7-Practice Problems.pdf

Chapter 7

Energy Principle

Problem 7.1

Air ßows through a rectangular duct of dimension 1 ft × 5 ft. The velocity proÞle

is linear, with a maximum velocity of 15 ft/s. Find the kinetic energy correction

factor.

Solution

The kinetic energy correction factor is given by

! = 1

# 3

# 3 $"

(1)

Area is 1 × 5 = 5 ft 2 % Since the proÞle is linear, the average velocity is half the

maximum velocity.

# = # MAX &2

= 7%5 ft/s

Pick a di!erential area with a height of $' and width of 5 ft.

$" = 5$'

CHAPTER 7. ENERGY PRINCIPLE

Eq. (1) becomes

! = 1

7%5

# 3 (5$')

"=1 ft

= 0%00237

[#(')] 3 $'

(2)

"=0

Since #(') is a straight line, it can be Þt with an equation of the form

# = ('+) , where ( is slope and ) is intercept. The result is

#(') = 15'

(3)

Combining Eqs. (2) and (3)

15 3

"=1 ft

! = 0%00237

' 3 $'

15 3

"=0

= 0%00237

(1&4)

! = 2

Problem 7.2

Water ßows out of a large tank through a 1-cm diameter siphon tube. The siphon

is terminated with a nozzle of diameter 3 mm. Determine the minimum pressure

in the siphon and determine the velocity of the water leaving the siphon. Assume

laminar ßow and assume all energy losses due to e!ects of viscosity are negligible.

Solution

Let location 1 be coincident with the free surface of the water in the tank. Let

location 2 be the exit of the siphon. The energy equation between 1 and 2 is

* 1

+ +! 1

2, +- 1 +. # = * 2

+ +! 2

2, +- 2 +. $ +. %

Now * 1 = * 2 = 0/# 1 ! 0, - 1 = 3 m, . # = . $ = . % = 0% Since the ßow is laminar,

! 2 = 2% The energy equation simpliÞes to

- 1 = ! 2

# 2

3 = 2

# 2

2×9%8

# 2 = 5%42 m/s

The minimum pressure will occur at the highest point in the siphon; let this be

location 3. The energy equation between 1 and 3 is

* 1

+ +! 1

2, +- 1 +. # = * 3

+ +! 3

2, +- 3 +. $ +. %

Dropping terms that are zero and simplifying gives

- 1 = * 3

+ +! 3

2, +- 3

(1)

To Þnd # 3 , use the continuity principle.

# 3 " 3 = # 2 " 2

# 3 = # 2

" 2

" 3

= 5%42 0%003 2

0%01 2

= 0%488 m/s

Substitute numbers into Eq. (1).

- 1 = * 3

+ +! 3

2, +- 3

3 =

9800 +2 0%488 2

2×9%81 +4

* 3

9800

"1%024 =

* 3 = "10%0 kPa

# 1

# 2

# 1

# 3

* 3

CHAPTER 7. ENERGY PRINCIPLE

Problem 7.3

A pump with an e"ciency of 70% pumps water at 60 & F in a four-in. pipe. Deter-

mine the power required by the pump. Neglect head loss, and assume all kinetic

energy correction factors are unity.

Solution

The energy equation between sections 1 and 2 is

* 1

+ +! 1

2, +- 1 +. # = * 2

+ +! 2

2, +- 2 +. $ +. %

By continuity, # 1 = # 2 / and so the velocity head terms cancel. Also, - 1 = - 2 % The

energy equation simpliÞes to

+ +. # = * 2

. # = * 2 "* 1

= (12/000"2000) lbf/ft 2

62%4 lbf/ft 3

= 160 ft

The ßow rate of water by weight is

+0 = +#"

1 ×0%3333 2

62%4 lbf/ft 2

(3 ft/s)

ft 2

= 16%3 lbf/s

Power is

2 = . # (+0)&3

= (160 ft)×(16%3 lbf/s)

0%7

= 3730 ft-lbf/s

# 1

# 2

* 1

Converting units to horsepower

2 = 3730 ft-lbf/s

550 ft-lbf/(s-hp)

= 6%78 hp

Problem 7.4

The following sketch shows a small, hand-held sprayer to be used by homeowners.

Water ßows through the 6-ft-long by 3/8-in.-diameter hose. The hose is terminated

with a 1/16 in. diameter nozzle and the water exits the nozzle with a speed of 25

ft/s. Air in the tank is pressurized to produce the given ßow. Determine the

air pressure. Head loss in the system is given by . % = 5%0(4&5)

# 2 &2,

Solution

From continuity, the water speed in the hose is

# hose = # nozzle (5 nozzle ) 2

(5 hose ) 2

= 25

= 0%694 ft/s

The energy equation between section 1 and section 2 is

* 1

+ +! 1

2, +- 1 +. # = * 2

+ +! 2

# 2

2, +- 2 +. $ +. %

/ where

4 = 6 ft is the length of the hose, 5 is the diameter of the hose, and # is the

average velocity in the hose. Assume all kinetic energy correction factors are unity.

# 1

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