HMT Frequently Asked Questions #3 Convective Heat

Transfer Corrections, Boiling & Condensation, and Mass

Transfer Fundamentals

-----Original Message-----

From: Wang Chi-Hwa

Subject: RE: Chapter 20 questions Dear Nevin:

Thanks for your e-mail.

(1) Fluid is inside the enclosure. By conduction only, q = -k T/x=-h (T).

Nu = hL/k=[k/x]*L/k;

L= x -> Nu=1.

(2) Eq 20-10.

(3) Yes. (4) EQUATIONS 19-25, 26: LAMINAR FLOW. EQUATION 20-31& 32 -> TURBULENT FLOW. (5) Yes. Chi-Hwa Wang -----Original Message-----

From: nevin nar [mailto:nevnahh@yahoo.com]

Subject: Chapter 20 questions

Dear Prof,

I've got a couple of ques to ask. Pls bear with me yar? Thks.

1) For rectangular enclosures, is the fluid inside the enclosure or flowing past it? Also,

why izzit that for case 1, when teta=180 degrees (upper surface is heated), heat transfer is

by conduction ie Nu=1?

2) For tut 5 q1, which eq is better to use? izzit eq 20-10 or eq 20-11? This is for the case

of horizontal cyclinders.

3) For eg 2 on pg 324, are we assuming steady state for heat transfer?

For spheres: WWWR Page 316 NuD = 2 + 0.43 RaD(1/3)

Ra -> 0, NuD = 2.

h T = k |dT/dr| = k* T/(D/2)

-> h D/ k = 2 = NuD

Hot

cold

Conduction only

4) On pg 326 of WWWR, for the case of flow parallel to plane surface, they apply

colburn analogy for turbulent flow in boundary layer. Why isn't Reynold's analogy used

since on pg 303, they mentioned that for turbulent flow, Pr is unity and Reynolds analogy

applies in the absence of form drag?

5) Lastly, I think the Cp values for tut 5 q 20.26 is wrong. The values should be

multiplied by 1000.

Thanks for your time.

Rgds,

Nevyn CN 5

CN2125 Heat and Mass Transfer Quiz #1 (March 4, 2020)

E-Quiz: Time: 10:00-11:00am.

Covering Range: Weeks 1-5 Materials

Your short answers should be provided under the question directly for submission

and grading. Additional supplementary answer involving formula or drawing

should be saved as the filename “Student Number.pdf” and this is to be submitted to

LumiNUS under the Folder “CN2125 Mid-term E-Quiz” immediately after the E-

Quiz. Late submission will be rejected from the system. Grading will be based on

both procedures and final answers.

Student Name: ___________

Student Number: __________

Your parameter value for (A, B) will be based on the last digit of your student

number.

e.g. If your metric card number is A0185796Y, please take the (A, B) value from

parameter set (6) below. A similar rule is applied to other cases. Suggested Solution

Q1

1. A square isothermal chip is of width w = Amm on a side and is mounted on a substrate

such that its side and back surfaces are well insulated, while the front surface is exposed

to the flow of a coolant at T = 15oC. From reliability considerations, the chip temperature

must not exceed T = BoC. If the coolant is air and the corresponding convection

coefficient is h = 200 W/m2-K, what is the maximum allowable chip power?

(0) A = 5.5, B = 85. (1) A = 5.5, B = 90. (2) A = 5.5, B = 80. (3) A = 10.5, B = 85.

(4) A = 10.5, B = 90. (5) A = 10.5, B = 80. (6) A = 15.5, B = 85. (7) A = 15.5, B = 90.

(8) A = 15.5, B = 80. (9) A = 20.5, B = 85.

Give your brief answer here:

Using the parameters (0) A = 5.5, B = 85 for the solutions:

T∞, h

Chip

Coolant

Assumptions: 1. System is at steady state.

2. Heat loss from the sides and bottom of the chip are negligible.

3. Temperature is uniform throughout the chip.

4. Heat transfer due to radiation from the chip surface is negligible.

From the energy balance, it can be found that the electric power dissipated in the chip is

equal to the heat energy transferred by convection to the coolant.

