2016 che2163 tutorial 5

CHE2163 Tutorial 5 (2016)

TUTOR: JANET LEONG

EMAIL: [email protected]

MONASHENGINEERING

Q1

Q4

Q5

2

Question 1

�� = 25 m/s

� = 5 mm

3

Question 1) a)

Schematic & Assumptions

� = 0.5 �Assumptions1. Steady-state2. 1D heat conduction (y-direction) in module3. Uniform � in module4. Negligible radiation heat transfer5. Rex,c = 5 x 105

�� = 25 m/s

� = 5 mm

4

Question 1) a)


� = 0.5 �

�� = 25 m/s

Energy balanceEin – Eout + Egen = Est

0 – qconv + qgen = 0h�AS(TS - T∞) = qV

q =��

�(T� T�)

� = 5 mm

6

Question 1) a) Calculate q

Determine if flow is laminar/turbulent at x = L = 0.7 m (700 mm)

�� =��

�

7


T� =T� + T�

2=

150 + 25

2= 87.5 ℃ (360K)

Table A.4 (air)

∴ Re� =u�L

ν=

(25)(0.7)

22 × 10��= 7.95 × 10�

Tf (K) 360ν.106 (m2/s) 22.02k.103 (W/mK) 30.76Pr 0.698

8


Flow is fully turbulent over module:

� �� =��

�= �. ��

��

��

�̅ =�� + ��

�= �

��

9


Re��

��

=(25)(0.7 +

0.052

)

22.02 × 10��= 8.23 × 10�

Nu��

��

= 0.0296(8.23 × 10�)�� (0.698)

��= 1418

h� =

Nu��

��

k

L +b2

=(1418)(30.76 × 10��)

0.7 +0.05

2

= 60.2 W /m�K

q =h�A�

VT� T� =

h�(bw )

(bwa)T� T� =

h�

aT� T�

∴ q =(60.2)

(0.005)150 25 = �. �� × �� /� �

10

Question 1) b) Find Tmax in module

11


12


13


14


15

Question 2

16

Question 2) a) Find initial rate of heat transfer if Ti = 300C

Schematic

17

Assumptions

1. Constant properties

2. Negligible radiation

3. Negligible effect of conveyor velocity on boundary layer development

4. Isothermal plate

5. Negligible heat transfer from sides of plate

6. Rex,c = 5 x 105


Assumptions

18


q��= 2q = 2h�A�(T� T�)

Find h – laminar/ turbulent flow?

Re� =u�L

νTable A.4 (air)

T� =T� + T�

2=

300 + 20

2= 160℃ (433K)

19


Table A.4 (air)

Re� =u�L

ν=

(10)(1)

(30.36 × 10��)= 3.29 × 10�

Tf (K) 433ν.106 (m2/s) 30.36k.103 (W/mK) 36.11Pr 0.687

20


21


Nu � = 0.664 Re��/�

Pr�/�

Nu � = 0.664 (3.29 × 10�)�/�

(0.687)�/�= 336

h� =k

LNu � =

36.11 × 10��

1336 = 12.1 W /m�K

q��= 2h�A� T� T� = 2(12.1)(1�)(300 20)�� = ��

22

Question 2) b) Find rate of change of plate temperature

Initial energy balance (t=0):

Ein – Eout + Eg = E��

0 2q�� + 0 = ρVC�

dT

dt

2hA� T� T� = ρVC�

dT

dt

dT

dt=

2hA�

ρVC�T� T�

23


Table A.1 (AISI 1010 carbon steel) @ TS = 573K

kp = 49.2 W/ mKCp = 548 J/kg.Kρ = 7832 kg/m3

At t =0,dT

dt=

2(12.1)(1�) 300 20

(7832)(1� × 0.006)(548)��

��= �. ��℃/�

24


Was the lumped capacitance model correct?

Bi =h�L�

k� where L� =

V

A�=

L�δ

L�= δ

Bi =h�δ

k�=

12.1 0.006

49.2= 0.00148 < 0.1

Therefore our assumption is correct.

25

An uninsulated steam pipe is used to transporthigh temperature steam from one building toanother. The pipe is of 0.5 m in diameter, has asurface temperature of 150 oC, and is exposed toambient air at -10 oC. The air moves in cross flowover the pipe with a velocity of 5 m s-1. What is theheat loss per unit length of the pipe? (Answer:3644 W/m)

Question 3


26

Question 3


Assumptions1.Steady-state2.Uniform TS

3.Negligible radiation

27

Question 3) Calculate heat transfer rate from uninsulated pipe

q�� = h�πDL(T� T�) (assume L = 1m)

T� = 0.5 T� + T� = 0.5 150 10 = 80℃ (353K)

Table A.4 (air) Tf (K) ≈ 350ν.106 (m2/s) 20.92k.103 (W/mK) 30Pr 0.7

28

Question 3) Calculate heat transfer rate from uninsulated pipe

Re� =u�D

ν=

(5)(0.5)

