJ

cve i

p

QT T

mc

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠= + = +⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠⎝ ⎠

To find the surface temperature of the collector, we need

( )s fQ hA T T= − s fQT ThA

→ = +

The surface temperature will vary along the length of the collector, since Tf varies along the length. The ratio is given as 400 W/m/Q A 2 in the problem statement. The minimum surface temperature will occur at the

entrance, where the air temperature is lowest. It is o o

,min 2 2W W400 155 15 C 17.6 Cm m KsT ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟⋅⎝ ⎠ ⎝ ⎠

Answer

The maximum temperature occurs at the exit o o

,max 2 2W W400 155 20.7 C 23.3 Cm m KsT ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟⋅⎝ ⎠ ⎝ ⎠

Answer

4 - 36

4-39 Water is siphoned from a waterbed into a bathtub through a 1 in. diameter hose. The top surface of the water in the bed is 26 in. above the exit of the hose. Assume the pressure in the waterbed is atmospheric and that the top surface recedes with negligibly small velocity. Also assume negligible frictional effects in the hose. What is the flow rate, in gal/min, of water into the bathtub?

Approach:

Apply the Bernoulli equation to find the velocity exiting the hose. Use the velocity to determine the flow rate.

Assumptions:

1. Density is constant. 2. The flow is frictionless and isothermal. 3. The velocity of the surface of the water in the

waterbed is negligible. 4. The pressure in the waterbed is atmospheric.

Solution: Taking station 1 to be the top of the water mattress and station 2 to be the exit of the hose, and applying Bernoulli’s equation:

2 21 1 2

1 22 2P Pgz gzρ ρ

+ + = + +V V2

Since the pressure in both the bathtub and the waterbed is atmospheric 1 2 atmP P P= =

Furthermore, the surface of the water in the mattress is receding at a very low velocity, ( ), so 1 0=V2

21 20

2gz g= − −

V z

Solving for 2V

( ) ( )2 1 2 2

ft 1ft ft2 2 32.17 26in. 11.8s 12in.

g z z⎛ ⎞⎛ ⎞= − = =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠V

s

The volumetric flow rate depends on velocity according to

22

2 2 3-3

gal1ft 0.5 galmin11.8 π ft 28.9fts 12 min2.228×10s

V A ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

V Answer

4 - 37

4-40 Water flows at 25 kg/s through a gradual contraction in a pipe. The upstream diameter is 8 cm and the downstream diameter is 5.6 cm. If the exit pressure is 60 kPa, find the entrance pressure. Assume frictionless flow.

Approach:

Apply the Bernoulli equation. Find velocities from the known mass flow rate.

Assumptions:

1. Density is constant. 2. The flow is frictionless and isothermal. 3. The system is at room temperature.

Solution: Taking station 1 to be the entrance of the contraction and station 2 to be the exit, and applying Bernoulli’s equation:

2 21 1 2

1 22 2P Pgz gzρ ρ

+ + = + +V V2

Since there are no elevation changes, 2 2

1 1 2 2

2 2P Pg g g gρ ρ

+ = +V V

Velocities can be determined using the mass flow rate, since

12 21

3

kg25 ms 4.99kg s997.5 (0.04) mm

m A

mA

ρ

ρ π

=

= = =⎛ ⎞⎜ ⎟⎝ ⎠

V

V

2 22 2

3

kg25 ms 10.2skg 0.056997.5 m

m 2

mAρ

π= = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

V

To find the pressure, rearrange the Bernoulli equation 2 2

1 2 2 2 12

3 21 3 2

( )

kg m60 10 Pa 997.5 (10.2 4.99 )m s

P P

P

ρ= + −

⎛ ⎞= × + −⎜ ⎟⎝ ⎠

V V

2

31 99.2 10 Pa 99.2 kPaP = × = Answer

4 - 38

4-41 A constriction in a water tube is used to provide suction on a submerged thin disk, as shown in the figure. Find the mass of the heaviest disk that can be supported.

Approach:

Apply the Bernoulli equation. Find velocities from the known mass flow rate.

Assumptions:

1. Density is constant. 2. The flow is frictionless and isothermal. 3. The system is at room temperature. 4. The pressure is constant across the throat of the

constriction. 5. There are negligible buoyancy forces on the disk.