Using Newton’s Law of Cooling:

No A B Power (W)

0 5.5 85 0.4235

1 5.5 90 0.4538

2 5.5 80 0.3933

3 10.5 85 1.5435

4 10.5 90 1.6538

5 10.5 80 1.4333

6 15.5 85 3.3635

7 15.5 90 3.6038

8 15.5 80 3.1233

9 20.5 85 5.8835

Suggested Solution Q2

2. Cylinders made of Carbon steel (density = 7832 kg/m3, k = 51.2 W/m-K, CP= 541

J/kg-K, = 1.21 x 10-5 m2/s) with the diameter of 0.1-m are heat treated in a gas-fired

furnace whose gases are at 1200K and provide a convective coefficient of AW/m2-K. If

the cylinders enter the furnace at 300K, how long must they remain in the furnace to

achieve a centerline temperature of BK? You may assume that the cylinders are very

long (L/D >> 10) and hence this heat transfer is treated as a one-dimensional radial

conduction problem.

(0) A = 101, B = 801 (1) A = 91, B = 801 (2) A = 81, B = 801 (3) A = 101, B = 751

(4) A = 91, B = 751 (5) A = 81, B = 751 (6) A = 101, B = 701 (7) A = 91, B = 701

(8) A = 81, B = 701 (9) A = 101, B = 651

Give your brief answer here:

Using the parameters (0) A = 101, B = 801 for the solutions:

Assumptions 1. Heat conduction is occurring only in one dimension (radial dimension)

2. Properties of carbon steel are uniform throughout the cylinders.

Determining the Biot Number:

Thus, the lumped capacitance method can be applied.

No A B Bi t (s)

0 101 801 0.04932 853

1 91 801 0.04443 947

2 81 801 0.03955 1064

3 101 751 0.04932 729

4 91 751 0.04443 809

5 81 751 0.03955 909

6 101 701 0.04932 619

7 91 701 0.04443 687

8 81 701 0.03955 771

9 101 651 0.04932 518

Suggested Solution Q3

3. Air enters a 25-cm-diameter, A-m long underwater duct at 50oC and 1 atm at a mean

velocity of 7 m/s, and is cooled by the water outside. If the average heat transfer

coefficient is B W/m2-oC and the tube temperature is nearly equal to the water

temperature of 20oC, determine the exit temperature of air and the rate of heat transfer.

Hints: (i) You might assume the average properties of air at 32.5oC to be given by the

following: = 1.16 kg/m3, Cp = 1007 J/kgoC. (ii) The temperature drop across of the

wall of underwater duct can be neglected. (iii) Initial guess of the outlet temperature of

air = 15oC.

State your assumptions, if any.

(0) A= 12.0; B = 85.0 (1) A= 11.5; B = 85.0 (2) A= 10.5; B = 85.0 (3) A= 13.0; B = 85.0

(4) A = 11.5; B = 84.5 (5) A= 12.5; B = 84.5 (6) A= 12.0; B = 84.5 (7) A= 13.5; B = 84.0

(8) A = 12.5; B = 85.5 (9) A= 12.0; B = 85.5

Assumptions: 1. The surface temperature across the duct is constant

2. Thermal resistance of the duct material is neglected.

Schematic Diagram:

Solution 1:

Initial guess of outlet temperature of air = 15 C

Hence, bulk mean temperature of air = (50 + 15)/2

= 32.5 C

The properties of air at the anticipated bulk mean temperature of air of 32.5 C are

= 1.16 kg/m3, Cp = 1007 J/kgoC

Diameter

= 0.25 m

Air

Ti (Inlet temperature) = 50C

V (Velocity) = 7 m/s

Ts (Tube temperature) = 20C

Air

Te (Exit temperature) = ?

Q (Rate of heat transfer) = ?

A m

The mass flow rate of air (m) = (πD2 / 4)V

= (1.16 kg/m3) (π (0.25 m)2)/4 ) (7 m/s)

= 0.39858 kg/s (2 Marks)

Area of the tube = π D L

= π (0.25 m) (12 m)

= 9.424 m2 (1 Marks)

From energy balance,

= Te = Ts + (Ti – Ts)

= Te = 20 + (50-20) (0.135)

= Te = 24.076 C

Since the exit temperature is not within 3 C of the initial guess, the new initial guess is 25

C

Hence, bulk mean temperature of air = (50 + 25)/2

= 37.5 C

The properties of air at the anticipated bulk mean temperature of air of 37.5 C are

= 1.14 kg/m3, Cp = 1007 J/kgoC

The mass flow rate of air (m) = (πD2 / 4)V

= (1.14 kg/m3) (π (0.25 m)2)/4 ) (7 m/s)

= 0.39171 kg/s

From energy balance,

= Te = Ts + (Ti – Ts)

= Te = 20 + (50-20) (0.13)

= Te = 23.93 C (3.5 Marks)

Since the exit temperature is within 3 C of the initial guess, the solution is assumed to

have converged.

