CN2125 Heat and Mass Transfer

Review #2

8a

1

3 40.40.62

v v L v fg pv

o v s sat

k g h C Th

D T T

Regime V: WWWR 21-7

**

11

12

Derivation of Equation WWWR 25-10

CN2125 Heat and Mass Transfer

2017-2018; Final Examination

Answer all questions.

I. (1), (2), (3).

II. (4), (5), (6).

III. (7), (8).

-------------------------------

IV. *****

V. *****

Open-Book Examination:

1. Textbooks and all references

2. Homework and Tutorial Solutions.

3. Certified calculators.

Hot Topics:

(i) Steady Heat Conduction: Basic definitions;

Differential equations and boundary conditions;

Thermal resistor models for composite walls. Critical

thickness of insulation. Uniform and non-uniform heat

generation and the resulting temperature profiles in

different coordinate systems. (ii) Unsteady Heat

Conduction: Lump parameter analysis; Temperature-

Time charts for simple geometrical shape (1-D)

(iii) Energy- and Momentum Transfer Analogies:

Application to pipe flow. (iv) Natural Convection:

Correlations for spheres and cylinders. (v) Natural

convection for vertical and horizontal cylinders. Forced

Convection: Laminar and Turbulent Pipe flows. Cross

flow past through spheres. (vi) Boiling and

Condensation: Nucleate and film boiling; Film

condensation on vertical plate; (vii) heat exchangers;

(viii) Mass Transfer Fundamentals: Estimation of gas

and liquid phase diffusivities. Pore diffusion.

To be addressed by Dr.

Praveen Linga

CN2125 Heat and Mass Transfer Quiz #1 (8 March, 2018)

Venue: MPSH2A. Time: 10:00-11:00am.

Covering Range: Week 1-5 Materials

Student Name:_________________ Matriculation Number:__________

1. You can estimate the time required to cook fish ball noodles by applying the theory of

unsteady heat conduction in CN2125 Heat and Mass Transfer. It is assumed that the

cooking is done when the center of the fish ball reaches the temperature of 95oC. The fish

balls are 3-cm in diameter, cooked in 100oC water from an initial temperature of 10oC. If

the convective heat-transfer coefficient between the fish-balls and water is 4,000 W/m2-K

and the thermal conductivity (k) and thermal diffusivity () of fish balls are 0.658 W/m-K

and 1.60 x 10-7 m2/s respectively. Construct you solution by referring to Charts for Solution

of Unsteady Transport Problems (see Appendix F3 or equivalent chart in any of the

following textbooks, WWWR, WRF or Wiley Custom Learning Version 2018) to

determine the temperature of the sphere. State your assumptions, if any.

Give your brief answer here:

Bi = hV/kA=3400/0.658*(0.015/3)= 30.395

To use the temperature-time chart (Figure F3) to solve this question.

Y = (100-95)/(100-10) = 0.056

n = 0

m = k/hx1= 0.658/(4000*0.015)= 0.0109

Referring to Figure F6, X = 0.39 = t/x12 (0.3<X<0.5 is accepted due to eye-ball

interpolation)

t = 548seconds.

2. A furnace wall consisting of 0.25m of fire clay brick (thermal conductivity kA = 1.13

W/m-K), LB m of kaolin (thermal conductivity kB = 1.45 W/m-K), and a 0.10m outer layer

of masonry brick (thermal conductivity kC = 0.66 W/m-K) is exposed to furnace gas at

1350K with air at 300K adjacent to the outside wall. The inside and outside convective

heat transfer coefficients are 115 and 23 W/m2-K, respectively. Determine the thickness

(LB) of kaolin such that the outside temperature of the masonry brick cannot exceed 320K.

Give your brief answer here:

Assume A=1 m2

𝑞 = ℎ𝑖𝑛𝐴(𝑇𝑜 − 𝑇𝑐𝑜𝑙𝑑 𝑎𝑖𝑟) = 23 × 1 × (320 − 300) 𝑊 = 460 𝑊

𝑅ℎ𝑖 =1

ℎ𝑖𝐴=

1

115 𝐾/𝑊 = 0.0087 𝐾/𝑊

𝑅ℎ𝑜 =1

ℎ𝑜𝐴=1

23 𝐾/𝑊 = 0.0435 𝐾/𝑊

𝑅𝑘𝐴 =𝐿𝐴𝑘𝐴𝐴

=0.25

1.13 𝐾/𝑊 = 0.221 𝐾/𝑊

𝑅𝑘𝐵 =𝐿𝐵𝑘𝐵𝐴

=𝐿𝐵1.45

𝐾/𝑊

𝑅𝑘𝐴 =𝐿𝐶𝑘𝐶𝐴

=0.1

1.13 𝐾/𝑊 = 0.221 𝐾/𝑊

𝑞 = ℎ𝑖𝑛𝐴(𝑇𝑜 − 𝑇𝑐𝑜𝑙𝑑 𝑎𝑖𝑟) = 23 × 1 × (320 − 300) 𝑊 = 460 𝑊

kA kB kC

LA =

0.25m

LB LC =

0.1m

Ti T0 < 320 K

Air

1350K,

h = 115

W/m2-K

Air

300K,

h = 23

W/m2-K

𝑞 =𝑇ℎ𝑜𝑡 𝑎𝑖𝑟 − 𝑇𝑐𝑜𝑙𝑑 𝑎𝑖𝑟

𝑅ℎ𝑖 + 𝑅ℎ𝑜 + 𝑅𝑘𝐴 + 𝑅𝑘𝐵 + 𝑅𝑘𝐶=

1050

0.4142 + 𝐿𝐵/1.45= 460

Solve LB=2.7 m

3. Determine the steady-state surface temperature of an electric cable, 25 cm in diameter,

which is suspended horizontally in still air in which heat is dissipated by the cable at a

rate of 27 W per meter of length. The air temperature is 20oC. Since the cable has a

length much greater than its diameter, you can consider a 1-D heat transfer in the radial

direction of the cable. The corresponding heat transfer coefficient (or Nusselt number)

for a horizontal cylinder can be estimated by the Churchill and Chu correlation.

Recommended initial guess for the question is given by 304K. .

Give your brief answer here:

𝐷 = 25 𝑐𝑚 ≡ 0.25𝑚

𝐿 = 10 𝑚 𝑞

𝐿= 25

𝑊

𝑚

𝑇∞ = 20𝑜𝐶 ≡ 293 𝐾 Let the initial guess be T = 304 K.

a) Find Tfilm (1 mark)

𝑇𝑓𝑖𝑙𝑚 =𝑇 + 𝑇∞2

𝑇𝑓𝑖𝑙𝑚 =304 𝐾 + 293 𝐾

2= 298.5 𝐾

(1 mark deducted if 𝑻𝒇𝒊𝒍𝒎 was not evaluated, 0.5 marks deducted for errors in

calculation)

b) Obtain values for 𝑔𝛽𝜌2

𝜇2, 𝑃𝑟 𝑎𝑛𝑑 𝑘 using the film temperature evaluated at a) (1 mark)

From Appendix I,

Using linear interpolation, (1 mark)

𝑔𝛽𝜌2

𝜇2= 1.3636 × 108 𝐾−1𝑚−3

𝑃𝑟 = 0.708375

𝑘 = 2.6122 × 10−2𝑊

𝑚.𝐾

Note: Values above will differ depending on the Tfilm obtained in a).

[1 mark deducted if interpolation was not done. If interpolation was done, 0.5 – 1

mark is deducted depending on the number of values above that is incorrect. A

maximum of 2 marks is deducted if there is no sign that an Appendix was used in

obtaining the values (taking into account the marks awarded for interpolation and

for correct values).]

c) Calculate Grashof Number, Gr (1 mark)

𝐺𝑟 =𝑔𝛽𝜌2

𝜇2𝐷3∆𝑇 = 1.3636 × 108 𝐾−1𝑚−3(0.25 𝑚)3(304 𝐾 − 293 𝐾) = 2.34 × 107

(1 mark deducted if the equation used to calculate the Grashof Number is incorrect,

0.5 marks deducted for errors in calculation)

d) Calculate Rayleigh Number, Ra (1 mark)

𝑅𝑎 = 𝐺𝑟. 𝑃𝑟 = 2.344 × 107(0.708375) = 1.66 × 107

(1 mark deducted if the equation used to calculate the Rayleigh Number is incorrect,

0.5 marks deducted for errors in calculation)

e) Use the correlation by Churchill and Chu provided to calculate the Nusselt Number, NuD

since Ra is within the required range, (1 mark)

𝑁𝑢𝐷 =

{

0.60 +0.387(𝑅𝑎𝐷)

1

6

[1 + (0.559

Pr)

9

16]

8

27

}

2

𝑁𝑢𝐷 =

{

0.60 +0.387(1.66 × 107)

1

6

[1 + (0.559

0.708375)

9

16]

8

27

}

2

𝑁𝑢𝐷 = 32.84

(0.5 - 1 mark deducted for errors in calculation, depending on the severity of the

error)

f) Use the Nusselt Number obtained to determine the heat transfer coefficient, h (1 mark)

𝑁𝑢𝐷 =ℎ𝐷

𝑘

ℎ =𝑁𝑢𝐷𝑘

𝐷=32.84(2.6122 × 10−2𝑊 𝑚−1𝐾−1)

0.25𝑚= 3.43 𝑊 𝑚−2𝐾−1

(1 mark deducted if the equation used to calculate the Grashof Number is incorrect,