20.92 × 10��= 1.196 × 10�

Nu � = 0.3 +0.62Re�

�� Pr

��

1 +0.4Pr

�/� �/�1 +

Re�

282000

��

�/�

Nu � = 242

h� = Nu �

k

D= 242

0.03

0.5= 14.5 W /m�K

�� = �̅�� = ��. � �. � (�) ��

�� = �� /�

29

Question 4

30

Question 4

Schematic

AS = 4 x 10-4 m2

31

Assumptions:

1. Steady-state conditions

2. Conditions over AS are uniform for both situations

3. Conditions over fin length are uniform

4. Flow over pin fin approximates cross-flow over a cylinder

Question 4

Assumptions

32

Question 4) a) Calculate maximum rate of heat transfer

33

q�,�� = M = hPk��A�

��θ�

q�,�� = h πD k��

πD�

4

��

θ�

q�,�� = hk��

π�D�

4

��

θ�


34


Find h and kfin

Choose simplest (Hilpertcorrelation) (Eq. 7.52)

Nu � = CRe��Pr

�� =

h��D

k

35


Choose simplest (Hilpertcorrelation) (Eq. 7.52)

Nu � = CRe��Pr

�� =

h��D

k

36


Find kfin and air properties in order to find h

T� = 0.5 T� + T� = 0.5 127 + 27 = 77℃ (350K)

Tf (K) 350ν.106 (m2/s) 20.92k.103 (W/mK) 30Pr 0.7

Tf (K) 350kfin (W/m.K) 15.60

Table A.1 – AISI 304 Table A.4 – air

37


Calculate ReD and determine C,m from Table 7.2

Re� =u�D

ν=

(5)(0.005)

20.92 × 10��= 1195

38

h�� =k

DCRe�

�Pr��

h�� =30 × 10��

5 × 10��0.683 1195 �.�� 0.7

��

h�� = 98.9 W /m�K

q�,�� = hπ�D�

4k��

�/�

θ�

q�,�� = 98.9π� 5 × 10��

415.66

��

(127 27)

��,�� = �. � �


39

L� =�.��

�(not in formula sheet)

L� = 2.65k��A�

hP

�/�

= 2.65k��(πD�/4)

h(πD)

�/�

L� = 2.65k��D

4h

�/�

= 2.65(15.66)(5 × 10��)

4(98.9)

�/�

L� = 0.0374 m or 37.4 mm

Question 4) b) How long is an ‘infinitely’ long rod?

40

Question 4) c) Calculate fin effectiveness, ��

ε� =q�

q�,�=

q�

h�A�,�θ�

41

q�� = h�A�θ� ∴ h� =q��

A�θ�

ε� =q�

q�,�=

q�

h�A�,�θ�=

q�

q��A�θ�

A�,�θ�

=q�

q��

A�

A�,�

�� =q�

q��

A�

A�,�=

2.2

0.5

0.02�

� 0.005 �/4= ��. �

Question 4) c) Calculate fin effectiveness, ��

42

% =q� q��

q��× 100%

q� = q� + q��

q� = 2.2 Wq�� = h��(A� A�,�)(T� T�)

h�� =q��

A�θ�=

0.5

4 × 10�� 127 27h�� = 12.5 W /m�K

Question 4) d) Calculate % increase in heat rate from AS with installed fin

43

q�� = 12.5 4 × 10��π 0.005 �

4(127 27)

q�� = 0.475 W

∴ q� = 2.2 + 0.475 = 2.675 W

% =q� q��

q��× 100%

% =2.675 0.5

0.5× 100% = ��%

Question 4) d) Calculate % increase in heat rate from AS with installed fin

44

A spherical, underwater instrument pod used to makesoundings and to measure conditions in the water has adiameter of 85 mm and dissipates 300 W.

(a) Estimate the surface temperature of the pod whensuspended in a bay where the current is 1 m/s and thewater temperature is 15 oC. (Answer: 18.8 oC)

(b) Inadvertently, the pod is hauled out of the water andsuspended in ambient air without deactivating the power.Estimate the surface temperature of the pod if the airtemperature is 15 oC and the wind speed is 3 m/s. (Answer:672 oC)

Question 5

45

Question 5

Schematic

Instrument podD = 85 mm

Pe = 300W

Pe = 300W

Ts,a = ?

Ts,w = ?

