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ME3122E - Tutorial Solution 4

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Problem Set 4-Solutions

1. Hot exhaust gases used in a finned-tube cross-flow heat exchanger heat 2.5kg/s of water from 35 to 85oC. The gases [cp = 1.09 kJ/kg.K] enter at 200oC and leave at 93oC. The overall heat-transfer coefficient is 180W/m2K. Calculate the area of the heat exchanger using (a) the LMTD approach and (b) the effectiveness – NTU method. (c) If the water flow rate is reduced by half, while the gas flow rate is maintained constant along with the fluid inlet temperatures. Calculate the percentage reduction in heat transfer as a result of this reduced flow rate. Assume that the overall heat transfer coefficient remains the same. [Ans: (a) 37.8 m2 (b) 37.8 m2 (c) 15%.]

Solution: a) 1 2 1 2 2 1 1 1 1 2 2 1

ln(

)

(200 85) (93 35)

ln(115 / 58)

83.3

85 35

0.303

(200 35)

200 93

2.14

(85 35)

From crossflow hx chart :

0.92

. .

2.5 4180 50 0.92 180

83.3

LMTD LMTD

T

T

T

T

T

C

t

t

P

T

t

T

T

R

t

t

F

Q

mC t

UA F T

A

A

  

 

 

2

37.8m

U=180W/m2K Water t1=350C m =2.5 Hot gases T1=2000C t2 =850C T2 =930C A T Th1 200 Th2 93 Tc1 85 Tc2 35 1 2

(2)

b) min max min max min 2

:

(200 93)

2.5 4180 (85 35)

4883

(

)

2.5 4180 10450

(

)

4883

0.467

2.5 4180

200 93

0.648

200 35

using

chart

1.4,

1.4 4883

37.98

180

g g g g w w g

Energy balance

m c

m c

C

gas

m c

C

water

C

C

UA

NTU

NTU

C

A

m

c) min max min min max 2 2

4883

2.5 4180 / 2 5225

1.4 (no change)

4883/ 5225 0.935

0.55

200 35

90.75 200

109.25

4883 90.75

( )

( )

4.18 2.5 50 488

( )

g g w w g g g g g g

m c

C

m c

C

UA

NTU

C

C

C

Cross flow hx chart

T

T

T

T

C

Q

C

T

Q a

Q c

Q a

 

 

 

3 90.75

15%

4.18 2.5 50

(3)

2. A shell-and-tube heat exchanger operates with two shell passes and four tube passes. The shell fluid is ethylene glycol, which enters at 140oC and leaves at 80oC with a flow rate of 4500 kg/h. Water flows in the tubes, entering at 35oC and leaving at 850C. The overall heat-transfer coefficient for this arrangement is 850 W/m2K. Calculate the flow rate of water required and the area of the heat exchanger. [Ans: 0.984 kg/s, 5.24 m2].

The flow rate of glycol to the exchanger is reduced in half with the entrance temperatures of both fluids remaining the same. What is the water exit

temperature under these new conditions, and by how much is the heat-transfer rate reduced? [Ans: 70.9oC, 28.2%]

Solution: 3 max min min max

)

4500 / 3600

/ ,

2.742

/

4500

2.742 (140 80) 205.7

3600

4.180 (85 35)

0.984

/

0.984 4.180 10

4113

4500

2742 3428

3600

3428

0.833

4113

140 80

0.

140 35

g g w w w g g

i

m

kg s

c

kJ kgK

Q

kW

m

m

kg s

C

C

C

C

C

C

min 2o 2

571

/

1.3

850W/m C.

1.30 3428 / 850 5.24

Hence

NTU chart gives

NTU

UA C

given U

A

m

 

A T 140 80 85 35 1 2 Ethylene glycol water

(4)

min min min max 1 2 2 2

)

4500 /(2) 3600

3428 / 2

1.30

2.6

0.5

0.833/ 2 0.417

0.82

0.82;

140 35

140

86.1

53.9

(3428 / 2) 86.1 147.6

( )

( )

205.7 147.

