2016 CHE2163 Tutorial 5
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Transcript of 2016 CHE2163 Tutorial 5
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)
Schematic & Assumptions
� = 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
Question 1) a) Calculate q
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
Question 1) a) Calculate q
Flow is fully turbulent over module:
� �� =���
�= �. ����� ��
����
��
�̅ =�� + ��� �
�= �
����
9
Question 1) a) Calculate q
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 = �. �� × ��� � /� �
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
Question 2) a) Find initial rate of heat transfer if Ti = 300C
Assumptions
18
Question 2) a) Find initial rate of heat transfer if Ti = 300C
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
Question 2) a) Find initial rate of heat transfer if Ti = 300C
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
21
Question 2) a) Find initial rate of heat transfer if Ti = 300C
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
Question 2) b) Find rate of change of plate temperature
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
Question 2) b) Find rate of change of plate temperature
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
Schematic & Assumptions
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
�� = �̅��� �� �� = ��. � �. � (�) ��� ��
�� = ���� � /�
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
33
q�,��� = M = hPk���A�
��θ�
q�,��� = h πD k���
πD�
4
��
θ�
q�,��� = hk���
π�D�
4
��
θ�
Question 4) a) Calculate maximum rate of heat transfer
34
Question 4) a) Calculate maximum rate of heat transfer
Find h and kfin
Choose simplest (Hilpertcorrelation) (Eq. 7.52)
Nu � = CRe��Pr
�� =
h��D
k
35
Question 4) a) Calculate maximum rate of heat transfer
Choose simplest (Hilpertcorrelation) (Eq. 7.52)
Nu � = CRe��Pr
�� =
h��D
k
36
Question 4) a) Calculate maximum rate of heat transfer
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
Question 4) a) Calculate maximum rate of heat transfer
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)
��,��� = �. � �
Question 4) a) Calculate maximum rate of heat transfer
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?
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
Question 5) a) Estimate Ts,w of pod when suspended in flowing water at 15C and 1m/s
Nu � = 2 + 0.4Re�
�� + 0.06Re�
�� Pr�.�
μ
μ�
�/�
All fluid properties evaluated at T∞
EXCEPT µS, which evaluated is at TS
50
Question 5) a) Estimate Ts,w of pod when suspended in flowing water at 15C and 1m/s
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
Question 5) a) Estimate Ts,w of pod when suspended in flowing water at 15C and 1m/s
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
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 1)
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
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 2)
Iteration 21. Guess TS = 578oC (851K)
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) 851
μ.107 (N.s/m2) 384.6
55
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 2)
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
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 3)
Iteration 31. Guess TS = 664oC (937K)
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) 937
μ.107 (N.s/m2) 407.9
57
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 3)
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
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 4)
Iteration 41. Guess TS = 670oC (943K)
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) 943
μ.107 (N.s/m2) 409.5
59
Question 5) b) Estimate Ts,a of pod when suspended in flowing air at 15C and 3m/s (ITERATION 4)
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.