Solution: Taking station 1 to be the entrance of the contraction and station 2 to be at the throat, and applying Bernoulli’s equation:

2 21 1 2 2

1 22 2P Pgz gρ ρ

+ + = + +V V z

From conservation of mass 1 2m m= → 1 1 2 2A Aρ ρ=V V The water is incompressible, therefore

2 2

1 21 22 2

D Dπ π⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠V V

2

12 1

2

DD

⎛ ⎞= ⎜ ⎟

⎝ ⎠VV

2m 13.1 cm3s 8.2 cm

⎛ ⎞⎛ ⎞=⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

m7.66s

=

There are no elevation changes, so and the Bernoulli equation reduces to 1 2z z=

2 2

1 22 1 2

P P ρ⎛ ⎞−

= + ⎜ ⎟⎝ ⎠

V V

2 2

3

m m3 7.66kg s s101,000Pa 997m 2

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎝ ⎠= +⎜ ⎟⎝ ⎠

376.2 10 Pa= ×

Let P3 be the pressure of the water in the tube just above the disk. The water in the tube is stagnant so

( )3 2 1 2P P g h hρ= + +

33 2

kg m 4.6 3.776.2 10 997 9.81 mm s 100

+⎛ ⎞⎛ ⎞⎛ ⎞= × +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

377.1 10 Pa= ×

Let P4 be the pressure of the water on the underside of the disk. 4 2atmP P ghρ= +

33 2

kg m 3.7101 10 997 9.81 mm s 100

⎛ ⎞⎛ ⎞⎛ ⎞= × + ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

3101.4 10= ×

From a force balance on the disk

3 3 4 3P A mg P A+ =

( )3 4 3A P P

mg

−=

( )2

2 3 3

2

0.015 m 101.4 10 77.1 10 Pa2

m9.81s

π ⎛ ⎞ × − ×⎜ ⎟⎝ ⎠=

0.438 kgm = Answer

4 - 39

4-42 Water issues from a hole in a large tank, as shown. Assuming frictionless flow, find L. Approach:

Apply the Bernoulli equation to find the velocity at the hole in the tank. Then use Newton’s second law on a fluid particle to find the time until it hits the surface. From the time of fall and the horizontal velocity, the distance L can be determined.

Assumptions:

1. Density is constant. 2. The flow is frictionless and isothermal. 3. The system is at room temperature. 4. The tank is large so that the water recedes with very

small velocity. 5. Water issues horizontally from the tank. 6. Aerodynamic drag on the water jet is negligible.

Solution: Taking station 1 to be the top surface of the water and station 2 to be the exit from the tank, and applying Bernoulli’s equation:

2 21 1 2 2

1 22 2P Pgz gρ ρ

+ + = + +V V z

Since and the water recedes with negligibly small velocity 1 2 atmP P P= =

( )2 1 22g z z= −V ( )2

m2 9.81 5 ms

⎛ ⎞= ⎜ ⎟⎝ ⎠

m9.9s

=

To find the time of flight, apply Newton’s second law to a fluid particle at height z2. We only consider forces in the vertical direction.

F ma=dVmg mdt

= =

2

2

d xmg mdt

=

Integrating once

1dx gt cdt

= +

Integrating again

2

1 22gtx c t c= + +

The boundary conditions are at 0t = 0x = 2 0c⇒ =

at 0t = 0idxdt

= =V 1 0c⇒ =

So that the solution is 2

2gtx = . Solving for time of drop

( )

2

2 3 m2m9.81s

xtg

= = 0.782 s=

The lateral distance traveled during this drop time is

( )2m9.9 0.782ss

L t ⎛ ⎞= =⎜ ⎟⎝ ⎠

V 7.72 m= Answer

4 - 40

4-43 A firefighter aims a jet of water at a window in a burning building, as shown below. The jet is horizontal when it enters the window. If the nozzle has a diameter of 2.5 in., what is the mass flow rate of the water? (Assume there is no aerodynamic drag on the jet and the water is at 50oF).

Approach:

Apply the Bernoulli equation to the free jet. Decompose the velocities at the nozzle and the window into horizontal and vertical components. Use the fact that the vertical component is zero at the window and that the horizontal component is the same at the nozzle and at the window to determine the velocity at the nozzle.

Assumptions: 1. Density is constant. 2. The flow is frictionless and isothermal. 3. The system is at 50oF. 4. Water strikes the building horizontally. 5. Aerodynamic drag on the water jet is negligible.