( )

g g g g g g o g g g

ii m

C

C

UA

NTU

C

C

C

chart

T

T

T

T

C

Q

C

T

kW

Q i

Q ii

Q i

 

 

1 1

6

28.2%

205.7

147.6

35.9

35

0.984 4.18

35 35.9 70.9

water w w

T

T

T

C

(5)

3. A small steam condenser is designed to condense 0.76 kg/min of steam at 85 kN/m2 with cooling water at 10oC. The exit water temperature is not to exceed 57oC. The overall heat-transfer coefficient is 3400 W/m2K. Calculate the area required for a double-pipe heat exchanger. Use both LMTD and effectiveness-NTU methods. [Ans: 0.145m2]. Solution: g 2 fg f 1 2 LMTD 1 2 2 i) Steam is condensed at 85KN/m h =2.27 MJ/kg 0.76 Q = m h 2270 = 28.75kW 60 T ln( / ) (95-10)-(85-57) = 58.3 ln(85 / 38) Q=UA LMTD 28.75 A= 0.145 3.4 58.3 T T T T C m                 2 1 1 1 max min c c c min min min ii) C , T T 57 10 = 0.553 T T 95 10

For heat exchanger with steam condensing, 1

=0.553=1 , giving NTU =0.805 UA NTU= 0.805, Q= (57 10) 28753 611.77 Su h NTU NTU steam C water e e C C C

                   2

bstituting qives A=0.145m

10 57 Steam Tsat =950C A 1 2 water A A T

(6)

4. A shell-and-tube heat exchanger consists of 135 thin-walled tubes in a double-pass arrangement, each of 12.5 mm diameter with a total surface area 47.5 m2. Water (the tube-side fluid) enters the heat exchanger at 15oC and 6.5 kg/s and is heated by exhaust gas entering at 200oC and 5 kg/s. The gas may be assumed to have the properties of atmospheric air, and the overall heat transfer coefficient is approximately 200 W/m2. What are the gas and water outlet temperatures? Assuming fully developed flow, what is the tube-side convection coefficient? [Ans: To(water) = 41.6oC, h = 2320 W/m2K.]

Solution: i) 2 2 min max min max

5

/ ;

1005 /

6.5

/ ;

4200 /

200 /

;

47.5

5 1005

6.5 4200

5 1005

0.184

6.5 4200

200 47.5

1.89

5 1005

Shell and tube Heat Exchanger

-NTU chart

g pg w pw g w

m

kg s

c

J kgK

m

kg s c

J kgK

U

W m K

A

m

C

C

C

C

C

C

NTU

 

2 2 2 1 1 g min

0.78

C (200

)

200

0.78

(200 15)

200 15

55.7

5 1005 (200 55.7) 6.5 4200 (

15)

41.6

g g g w w

T

T

C

T

C

Q

T

T

C

 

  A T 2000C Tg2 Tw1 15 1 2 Gas water

(7)

ii) bw 2 3 6 0.8 0 w

Water in tube: properties are evaluated at

15 41.6

T

28.3

2

u

Re

;

4

4

4 (6.5 /135)

Re

(12.5 10 ) 855 10

=5736>2000 turbulent tube flow

hd

Nu=

0.023Re Pr

k

d d

C

d

d

m

u

m

d

  

   

.4 w 2

Substituting for Pr=5.83, k

0.613 /

,

0.0125

gives h=2320W/m

W mK d

m

K

(8)

5. A single-pass cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30oC to 80oC at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225oC and 100oC, respectively. If the overall heat transfer coefficient is 200W/m2K, estimate the required surface area. [Ans: 35.2m2].

Solution:

2 1 LMTD

3 / ; 4.2 / ,

1.005 /

Use LMTD Method, counter flow equivalent all the temperatures are known

q=m ( ) 3 4.20 (80 30) 630kJ/s (225-80)-(100-30) T = 103 ln(145/70) R(or Z)= w w g w w w w m kg s c KJ kgK c KJ kgK c T T C                LMTD 2 min 225-100 2.5 80-30 80 30 0.26 225 30

for single pass, crossflow hx, 1mix, 1unmixed, correction factor F=0.92

Hence

q=630=UAF T =0.200 A 0.92 103

giving A=33.2m

method just as straight forward, C P NTU

          1 2 1 2 max min max , C C Calculate C 225 110 0.64 225 30 , , gas g g water w w g g g w C m c C m c T T T T

from chart get NTU hence A

            A T 2250C 100 80 30 1 2 Gas water

(9)