0.5 marks deducted for errors in calculation)

g) Using Newton’s Law of Cooling and the heat transfer coefficient obtained to validate if

the initial guess is accurate (1 mark)

𝑞 = ℎ𝐴𝑆(𝑇 − 𝑇∞) 𝑞 = ℎ(𝜋𝐷𝐿)(𝑇 − 𝑇∞) 𝑞

𝐿= 𝜋𝐷ℎ(𝑇 − 𝑇∞)

𝑇 =

𝑞

𝐿

𝜋𝐷ℎ+ 𝑇∞ =

27𝑊𝑚−1

𝜋(0.25𝑚)(3.43 𝑊 𝑚−2𝐾−1)+ 293𝐾 = 303.02 𝐾

(1 mark deducted if the equation used to calculate the surface temperature is

incorrect, 0.5 marks deducted for errors in calculation)

Since the surface temperature calculated is quite close to the initial guess (within a

difference of 3K or within 5% deviation), the initial guess is valid and the steady state

surface temperature is 304 K.

(1 mark is deducted if there is no satisfactory conclusion provided)

(A base score of 2 marks is awarded if a general flow of the methods above can be

seen in their answer)

Grader’s Remarks (Shen Ye, Office E4-04-11 / Phone: 93966915, E-mail:

shen_ye@u.nus.edu; Teoh Jia Heng Office E4-04-11 / E-mail: e0225106@u.nus.edu)

Average = 27.84

Standard Deviation = 2.72

Question 1:

Question 1 tests the knowledge of unsteady-state heat transfer. Students need to evaluate

three dimensionless numbers. From eye-ball interpolation on Appendix F3, the reading of

Relative time X can then be calculated.

Common mistakes:

1. Some students calculated the wrong Bi number.

2. Some students failed to get the correct number by eye-ball interpolation. Students should

be aware of the y-axis of figure. X reading between 0.3 to 0.5 is accepted in this question.

3. For sphere, x1 is the radius instead of diameter.

Question 2:

Question 2 is about using thermal circuit-in-series model to calculate the thickness for a

layer material (kaolin) to keep the outside temperature below 320K. Some students

formulated part of the five resistances but did not finish the questions completely to present

the values of LB. Some students had made mistakes in the calculation and obtain wrong

value in LB.

Question 3:

For Question 3, the method was mainly focused on during the grading process. Students

were awarded a large bulk of the marks if they could show the steps to estimate the surface

temperature in an iteration. Students were penalized little for incorrect interpolation,

absence of interpolation, incorrect values and absence of values.

Common mistakes include:

1. Using T = 20 C (temperature of the surrounding air) to determine the properties of air.

The properties of air should be determined at the film temperature.

2. Using the length of the wire and not the diameter of the wire as the characteristic length

when determining the Grashof Number and Nusselt Number.

3. Using the cross-sectional area of the wire and not the surface area of the wire when

applying Newton’s Law of Cooling.

4. Neglecting the fact that the value of gβρ2

μ2 directly obtained from the table in the

Appendix should be further multiplied with 108. There were also instances whereby the

thermal conductivity values (k) directly obtained was not further multiplied with 10-2.

5. Most students used linear interpolation to determine the properties of air at the film

temperature but some students referred to the properties of air at T = 300K instead, which

will yield different values, more so for 𝑔𝛽𝜌2

𝜇2. However, this is a minor issue.

Overall, majority of the students have done well in this question and were able to list down

all the steps required to estimate the surface temperature.

Dear CN2125 Students:

CN2125 Homework Assignment # 2 (Due 11:45am, March 13, 2018)

Attached please find the list of three problems for the second assignment. Please take

note of the extra information and hints given at the end of each question. We will

collect the assignment at the end of our lecture on March 13, 2018@ 11:35am.

Sincerely yours,

Chi-Hwa Wang, Co-instructor CN2125

(1) WWWR 19.26

Water at 60 oF enters a 1-in ID tube which is used to cool a nuclear reactor. The water flow

rate is 30 gal/min. Determine the total heat transfer, exiting water temperature, and the wall

temperature at the exit of a 15-ft long tube if the tube wall condition is one of uniform heat

flux of 500 Btu/hr ft2.