Solution: Taking station 1 to be the exit of the nozzle and station 2 to be at the window, and applying Bernoulli’s equation:

2 21 1 2 2

1 22 2P Pgz gρ ρ

+ + = + +V V z

2

In a free jet, pressure is atmospheric everywhere, so 1 2P P=The two velocities may be decomposed into horizontal and vertical components

2 21 1 1x z= +V V V

2 22 2

22x z= +V V V

Substituting these into the Bernoulli equation 2 2 2 2

1 1 2 21 22 2

x z x zgz g⎛ ⎞ ⎛+ +

+ = +⎜ ⎟ ⎜⎝ ⎠ ⎝

V V V Vz

⎞⎟⎠

Since the jet is horizontal at point 2, . There are no forces in the x direction (no aerodynamic drag), so 2 0z =V

1 2x x=V V and the Bernoulli equation becomes

( )2

11 20

2z g z z= + −

V

( )1 2 12z g z z= −V ( )2

ft2 32.17 28 3 fts

⎛ ⎞= −⎜ ⎟⎝ ⎠

40.1 ft s=

Now find the magnitude of the velocity at the nozzle exit from

( )1

1

sin 70z = °VV

( )140.1ft s 42.7 ft ssin 70

= =°

V

The mass flow rate is

1m Aρ= V( )( )

2

23

lbm ft 2.562.4 42.7 ftft s 2 12

π⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

90.8 lbm s= Answer

4 - 41

4-44 Water flowing in a horizontal pipe branches into two pipes as shown and issues into the atmosphere. Neglecting all viscous effects, find the volumetric flow rate in each pipe and the diameter, D3.

Approach:

Apply the Bernoulli equation between points 1 and 2 and also between points 1 and 3. Use these two equations and conservation of mass to find the unknown diameter.

Assumptions:

1. Density is constant. 2. The flow is frictionless and isothermal. 3. The system is at room temperature.

Solution:

Apply Bernoulli’s equation between points 1 and 2 to get

2 2

1 1 21 22 2

P Pgz gzρ ρ

+ + = + +V V2

2

There are no elevation changes, so . To find , 1z z= 1V

1 11 2

1 1

2

V VA Dπ

= =⎛ ⎞⎜ ⎟⎝ ⎠

V

3

2 22

m0.02 ms 7.07s6 1mcm

2 10 cmπ

= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

Rearranging the Bernoulli equation

2

1 1 22 2

2P P

ρ⎛ ⎞−

= +⎜ ⎟⎝ ⎠

VV

2

3

1000 Pam (120 101)kPa7.07 1kPas2 kg2 998m

⎡ ⎤⎛ ⎞⎛ ⎞ −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥= +

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

m9.39s

=

The volumetric flow rate is

2

22 2 2 2 2

DV A π ⎛ ⎞= = ⎜ ⎟⎝ ⎠

V V2 2

24 2

m 2.5 1m9.39 cms 2 10 cm

π⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3m0.00461s

=

By conservation of mass 1 2m m m= + 3 3→ 1 2V V Vρ ρ ρ= +

V V= −

Since water is incompressible

3 1 2V 0.02 0.00461= −3m0.0154

s=

To find the diameter, apply the Bernoulli equation again between points 1 and 3 to get

22

3 31 11 32 2

PP gz gzρ ρ

+ + = + +VV

2

1 313 2

2P P

ρ⎛ ⎞−

= +⎜ ⎟⎝ ⎠

VV ( )27.07 (120 101)(1000)22 998

⎡ ⎤−= +⎢ ⎥

⎢ ⎥⎣ ⎦2

m9.39s

= = V

2

33 3 3 3 2

DV A π ⎛ ⎞= = ⎜ ⎟

⎝ ⎠V V

33

3

2V

Dπ

=V

3m0.01542 0.0457 m 4.57cm

m9.39s

s

π= = =

⎛ ⎞⎜ ⎟⎝ ⎠

Answer

4 - 42

4-45 Water at 20oC issues from the bottom of a large tank which is filled to a height of 4 m. The air pressure above the water is atmospheric. The water flows over an aluminum rod of diameter 1.1 cm and length 3 cm. The heat transfer coefficient between the water and the rod depends on velocity and is given by

0.8160h = V

where h is in W/m2·K and is in m/s. The rod is initially at 65V oC. How long will it take for the rod to cool to 30oC? Assume the surface of the water recedes very slowly, so that the depth of the water remains at 4 m throughout the process.

