A sample of the teaching work I do.

1. CHE2163 Tutorial 5 (2016)
TUTOR: JANET LEONG
EMAIL: JANET.LEONG@MONASH.EDU
MONASH
ENGINEERING
Q1
Q4
Q5

2. 2
Question 1
= 25 m/s
= 5 mm

3. 3
Question 1) a)
Schematic & Assumptions
= 0.5
Assumptions
1. Steady-state
2. 1D heat conduction (y-direction) in module
3. Uniform in module
4. Negligible radiation heat transfer
5. Rex,c = 5 x 105
= 25 m/s
= 5 mm

4. 4
Question 1) a)
Schematic & Assumptions
= 0.5
= 25 m/s
Energy balance
Ein – Eout + Egen = Est
0 – qconv + qgen = 0
hAS(TS - T∞) = qV
q = (T T )
= 5 mm

5. 5

6. 6
Question 1) a) Calculate q
Determine if flow is laminar/turbulent at
x = L = 0.7 m (700 mm)
=

7. 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) 360
ν.106 (m2/s) 22.02
k.103 (W/mK) 30.76
Pr 0.698

8. 8
Question 1) a) Calculate q
Flow is fully turbulent over module:
= = .
̅ =
+
=

9. 9
Question 1) a) Calculate q
Re =
(25)(0.7 +
0.05
2
)
22.02 × 10
= 8.23 × 10
Nu = 0.0296(8.23 × 10 ) (0.698) = 1418
h =
Nu k
L +
b
2
=
(1418)(30.76 × 10 )
0.7 +
0.05
2
= 60.2 W/m K
q =
hA
V
T T =
h(bw)
(bwa)
T T =
h
a
T T
∴ q =
(60.2)
(0.005)
150 25 = . × /

10. 10
Question 1) b) Find Tmax in module

11. 11
Question 1) b) Find Tmax in module

12. 12
Question 1) b) Find Tmax in module

13. 13
Question 1) b) Find Tmax in module

14. 14
Question 1) b) Find Tmax in module

15. 15
Question 2

16. 16
Question 2) a) Find initial rate of heat transfer if Ti = 300C
Schematic

17. 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. 18
Question 2) a) Find initial rate of heat transfer if Ti = 300C
q = 2q = 2hA (T T )
Find h – laminar/ turbulent flow?
Re =
u L
ν
Table A.4 (air)
T =
T + T
2
=
300 + 20
2
= 160℃ (433K)

19. 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) 433
ν.106 (m2/s) 30.36
k.103 (W/mK) 36.11
Pr 0.687

20. 20
Question 2) a) Find initial rate of heat transfer if Ti = 300C

21. 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
L
Nu =
36.11 × 10
1
336 = 12.1 W/m K
q = 2hA T T = 2(12.1)(1 )(300 20)
=

22. 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. 23
Question 2) b) Find rate of change of plate temperature
Table A.1 (AISI 1010 carbon steel) @ TS = 573K
kp = 49.2 W/mK
Cp = 548 J/kg.K
ρ = 7832 kg/m3
At t =0,
dT
dt
=
2(12.1)(1 ) 300 20
(7832)(1 × 0.006)(548)
= . ℃/

24. 24
Question 2) b) Find rate of change of plate temperature
Was the lumped capacitance model correct?
Bi =
hL
k
where L =
V
A
=
L δ
L
= δ
Bi =
hδ
k
=
12.1 0.006
49.2
= 0.00148 < 0.1
Therefore our assumption is correct.

25. 25
An uninsulated steam pipe is used to transport
high temperature steam from one building to
another. The pipe is of 0.5 m in diameter, has a
surface temperature of 150 oC, and is exposed to
ambient air at -10 oC. The air moves in cross flow
over the pipe with a velocity of 5 m s-1. What is the
heat loss per unit length of the pipe? (Answer:
3644 W/m)
Question 3
Schematic & Assumptions

26. 26
Question 3
Schematic & Assumptions
Assumptions
1.Steady-state
2.Uniform TS
3.Negligible radiation

27. 27
Question 3) Calculate heat transfer rate from uninsulated
pipe
q = hπDL(T T ) (assume L = 1m)
T = 0.5 T + T = 0.5 150 10 = 80℃ (353K)
Table A.4 (air) Tf (K) ≈ 350
ν.106 (m2/s) 20.92
k.103 (W/mK) 30
Pr 0.7

28. 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 +
0.4
Pr
/ /
1 +
Re
282000
/
Nu = 242
h = Nu
k
D
= 242
0.03
0.5
= 14.5 W/m K
= ̅ = . . ( )
= /

29. 29
Question 4

30. 30
Question 4
Schematic
AS = 4 x 10-4 m2

31. 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. 32
Question 4) a) Calculate maximum rate of heat transfer