The logarithmic mean temperature difference

∆Tln =

= 12.826 C

According to Newton’s Law of Cooling,

Q = hA∆Tln

= (85 W/m2.oC)(9.424 m2)(12.826)

= 10274.13 W

= 10.274 kW (3.5 Marks)

Solution 2:

The solution is the same as above till the calculation of the exit temperature.

Q = mCp∆T

= (0.39171 kg/s)(1007 J/kgoC)(23.93 – 50)

= 10283.36 W

= 10.283 kW

Solution 3:

Dynamic viscosity at 37.5 C = 1.82 x 10-5 kg/m.s

Kinematic viscosity at 37.5 C = 1.51 x 10-5 m2/s

Conductivity at 37.5 C = 0.0255 W/m.K

Re = VD/ѵ

= 115894.03

The flow is turbulent.

Pr = µCp/k

= 0.7187

Analogies that can be used to find Stanton Number 1. Reynolds Analogy

2. Colburn Analogy

3. Prandtl Analogy

4. Von Karman Analogy

Dittus-Boelter, Colburn, Sieder and Tate correlations cannot be used as L/D<60.

Final Answers

Matric No. A B Area (m2)

Te ∆Tln Q =

hA∆Tln

(kW)

Q =

mCp∆T

(kW)

0 12 85 9.424 23.93 12.826 10.274 10.283

1 11.5 85 9.032 24.28 13.208 10.140 10.145

2 10.5 85 8.246 25.07 14.022 9.828 9.833

3 13.0 85 10.210 23.32 12.120 10.518 10.523

4 11.5 84.5 9.032 24.33 13.261 10.120 10.125

5 12.5 84.5 9.817 23.66 12.520 10.385 10.389

6 12.0 84.5 9.424 23.98 12.881 10.257 10.263

7 13.5 84.0 10.602 23.13 11.888 10.587 10.598

8 12.5 85.5 9.817 23.57 12.416 10.421 10.425

9 12.0 85.5 9.424 23.88 12.770 10.289 10.303

First Iteration Te (using 15 C as initial guess)

Metric No Te

0 24.076

1 24.430

2 25.232

3 23.452

4 24.480

5 23.797

6 24.124

7 23.261

8 23.705

9 24.029

CN2125 Mid-term Quiz Statistics: Average = 28.46 High = 30/30 Low = 11/30 Feedback for your performance in the Mid-term Quiz is available in LumiNUS. Please refer to "Files" -> "CN2125 Mid-term E-Quiz" to download your workings containing markings and feedback. Students who have mistakenly uploaded their workings in "CN2125 Assignment #2" can view their workings containing markings and feedback in that folder. Grader’s Remark Q1

Grader: Teoh Jia Heng Email: E0225106@u.nus.edu

For Question 1, students were required to determine the maximum allowable chip power

such that the temperature of the chip does not exceed a certain value. Most students were

able to use or derive Newton’s Law of Cooling or the thermal resistance model to

determine the heat energy dissipated by the flowing coolant on the surface of the chip via

convection and from there determine the maximum allowable chip power.

Common mistakes include 1. Some students either substituted the width of the chip instead of the area in Newton’s Law of

Cooling while others did not convert the width given to S.I. units (mm to m) before

substituting it into the equation. There are also students who incorrectly converted the width

given.

Grader’s Remark Q2

Grader: Teoh Jia Heng Email: E0225106@u.nus.edu

For Question 2, students had to determine the time required for a carbon steel cylinder to

reach a desired temperature at its center after being placed in a furnace. Most students

were able to determine that the lumped parameter analysis is applicable in this case after

checking for the Biot Number and from there, there were able to calculate the time

required.

Common mistakes include:

1. Some students did manage to calculate the Biot number but did not specify the justification

on the use of the lumped parameter analysis using their calculated Biot number.

2. Some students did not use the lumped parameter analysis and instead used graphical methods

(eg: Figure F.5 from Appendix F). Most of those who did so weren’t able to correctly

interpolate the log scale of the y-axis and presented answers that were far from the correct one.

Presumably, students who used the graphical method either did not check the Biot number or

incorrectly calculated the Biot number and ended up with a value greater than 0.1. However,

marks are still awarded if students did use graphical methods.