Approach:

Apply the Bernoulli equation to find the velocity of water issuing from the tank. From the given formula, determine the heat transfer coefficient. Use the lumped system approximation to calculate the time required for the rod to cool to 30oC.

Assumptions:

1. Density is constant. 2. The flow in the tank is frictionless and

isothermal. 3. The heat transfer coefficient is uniform

over the surface of the rod and independent of temperature.

4. The surface of the water in the tank recedes very slowly.

Solution:

Apply Bernoulli’s equation between point 1 (the upper surface of water in the tank) and point 2 (the place where water issues from the tank), to get

2 2

1 1 21 22 2

P Pgz gzρ ρ

+ + = + +V V2

The pressure is atmospheric at both locations and the surface recedes with negligibly small velocity, so 2

21 2 2

gz gz= +V → ( ) ( )2 1 2 2

m m2 2 9.81 4m 8.86s s

g z z ⎛ ⎞= − = =⎜ ⎟⎝ ⎠

V

From the given formula

( )0.80.82

W160 160 8.86 916m K

h = = =⋅

V Check to see if the lumped system approximation can be used (find thermal conductivity from Table A-2).

charhLBi

k= where ( )( )

( ) ( )2

2

1.1cm 3cm0.402cm

2 2 2 2 2 3cm 2 1.1cmcharV r L rLLA rL r L r

ππ π

= = = = =+ + +

( )2

W 1m916 0.402cmm K 100cm

0.0156W237

m K

Bi

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⋅⎝ ⎠ ⎝ ⎠= =

⎛ ⎞⎜ ⎟⋅⎝ ⎠

Since the lumped system approximation is valid and the time of cooling can be calculated as (with density and specific heat from Table A-2)

0.1,Bi <

( ) 3

2

kg J 0.4022702 903 mm kg K 100 30 20ln ln 16.1sW 65 20916

m K

p char f

i f

c L T t Tt

h T Tρ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎡ ⎤− ⋅ −⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎝ ⎠= − = − =⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎢ ⎥⎣ ⎦

⋅

Answer

4 - 43

4-46 Air at 17oC, 100 kPa flows in a duct. A stagnation tube connected to a U-tube manometer filled with mercury is placed in the duct. Using data on the figure below, find the air velocity. Assume atmospheric pressure is 100 kPa.

Approach:

Apply the Bernoulli equation to find the velocity of the air in terms of P1 and P2 (See figure). Use hydrostatics to determine the relationship between P3 and P4. Recognizing that and allows the solution to be completed.

2P P= 3 41P P=

Assumptions:

1. The air and the mercury are both incompressible.

2. The density of the air is much lower than the density of mercury.

Solution: Apply Bernoulli’s equation between point 1 (the flow at the entrance upstream from the tube) and point 2 (the place where the air strikes the stagnation tube), to get

2 2

1 1 21 22 2air air

P Pgz gzρ ρ

+ + = + +V V2

Point 2 is a stagnation point where the velocity is zero, and the elevation does not change, therefore,

2 11 2

air

P Pρ

⎛ ⎞−= ⎜ ⎟

⎝ ⎠V

The pressure is the same at point 2 and point 3, since the tube is filled with a stagnant column of air, that is Furthermore, the pressure at point 4 is atmospheric, as is the pressure at point 1, that is Therefore

2 3.P P= 1 4.P P=

3 41 2

air

P Pρ

⎛ ⎞−= ⎜ ⎟

⎝ ⎠V

From hydrostatics 3 4 HgP P ghρ= + → 3 4 HgP P ghρ− =

Using mercury density from Table A-6 and air density from Table A-7,

3 2

1

3

kg m 302 13,586 9.81 mmm s 10002 80.8kg s1.224

m

Hg

air

ghρρ

⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠⎝ ⎠= = =⎜ ⎟

⎝ ⎠V Answer

4 - 44

4-47 Oil (SG = 0.77) flows in a pipe with a sudden contraction, as shown in the figure. A stagnation tube open to the atmosphere is placed in the upstream section. If the oil in the stagnation tube rises to a height, h = 22 cm, find the velocity at the exit of the pipe.

Approach:

Apply the Bernoulli equation between points 1 and 2 and also between points 1 and 3. Use hydrostatics to express P3 in terms of P4. Combine equations, recognizing that P2 = P4 and solve for the unknown velocity.