6. The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at 30oC and 0.53 kg/s. Oil at 0.026 kg/s enters at 75oC and flows through a tube of 10 mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil-side heat transfer coefficient. If the overall convection coefficient is 53 W/m2K and the total heat transfer area is 1 m2, determine the effectiveness. What is the exit temperature of the oil? [Ans: 46.2oC]

Solution:

2 2

o2

cross-flow hx, fluids unmixed

oil inside tube, 75 ( ), 0.026kg/s air, 30 (inlet), 0.53kg/s

U=53W/m , 1

Constant heat flux for tube, find h(oil side)

Another case of T (outlet) unknown assume C inlet C K Am   o oil o2 b b 3 4 2 T 45 T (75 45) / 2 60

(Other procedure may lead to non-integer value of T ) Tables: engine oil @ 60 :

864 / , 2.047 / , 0.839 10 / , 0.14 / Pr 1050 p C C C kg m C kJ kgK m s k W mK                 

(10)

4 2 h h min c c m

4

Re

4 0.026

0.01 864 (0.839 10 )

45.7 2000

flow inside tube is laminar, constant heat flux

hd

0.01

Nu= 4.36 =

;

k

0.14

61.04 /

oil: m

= 0.026×2047=53.2=C

air: m

= 0.53×1005=533=C

oil

ud

m

d

h

h

W m K

c

c

 

1 1 1 ax min max min min min

C

0.10 ;

C

53 1

1.0

C

53.2

From ε-NTU chart for cross flow hx,both fluid unmixed

=0.64

C (

)

75

=0.64=

C (

)

75 30

75 0.64 45 46.2

(not far from assumed value of 45)

o o o h h h h c h

UA

NTU

T

T

T

T

T

T

C

(11)

7. Water flows over a 3-mm-diameter sphere at 6 m/s. The free-stream temperature is 38oC, and the sphere is maintained at 93oC. Calculate the heat-transfer rate.

[Ans: 84.9W].

Solution:

o

3 3

at T =38 C, properties for water =993kg/m , 0.682 10 / 0.63 / , Pr 4.55 kg ms k W mK        

Should not be evaluated at Tf =38+93/2=65.50C because Nu eqn is evaluated at free steam condition for liquid (water)

Water u=6m/s T=380C d=3mm Ts=930C 3 0.25 -0.3 0.54 4 w 2 2 993 0.003 6 Re 26, 210 2000 0.682 10 appropriate Nu expression is NuPr 1.2 0.53Re 130.09 μ evaluated at 93 , 3.06 10 / 250.4 0.63 250.42 52589 / 0.003 (4 ) (93 w w C kg ms hd Nu k h W m K Q r h                              38) 84.9W

(12)

8. Air at 90oC and 1 atm flows past a heated 1/16-in-diameter wire at a velocity at 6 m/s. The wire is heated to a temperature of 150oC. Calculate the heat transfer per unit length of wire. [Ans: 59.86 W/m].

Solution: f 5 3 Air at 1 atm T 90 , 6 /

First determine Re of the flow

Air properties are evaluated at the film temperature 90 150 T 120 393 2 2 1.01325 10 0.898 / 287 393 1.013 / ; 2.25 f w f f p f C u m s T T C K P kg m RT c kJ kgK                       5 3 d 5 1/3 0.466 1/3 6 10 / 0.03314 / ; Pr 0.690 0.898 6 (1.5875 10 ) Re 379.1 2.256 10 : 0.683, 0.466 Pr 0.683(Re ) Pr convection heat transfer coefficient

.0. f n d d kg ms k W mK u d From table C n u d hd Nu C k k h d                              0.466 1/3 0.466 0.33 3 2 " w 3 683(Re ) Pr 0.03314 0.683 (379.1) 0.690 1.5875 10 200.5 /

Heat transfer from wire to air

( ) .( )( ) .( )(T ) 200.5 (1.5875 10 )(150 90) 60.0 / d w w W m K q hA T T h dl T T q q h d T l W m                            

References

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