3-Practice Problems.pdf

Chapter 3

Fluid Statics

Problem 3.1

For a lake, Þ nd the depth ! at which the gage pressure is 1 atmosphere. The

speci Þ c weight of water is 62.3 lbf/ft 3 .

Solution

At the free surface of the lake, pressure will be " surface = 1#0 atm absolute or 0.0

atm gage. At a depth ! , the pressure will be 1 atm gage.

In a static ß uid of constant density, the piezometric head ("$% + &) is constant.

Thus

" surface

+& surface = 1 atm

+(& surface !!)

(1)

CHAPTER 3. FLUID STATICS

Since " surface = 0 atm gage, Eq. (1) becomes

! = 1 atm

= (14#7 lbf/in 2 )(144 in 2 $ ft 2 )

62#3 lbf/ft 3

= 34.0 ft

Problem 3.2

A tank that is open to the atmosphere contains a 1.0-m layer of oil ( ' = 800 kg/m 3 )

ß oating on a 0.5-m layer of water ( ' = 1000 kg/m 3 ). Determine the pressure at

elevations ( , ) , *+ and , . Note that ) is midway between ( and * .

Solution

At a horizontal interface of two ß uids, pressure will be constant across the inter-

face. Thus the pressure in the oil at ( equals the pressure in the air (atmospheric

pressure).

" ! = " "#$

= 0 kPa gage

Since the oil layer is a static ß uid of constant density, the piezometric pressure is

constant

" ! +% %&' & ! = " ( +% %&' & ( = " ) +% %&' & ) = constant

(1)

where & denotes elevation. Let & ! = 0+ & ( = !0#5 m, & ) = !1#0 m. Then, Eq.

(1) becomes

" ! = " ( +% %&' (!0#5 m ) = " ) +% %&' (!1#0 m )

" ( = " ! +% %&' (0#5 m )

= " "#$ +(800)(9#81)(0#5)$1000

= 3.92 kPa-gage

Similarly

" ) = " ! +% %&' (1#0 m )

= " "#$ +(800)(9#81)(1#0)$1000

= 7.85 kPa-gage

At elevation C, pressure in the oil equals pressure in the water. Since the piezo-

metric pressure in the water is constant, we can write

" ) +% *"#+, & ) = " - +% *"#+, & -

" - = " ) +% *"#+, (& ) !& - )

= 7#85+(1000)(9#81)(0#5)$1000

= 12#8 kPa-gage

Problem 3.3

A U-tube manometer contains kerosene, mercury and water, each at 70 % F. The

manometer is connected between two pipes ( ( and ) ), and the pressure di ! erence,

as measured between the pipe centerlines, is " ( !" ! = 4#5 psi. Find the elevation

di ! erence & in the manometer.

Solution

Apply the manometer equation (Eq. 3.17 in the 8th edition). Begin at location )

and add pressure di ! erences until location ( is reached.

" ( +(2+&)% *"#+, !&% ./ !2% 0+,% = " !

Rearranging

" ( !" ! = 2(% 0+,% !% *"#+, )+&(% ./ !% *"#+, )

(1)

Looking up values of speci Þ c weight and substituting into Eq. (1) gives

[4#5×144] lbf/ft 2 =

(2 ft )(51!62#3) lbf/ft 3 +(& ft )(847!62#3) lbf/ft 3

CHAPTER 3. FLUID STATICS

& = (648+22#6)

784#7

= 0#855 ft

Problem 3.4

A container, Þ lled with water at 20 % C, is open to the atmosphere on the right

side. Find the pressure of the air in the enclosed space on the left side of the

container.

Solution

The pressure at elevation 2 is the same on both the left and right side.

" 2 = " atm +%(0#6 m )

= 0+

(0#6 m )

= 5#89 kPa

Since the piezometric head is the same at elevations 1 and 2

" % +& 1 = " % +& 2

" 1 = " 2 +%(& 2 !& 1 )

= (5#89 kPa )+

9#81 kN $ m 3

(!1#0 m )

= !3#92 kPa gage

9#81 kN $ m 3

Problem 3.5

A rectangular gate of dimension 1 m by 4 m is held in place by a stop block at

) . This block exerts a horizontal force of 40 kN and a vertical force of 0 kN. The

gate is pin-connected at ( , and the weight of the gate is 2 kN. Find the depth !

of the water.

Solution

A free-body diagram of the gate is

where - is the weight of the gate, . is the equivalent force of the water, and / is

the length of the moment arm. Summing moments about ( gives

) 1 (1#0SIN60 % )!. ×/ +-(0#5COS60 % ) = 0

. ×/ = ) 1 SIN60 % +-(0#5COS60 % )

= 40+000SIN60 % +2000(0#5COS60 % )

(1)

= 35+140 N-m

The hydrostatic force . acts at a distance 0$1( below the centroid of the plate.

Thus the length of the moment arm is

/ = 0#5 m + 0

(2)

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