Assumptions:

1. Neglect any elevation difference between points 1 and 2.

2. The density of the oil is constant. 3. Frictional effects are small.

Solution: Apply the Bernoulli equation between points 1 and 2

2 2

1 1 21 22 2

P Pgz gzρ ρ

+ + = + +V V2

This simplifies to

2 2

1 1 2 2

2 2P Pρ ρ

+ = +V V

Also apply the Bernoulli equation between points 1 and 3

22

3 31 11 32 2

PP gz gzρ ρ

+ + = + +VV

Because point 3 is a stagnation point, and 3 0,=V

2

31 1

2PP

ρ ρ+ =

V

Combining the two simplified Bernoulli equations

2

3 2 2

2P Pρ ρ

= +V

In the tube 3 4P P ghρ− = but , so 4 2 atP P P= = m

3 2P P ghρ− = Combining equations

2

2

2ghρρ

=V

2 2gh=V ( )2

m 1(2) 9.81 22cms 100

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

mcm

m2.08s

= Answer

Comment:

Frictional effects in a sudden contraction are often significant, and a more advanced analysis should take them into account.

4 - 45

4-48 Hydraulic fluid at 80°F flows through a Venturi meter. The diameter at the entrance is 8.1 in. and at the throat it is 5.2 in. The pressure at the entrance is 14.7 psia. If the pressure at the throat is measured to be 10.8 psia, find the velocity at the entrance.

Approach:

Determine the volumetric flow rate from the equation for a Venturi meter. Knowing the area and volumetric flow rate, calculate the velocity.

Assumptions

1. The density of the hydraulic fluid is constant. 2. Frictional effects are small.

Solution: The areas of the entrance and throat are

221

1 51.5 in.2DA π ⎛ ⎞= =⎜ ⎟

⎝ ⎠

222

2 21.2 in.2

DA π ⎛ ⎞= =⎜ ⎟⎝ ⎠

From Table B-6 at 80°F, 3

lbm52.5 .ft

ρ = For a Venturi meter

( )( )

1 22 2

2 1

2 11 /

P PV A

A Aρ

⎛ ⎞−⎜ ⎟=⎜ ⎟−⎝ ⎠

( )2

2 2 222

22

3

lbf 32.17 lbm ft 144 in.2 14.7 10.8in. 1 lbf s 1 ft1ft 121.2in. lbm144in. 21.252.5 1ft 51.5

V

⎧ ⎫⎧ ⎫⎛ ⎞⎛ ⎞⋅−⎪ ⎪⎜ ⎟⎜ ⎟ ⎪ ⎪⋅⎛ ⎞ ⎪ ⎪⎝ ⎠⎝ ⎠= ⎨ ⎬⎜ ⎟

⎛ ⎞⎝ ⎠ ⎪ ⎪ − ⎜ ⎟⎪ ⎪ ⎝ ⎠⎩ ⎭⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪

3

1 221

2

ft4.25s1 ft51.5 in.

144 in.

VA

= =⎛ ⎞⎜ ⎟⎝ ⎠

V ft11.9s

= Answer

4 - 46

4-49 A block of mass 4 kg is propelled along a flat surface by a water jet, as shown. It moves to the right at a constant velocity of 2.5 m/s. The coefficient of friction between block and surface is µ = 0.33. If the inlet jet area is 6.7 cm2, find the inlet and outlet velocities of the water. Neglect wall shear and elevation changes.

Approach:

Determine the relative velocity between the block and the jet. Then apply Bernoulli’s equation and conservation of momentum.

Assumptions

1. The density of the water is constant. 2. Frictional effects are small. 3. Neglect the weight of the water.

Solution: Consider a control volume attached to the block as shown in the figure to the right. This control volume moves at speed to the right. Then bV

r j= −V V Vb where the r subscript designates relative velocity between jet and block. Since the moving control volume is not accelerating, we may analyze the system as if it were a stationary block with an impinging jet of velocity Applying conservation of mass to this system

.rV

1 2jm m m= +

1 1 2 2j rA A Aρ ρ ρ= +V V V By symmetry (if we neglect the weight of the water)

1 2 1 2A A= =V V

( )1 1 2 1 12j rA A A= + =V V V V From the Bernoulli equation:

2

02

j rj

Pgz

ρ= + +

V 20 1 1

12P gzρ

⎛ ⎞− + +⎜ ⎟

⎝ ⎠

V 0

2

02

j rj

Pgz

ρ

⎛ ⎞⎜ ⎟⎝ ⎠

= + +V 2

0 2 222

P gzρ

⎛ ⎞− + +⎜ ⎟

⎝ ⎠

V 0⎛ ⎞⎜ ⎟⎝ ⎠

since 1 2j atmP P P P= = = 1 2r∴ = =V V V

From conservation of momentum:

( ), ,

, ,but 0 and so

,x x cv e x e i x i

x e x i r

j r r

dF B m mdt

mg Aµ ρ

= + −

= =

− = −

∑ V V

V V VV V

( )( )( )

( )( )

2

23 2

4 2

0.33 4kg 9.81 m s1m998.2kg m 6.7cm

10 cm

rj

mgA

µρ

= =V 1 2m4.4s

= = =V V Answer

The inlet velocity is m4.4 2.5 6.9sj r b= + = + =V V V Answer

4 - 47

4-50 Water at 20°C flows through a pipe as shown below. At a point 0.8 meter above the pipe bend, the velocity is 6 m/s. The exit pressure is 101 kPa. Assuming frictionless, incompressible, steady flow, find the magnitude and direction of the anchoring force.

Approach:

Apply the Bernoulli equation to find the inlet pressure. Use conservation of linear momentum to determine anchoring force.

Assumptions:

1. The water is incompressible. 2. The flow is frictionless. 3. The system is at room temperature.

Solution: From conservation of mass,

1 1 2 2A Aρ ρ=V V

( )( )

21 1

2 22

6 9 / 219.4m s

5 / 2A

Aπ

π= = =

VV

From the Bernoulli equation:

2 2

1 1 2 21 22 2

P Pgz gρ ρ

⎛ ⎞ ⎛+ + = + +⎜ ⎟ ⎜

⎝ ⎠ ⎝

V V z⎞⎟⎠

( )2 2

2 1 21 22

PP gρρ

⎧ ⎫−= + −⎨ ⎬

⎩ ⎭

V V1z z−

( ) ( )

( )2 2

1 3 2

3

1000 Pa101kPa19.4 6kg m1 kPa998.2 9.81 0.8 mkgm 2 s998.2m

P

⎧ ⎫⎛ ⎞⎜ ⎟⎪ ⎪−⎪ ⎪⎛ ⎞ ⎛ ⎞⎝ ⎠= + −⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎪ ⎪⎪ ⎪⎩ ⎭

263 kPa=

Perform a momentum balance in the x-direction: ,x e x eF m=∑ V , 2 2 2x aF P A m− = V2

2, 2 2 2 2x aF P A Aρ= + V

Gage pressure must be used in this equation ( )2 0psigP =

2 22

2, 3 4 2

kg 5 1 m m998.2 cm 19.4m 2 10 cm sx aF π

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎟⎠

737 N= ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝⎝ ⎠

=

A momentum balance in the y-direction gives

iy iF m= −∑ V

2, 1 1 1y a 1F P A Aρ− = − V

2, 1 1 1y aF P Aρ⎡ ⎤= −⎣ ⎦V ( )

2 2 22

3 4

1000 Pa kg m 9 1 m263 101 kPa 998.2 6 cm1 kPa m s 2 10 cm

π⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= − −⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦2 802 N=

where gage pressure has again been used. The magnitude of the resultant force is 2 2 1090 Nx yF F F= + = Answer

The direction is

,

,

802arctan arctan737

y a

x a

FF

θ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

47.4= ° Answer

4 - 48

4-51 A water jet moving at 18 ft/s strikes a vane and is turned through 120o as shown. If the flow rate is 27 gal/min, and the flow is frictionless, find the magnitude and direction of the anchoring force needed to hold the vane in place.

Approach:

Apply conservation of momentum in both the x and y directions.