Additional information:

1 gal = 0.133681 ft3 or 1 ft3 = 7.48 gal

At 60 oF,

cp of water = 1.0 Btu/lboF

density of water = 62.3 lbm/ft3

viscosity of water = 0.76 10-3 lbm/ft sec

Pr = 8.07

Hints:

Take the initial guess of wall temperature = 60.6 oF

smooth pipe (roughness, e = 0.0) may be assumed

Prandtl analogy may be applied for turbulent flow

q = 500(15/12) = 1960 Btu/hr

q = mcpT = 1960

F131.0

0.1603.6248.7

30

1960T o

Tw,exit = 60.13 oF

T (bulk mean) = (60.13 + 60)/2 = 60.065 oF

Take the initial guess of wall temperature = 60.6 oF

Film temperature = (60.065 + 60.6)/2 = 60.33 60 oF

83731

1076.0

6014144

48.730

1213.62

Re3

(turbulent)

From Moody diagram, friction factor Cf = 0.00465

Using Prandtl analogy,

4

f

f 106.8107.82/00465.051

2/00465.0

1Pr2/C51

2/CSt

Fhrft/Btu2363106.80.148.7

604144303.62Stcvh 24

p

Twall (average) = 60.065 + 500/2363 = 60.27 60.3 oF

Twall (exit) = 60.13 + 500/2363 = 60.34 (assuming constant h)

(2) WWWR 21.15

A circular pan has its bottom surface maintained at 200 oF and is situated in saturated steam

at 212 oF. Construct a plot of condensate depth in the pan vs. time up to 1 hr for this

situation. The sides of the pan may be considered non-conducting.

Additional information:

At 212 oF,

thermal conductivity of water = 0.393 Btu/hr ft oF

density of water = 59.8 lbm/ft3

specific latent heat of vaporization = 970 Btu/lbm

At 206 oF,

thermal conductivity of water = 0.392 Btu/hr ft oF

density of water = 60.0 lbm/ft3

specific latent heat of vaporization = 974 Btu/lbm

L fg

q T dyk h

A y dt

; Where T = Tsat –Ts

0 0

y t

L

fg

k Tydy dt

h

2

2

L

fg

k Tyt

h

t1061.1t9700.60

12392.02y 42

t (hr) y (ft 102) y (in)

0 0 0

0.2 0.569 0.0682

0.4 0.804 0.0965

0.6 0.985 0.118

0.8 1.14 0.136

1.0 1.27 0.153

(3) WWWR 22.12

A shell-and-tube heat exchanger having one shell pass and eight tube passes is to heat

kerosene from 80 to 130 oF. The kerosene enters at a rate of 2500 lbm/hr. Water, entering

at 200 oF and a rate of 900 lbm/hr, is to flow on the shell side. The overall heat-transfer

coefficient is 260 Btu/hr ft2 oF. Determine the required heat-transfer area.

Additional information:

cp of water = 1.0 Btu/lbm oF

cp of kerosene = 0.51 Btu/lbm oF

Hint:

WWWR Figure 22.9a may be useful for your solution

mk = 2500 lbm/hr, mw = 900 lbm/hr

U = 260 Btu/hr ft2 oF

8.70

900

51.0801302500Tw

Tw,out = 129 oF

F9.58

4970ln

4970LMTD o

16.42609.58

51.0502500

LMTDU

qA

From Figure 22.9a,

4.150

71

80130

129200Z

416.0120

50

80200

80130Y

F = 0.83

A = 4.16/0.83 = 5.01 ft2

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.0 0.2 0.4 0.6 0.8 1.0 1.2

t (hr)

y (

in)

Grader Remarks: (Teoh Jia Heng, Office E4-04-11 / E-mail: e0225106@u.nus.edu;

Li He, Office E4-04-11 / E-mail: e0225080@u.nus.edu)

Class average: 29.25

Standard deviation: 3.864

Q1: 10 marks

Q2: 10 marks

Q3: 10 marks

Question 1:

This question involves convective heat transfer between the wall of a pipe and the water

flowing in this pipe. Most students were able to provide each requirement of the question.

For the third requirement, students either showed the average wall temperature or the

wall temperature at the exit, both of which are correct as the difference between these two

values are negligible.

Common mistakes include: 1. Providing a friction factor that is incorrect.

2. Miscalculating the Reynolds Number and from the incorrect value, deducing that the flow in

the pipe is laminar. The flow should in fact be turbulent.

3. Neglecting to use the Prandtl Analogy that is stated in the question. Since the flow is

turbulent, the use of the Prandtl Analogy is justified.

4. Substituting wrong values into equations and obtaining incorrect values even though their

equations and substitutions are correct. Very little marks are deducted from this error.

Question 2:

This question tests students’ knowledge of film condensation. Most students solve it

correctly. Some of them did not use the property at the right reference temperature. One

common mistake found during grading was the inappropriate drawing of the plot. Some

penalty marks were given to this type of mistake.

Question 3:

Question 3 is a shell-and-tube heat exchanger problem. This question tests students’

knowledge of heat exchanger design. Many students failed to get the correct log mean

temperature difference. They used the temperature difference between the two ends of each

stream instead of the temperature difference of the two streams at each end. One common

mistake was found during the grading: When calculating log mean temperature, Δ𝑇1 and

Δ𝑇2 are the temperature difference of the two fluids at respective position, as shown in the

following figure.

Figure 22.7 WWWR