T∞=15 oC (288K)V = 1 m/s

T∞=15 oC (288K)V = 3 m/s

qcv

qcvWATER

AMBIENTAIR

46

Assumptions

1. Steady-state

2. Constant properties

3. Flow over a smooth sphere

4. Uniform surface temperature

5. Negligible radiation

Question 5

Assumptions

47

Question 5) a) Estimate Ts,w of pod when suspended in flowing water at 15C and 1m/s

Energy balance on spherical pod:E�� E�� + E�� = E��

0 q�� + P� = 0

P� = h�A� T� T� where A� = πD�

T� =P�

h�A�

+ T�

48


Nu � = 2 + 0.4Re�

�� + 0.06Re�

�� Pr�.�

μ

μ�

�/�

All fluid properties evaluated at T∞

EXCEPT µS, which evaluated is at TS

49


50


1. Guess TS = 20oC (293K)2. What is the bulk fluid? Saturated Water, liquid

Table A.6 (Properties of saturated water)

T∞ (K) 288vf x 103 (m3/kg) 1.000µf x 106 (N.s/m2) 1138kf x 103 (W/mK) 594.8Prf 8.06

TS (K) 293µS x 106 (N.s/m2) 1007.4

51


Re� =ρu�D

μ=

u�D

μν�=

(1)(85 × 10��)

(10��)(1138 × 10��)= 7.469 × 10�

Nu � = 2 + 0.4 7.469 × 10� �/� + 0.06 7.469 × 10� �/� (8.06)�.�1138 × 10��

1007.4 × 10��

�/�

Nu � = 514.5

h� = Nu �

k

D= 514.5

594.8 × 10��

85 × 10�� = 3600 W /m�K

T� =P�

h�A�

+ T� =300

(3600)(π 85 × 10��) �)+ 15 = 18.67℃

Initially, we guessed TS = 20oC. The calculated value of TS is 18.67oC.Iterate 5 times TS = 18.8oC.

52

Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 1)

Iteration 11. Guess TS = 100oC (373K)2. What is the bulk fluid? Air

Table A.4 (Properties of air)

T∞ (K) 288μ.107 (N.s/m2) 178.6ν.106 (m2/s) 14.82k.103 (W/mK) 25.34Pr 0.710

TS (K) 373

μ.107 (N.s/m2) 218.3

53


Re� =u�D

�=

(3)(85 × 10��)

(14.8 × 10��)= 1.723 × 10�

Nu � = 2 + 0.4 1.723 × 10� �/� + 0.06 1.723 × 10� �/� (�. ��)�.�178.6 × ��

218.3 × ��

�/�

Nu � = 78.74

h� = Nu �

k

D= 78.74

25.34 × 10��

85 × 10�� = 23.47 W /m�K

T� =P�

h�A�

+ T� =300

(23.47)(� 85 × 10��) �)+ 15 = ��℃

Initially, we guessed TS = 100oC. The calculated value of TS is 578oC. We guess this TS value next.

54


Iteration 21. Guess TS = 578oC (851K)



TS (K) 851

μ.107 (N.s/m2) 384.6

55


Re� =u�D

ν=

(3)(85 × 10��)

(14.8 × 10��)= 1.723 × 10�

Nu � = 2 + 0.4 1.723 × 10� �/� + 0.06 1.723 × 10� �/� (0.710)�.�178.6 × 10��

390.6 × 10��

�/�

Nu � = 68.34

h� = Nu �

k

D= 68.34

25.34 × 10��

85 × 10�� = 20.38 W /m�K

T� =P�

h�A�

+ T� =300

(20.38)(π 85 × 10��) �)+ 15 = 664℃

Initially, we guessed TS = 578oC. The calculated value of TS is 664oC. We guess this TS next.

56





TS (K) 937

μ.107 (N.s/m2) 407.9

57


Re� =u�D

ν=

(3)(85 × 10��)

(14.8 × 10��)= 1.723 × 10�

Nu � = 2 + 0.4 1.723 × 10� �/� + 0.06 1.723 × 10� �/� (0.710)�.�178.6 × 10��

407.9 × 10��

�/�

Nu � = 67.63

h� = Nu �

k

D= 67.63

25.34 × 10��

85 × 10�� = 20.16 W /m�K

T� =P�

h�A�

+ T� =300

(20.16)(π 85 × 10��) �)+ 15 = 670℃

Initially, we guessed TS = 664oC. The calculated value of TS is 670oC. We guess this TS next.

58





TS (K) 943

μ.107 (N.s/m2) 409.5

59


Re� =u�D

ν=

(3)(85 × 10��)

(14.8 × 10��)= 1.723 × 10�

Nu � = 2 + 0.4 1.723 × 10� �/� + 0.06 1.723 × 10� �/� (0.710)�.�178.6 × 10��

409.5 × 10��

�/�

Nu � = 67.57

h� = Nu �

k

D= 67.57

25.34 × 10��

85 × 10�� = 20.14 W /m�K

T� =P�

h�A�

+ T� =300

(20.14)(π 85 × 10��) �)+ 15 = 671℃

Initially, we guessed TS = 670oC. The calculated value of TS is 671oC.

60

Question 5) Summary

2016 che2163 tutorial 5

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