33. 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. 34
Question 4) a) Calculate maximum rate of heat transfer
Find h and kfin
Choose simplest (Hilpert correlation) (Eq. 7.52)
Nu = CRe Pr =
h D
k

35. 35
Question 4) a) Calculate maximum rate of heat transfer
Choose simplest (Hilpert correlation) (Eq. 7.52)
Nu = CRe Pr =
h D
k

36. 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) 350
ν.106 (m2/s) 20.92
k.103 (W/mK) 30
Pr 0.7
Tf (K) 350
kfin (W/m.K) 15.60
Table A.1 – AISI 304 Table A.4 – air

37. 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. 38
h =
k
D
CRe Pr
h =
30 × 10
5 × 10
0.683 1195 . 0.7
h = 98.9 W/m K
q , = h
π D
4
k
/
θ
q , = 98.9
π 5 × 10
4
15.66 (127 27)
, = .
Question 4) a) Calculate maximum rate of heat transfer

39. 39
L =
.
(not in formula sheet)
L = 2.65
k A
hP
/
= 2.65
k (πD /4)
h(πD)
/
L = 2.65
k 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. 40
Question 4) c) Calculate fin effectiveness,
ε =
q
q ,
=
q
h A , θ

41. 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. 42
% =
q q
q
× 100%
q = q + q
q = 2.2 W
q = h (A A , )(T T )
h =
q
A θ
=
0.5
4 × 10 127 27
h = 12.5 W/m K
Question 4) d) Calculate % increase in heat rate from AS with
installed fin

43. 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. 44
A spherical, underwater instrument pod used to make
soundings and to measure conditions in the water has a
diameter of 85 mm and dissipates 300 W.
(a) Estimate the surface temperature of the pod when
suspended in a bay where the current is 1 m/s and the
water temperature is 15 oC. (Answer: 18.8 oC)
(b) Inadvertently, the pod is hauled out of the water and
suspended in ambient air without deactivating the power.
Estimate the surface temperature of the pod if the air
temperature is 15 oC and the wind speed is 3 m/s. (Answer:
672 oC)
Question 5

45. 45
Question 5
Schematic
Instrument pod
D = 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
AMBIENT
AIR

46. 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. 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 = hA T T where A = πD
T =
P
hA
+ T

48. 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

49. 49
Question 5) a) Estimate Ts,w of pod when suspended in
flowing water at 15C and 1m/s

50. 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) 288
vf x 103
(m3
/kg) 1.000
µf x 106
(N.s/m2
) 1138
kf x 103
(W/mK) 594.8
Prf 8.06
TS (K) 293
µS x 106
(N.s/m2
) 1007.4

51. 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
hA
+ 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. 52
Question 5) b) Estimate Ts,a of pod when suspended in
flowing air at 15C and 3m/s (ITERATION 1)
Iteration 1
1. Guess TS = 100oC (373K)
2. What is the bulk fluid? Air
Table A.4 (Properties of air)
T∞ (K) 288
μ.107 (N.s/m2) 178.6
ν.106 (m2/s) 14.82
k.103 (W/mK) 25.34
Pr 0.710
TS (K) 373
μ.107 (N.s/m2) 218.3

53. 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
hA
+ 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. 54
Question 5) b) Estimate Ts,a of pod when suspended in
flowing air at 15C and 3m/s (ITERATION 2)
Iteration 2
1. Guess TS = 578oC (851K)
Table A.4 (Properties of air)
T∞ (K) 288
μ.107 (N.s/m2) 178.6
ν.106 (m2/s) 14.82
k.103 (W/mK) 25.34
Pr 0.710
TS (K) 851
μ.107 (N.s/m2) 384.6

55. 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
hA
+ 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. 56
Question 5) b) Estimate Ts,a of pod when suspended in
flowing air at 15C and 3m/s (ITERATION 3)
Iteration 3
1. Guess TS = 664oC (937K)
Table A.4 (Properties of air)
T∞ (K) 288
μ.107 (N.s/m2) 178.6
ν.106 (m2/s) 14.82
k.103 (W/mK) 25.34
Pr 0.710
TS (K) 937
μ.107 (N.s/m2) 407.9

57. 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
hA
+ 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. 58
Question 5) b) Estimate Ts,a of pod when suspended in
flowing air at 15C and 3m/s (ITERATION 4)
Iteration 4
1. Guess TS = 670oC (943K)
Table A.4 (Properties of air)
T∞ (K) 288
μ.107 (N.s/m2) 178.6
ν.106 (m2/s) 14.82
k.103 (W/mK) 25.34
Pr 0.710
TS (K) 943
μ.107 (N.s/m2) 409.5

59. 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
hA
+ T =
300
(20.14)(π 85 × 10 ) )
+ 15 = 671℃
Initially, we guessed TS = 670oC.
The calculated value of TS is 671oC.

60. 60
Question 5) Summary