Assumptions:

1. There are no components of velocity in the z direction.

2. The system is at room temperature.

Solution: Apply conservation of momentum in the x-direction

( ), , ,x x cv e x e i x idF B m mdt

= + −∑ V V

The mass flow rate is 3

3

gal lbm 0.1337ft 1min27 62.2min ft 1gal 60se im m m

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

lbm3.74s

=

( ) (, , ,2 ,x x e x i xF m m= − = −∑ V V V V )1x The x-component of the exiting velocity is

( )o,2 2

ft ftcos 18 cos 60 9s sx θ ⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠V V

Substituting values

( )2

lbm ft 1lbf3.74 9 18lbm fts s 32.174

s

xF

⎛ ⎞⎜ ⎟⎛ ⎞= − −⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎣ ⎦ ⋅⎝ ⎠ ⎜ ⎟⎜ ⎟⎝ ⎠

∑ 3.14lbf= −

Apply conservation of momentum in the y-direction

( ), ,y y cv e y e idF B m mdt

= + −∑ V V ,y i

( ) ( )o 13.74 18sin 60 1.81lbf32.174yF ⎛ ⎞⎡ ⎤= =⎜ ⎟⎣ ⎦ ⎝ ⎠

∑The magnitude of the force may be calculated from the components

( ) ( )2 23.14 1.81 3.63 lbfF = + = Answer The direction is given by

1.81sin 0.4993.63

α = =

o30α = Answer

4 - 49

4-52 A jet engine on a commercial aircraft exhausts combustion gases at a rate of 8 kg/s. Upon landing, a thrust reverser blocks the exhaust and redirects the flow forward, as shown. This aids in braking the plane. The exhaust gases may be assumed to flow at 200 m/s relative to the plane and have properties very similar to those of air. Find the anchoring force needed to hold the thrust reverser on the back of the engine.

Approach:

Apply conservation of momentum in the x direction. Assumptions:

1. There are no components of velocity in the z direction.

2. The flow is symmetric about the x axis.

Solution: Forces in the y direction cancel, so we only need to consider the x – direction momentum equation , ,x e x e i x iF m m= −∑ ∑ ∑V V 2 ,2 3 ,3 1 1,R x xF m m m= + −V V V x Apply the Bernoulli equation between points 1 and 2

2 2

1 1 2 21 22 2

P Pgz gρ ρ

+ + = + +V V z

Since and , 1 2P P= 1 2z z=

1 2∴ =V V

1,m200sx =V

( )o2,

m200 sin 20sx

⎡ ⎤= −⎢ ⎥⎣ ⎦V

By symmetry

( )o3,

m200 sin 20sx

⎛ ⎞= −⎜ ⎟⎝ ⎠

V

The mass flow rates can be found using symmetry. 1 8kg sm =

2 3 11 4kg s2

m m m= = =

Substituting values

( )( ) ( )( )kg m4 200 sin 20 4 200 sin 20 8 200s sRF ⎛ ⎞⎛ ⎞= − + − −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

2147 NRF = − Answer

4 - 50

4-53 A jet of water of area 1.6 in.2 and velocity 22 ft/s strikes a plate and is deflected into two symmetrical streams, as shown. If θ = 42o, find the anchoring force necessary to hold the plate in place.

Approach:

Apply conservation of momentum in the x direction.

Assumptions:

1. There are no components of velocity in the z direction.

2. The flow is symmetric about the x axis.

Solution: Forces in the y direction cancel, so we only need to consider the x-direction momentum equation , ,x e x e i x iF m m= −∑ ∑ ∑V V 2 ,2 3 ,3 1 1,R x xF m m m= + −V V V x

To find the velocities, apply the Bernoulli equation between points 1 and 2

2 2

1 1 2 21 22 2

P Pgz gρ ρ

+ + = + +V V z

Since and , 1 2P P= 1 2z z=

1 2∴ =V V By symmetry 3 2=V VBy conservation of mass, noting that , 2 3m m=

( )1 2 3 2

22

1 1 1 3 2

2

lbm ft 1ft lbm62.2 22 1.6in. 15.2ft s 144in. s

m m m m

m Aρ

= + =

⎛ ⎞⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

V

By symmetry

2 3 11 l7.62 s

m m m= = =bm

Applying conservation of momentum in the x direction: 2 ,2 3 ,3 1 1,R x xF m m m= + −V V V x

( )o

2

lbm ft 1lbf7.6 22 cos 42lbm fts s 32.174

s

⎛ ⎞⎜ ⎟⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠ ⎜ ⎟⎜ ⎟⎝ ⎠

( ) ( ) ( ) ( )( )o 1 17.6 22 cos 42 15.2 2232.174 32.174

⎛ ⎞+ − − ⎜ ⎟⎝ ⎠

18.1 lbfxF = − Answer

4 - 51