2-Practice Problems.pdf

Chapter 2

Fluid Properties

Problem 2.1

Calculate the density and speci Þ c weight of nitrogen at an absolute pressure of

1 MPa and a temperature of 40 ! C.

Solution

Ideal gas law

! = "

From Table A.2, # = 297 J/kg/K. The temperature in absolute units is $ =

273+40 = 313 K.

! = 10 6 N/m 2

297 J/kgK ×313 K

= 10%75 kg/m 3

The speci Þ c weight is

& = !'

= 10%76 kg/m 3 ×9%81 m/s 2

= 105%4 N/m 3

CHAPTER 2. FLUID PROPERTIES

Problem 2.2

Find the density, kinematic and dynamic viscosity of crude oil in traditional units

at 100 o F.

Solution

From Fig. A.3, ( = 6%5×10 !5 ft 2 ) s and * = 0%86%

The density of water at standard condition is 1.94 slugs/ft 3 + so the density of crude

oil is 0%86×1%94 = 1%67 slugs/ft 3 or 1%67×32%2 = 53%8 lbm/ft 3 %

The dynamic viscosity is !( = 1%67×6%5×10 !5 = 1%09×10 !4 lbf · s/ft 2 %

Problem 2.3

Two parallel glass plates separated by 0.5 mm are placed in water at 20 o C. The

plates are clean, and the width/separation ratio is large so that end e ! ects are

negligible. How far will the water rise between the plates?

Solution

The surface tension at 20 o C is 7%3 × 10 !2 N/m. The weight of the water in the

column , is balanced by the surface tension force.

-,.!' = 2-/COS0

where - is the width of the plates and . is the separation distance. For water

against glass, COS0 ' 1% Solving for , gives

0%5×10 !3 m ×998 kg/m 3 ×9%81 m/s 2

= 0%0149 m = 29%8 mm

.!' = 2×7%3×10 !2 N/m

, = /

Problem 2.4

The kinematic viscosity of helium at 15 o C and standard atmospheric pressure (101

kPa) is 1.14 ×10 !4 m 2 ) s. Using Sutherland’s equation, Þ nd the kinematic viscosity

at 100 o C and 200 kPa.

Solution

From Table A.2, Sutherland’s constant for helium is 79.4 K and the gas constant is

2077 J/kgK. Sutherland’s equation for absolute viscosity is

1 ! =

$ !

3"2 $ ! +*

$ +*

The absolute viscosity is related to the kinematic viscosity by 1 = (! . Substituting

into Sutherland’s equation

! ! ( ! =

$ !

3"2 $ ! +*

$ +*

( ! = ! !

$ !

3"2 $ ! +*

$ +*

From the ideal gas law

! !

! = " !

$ !

( ! = " !

$ !

5"2 $ ! +*

$ +*

The kinematic viscosity ratio is found to be

( ! = 101

373

288

5"2 288+79%4

373+79%4

200

= 0%783

The kinematic viscosity is

( = 0%783×1%14×10 !4 = 8%93×10 !5 m 2 ) s

(

CHAPTER 2. FLUID PROPERTIES

Problem 2.5

Air at 15 o C forms a boundary layer near a solid wall. The velocity distribution

in the boundary layer is given by

3 = 1!EXP(!2 4

5 )

where 3 = 30 m/s and 5 = 1 cm. Find the shear stress at the wall ( 4 = 0)%

Solution

The shear stress at the wall is related to the velocity gradient by

6 = 1 .2

.4 | #=0

Taking the derivative with respect to 4 of the velocity distribution

.4 = 2 3

5 EXP(!2 4

5 )

Evaluating at 4 = 0

.4 | #=0 = 2 3

5 = 2× 30

0%01 = 6×10 3 s !1

From Table A.2, the density of air is 1.22 kg/m 3 + and the kinematic viscosity is

1.46 ×10 !5 m 2 ) s. The absolute viscosity is 1 = !( = 1%22×1%46×10 !5 = 1%78×10 !5

N · s/m 2 % The shear stress at the wall is

6 = 1 .2

.4 | #=0 = 1%78×10 !5 ×6×10 3 = 0%107 N/m 2

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