Grader’s Remark Q3

Grader: Vishnu Sunil Email: Vishnu.s@u.nus.edu

For Question 3, students had to determine the exit temperature of air and the heat transfer

rate. Most students were able to determine the exit temperature using the energy balance

through iterations until the exit temperature was within 3 degrees from the initial guess.

Most of the students have used q= mCp∆T or Newton’s law of cooling to find out the heat

transfer rate.

Common mistakes include: 1. Most students did not use the logarithmic mean temperature difference when using Newton’s

Law of cooling. Some students used the wrong ∆T in q= mCp∆T equation.

2. Some students used the correlations for turbulent flow to calculate Stanton number.

Correlations should be when the average heat transfer coefficient is not provided.

Furthermore, corrected h value need not be calculated as the appropriate average heat transfer

coefficient is provided for the given L/D.

3. Some students did not continue with the second iteration when their exit temperature was not

within 3 degrees from their initial guess.

Dear CN2125 Students:

CN2125 Assignment # 2 (Due 12:00noon, March 11, 2020)

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.

Due 12:00noon, March 11, 2020 by submission to LumiNUS under the Folder

CN2125 Assignment #2.

The graded assignments will be returned to you by e-mail roughly 2 weeks after

your submission.

Sincerely yours,

Chi-Hwa Wang, Co-instructor CN2125

(1) Problem 1: Modified from WRF Chapter 19 Question (11)

A cooking oven has a top square (L m x L m) surface temperature of 45oC when exposed

to still air. At this condition, the room air temperature is 20oC, and heat is transferred

from the top surface at 40W. To reduce the surface temperature, as required by safety

regulations, room air is blown across the top with a velocity of 20m/s under the same heat

transfer rate 40W. Conditions inside the oven may be considered unchanged.

(a) What is the size of the square surface?

(b) What will be the top surface temperature for this new operating condition (forced

convection case)? Suggested initial guess of surface temperature = 23oC.

Hints: (A) Please solve part (a) as a natural convection problem and then use the

MacAdam Correlation for hot horizontal surface facing up (105 < RaL < 2 x 107) to

predict the heat transfer coefficient. This can result in an energy balance equation for

solving the value of L. (B) For part (b), the value for Re is actually within the transition

zone (2 x 105 < Re < 3 x 106). It is assumed that the following equation is okay to predict

the average heat transfer coefficient, NuL= 0.664 ReL^(1/2)Pr^(1/3).

Grader’s Remark Q1 Grader: LI He Email: e0225080@u.nus.edu

Comment:

Most students did well in this question. Students can get partial marks if they can show

correct equations of natural convection in Question a. Student still can get partial marks when

they can provide correct equations and calculations Question b even if their answer in

Question a is wrong. Students can obtain full marks if the final result is correct and accurate.

Common Mistakes:

1. Many students failed to use the correct length scale in the equations. In Question a,

the problem is natural convection and length scale is equal L/4. However, Question b

is forced convection and length scale is not applicable in Question b.

2. Several students used wrong equations for Nusselt in Question 1 because of wrong

Ra number. Students would get a wrong Ra number if they failed to use correct

length scale.

3. Several students failed to realize that the Question b is forced convection problem. In

forced convection problem, Mcadam Correlation is not applicable, and L is the L in

the statement.

Solution:

(a) Natural convection and Horizontal hot surface facing up

For Appendix I, physic properties of air @ 300K

Using Mcadam Correlations for 105<RaL<2x107

Here L is length scale which is equal to . This length scale will be represented

as .

(b) Guess ,

From Appendix I,

Transition Regime / Forced Convection

Assuming laminar B.L.

Using

(2) Problem 2: WWWR Chapter 19 Question (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

Take the initial guess of wall temperature = 60.6 oF

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

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

Prandtl analogy may be applied for turbulent flow

Grader’s Remark Q2 Grader: LI He Email: e0225080@u.nus.edu

Comment:

A large number of students did well in this problem. Students can obtain partial marks if they

are able to solve this problem by using the Prandtl analogy. The rest of the marks will be

given if the student can calculate the Re number and temperature correctly.

Common Mistakes:

1. Some students did not use Prandtl analogy as required in the statement.

2. In the calculation of Re number, some students may get wrong physical properties

mistakenly.

3. When using moody diagram, a few students found the diagram online instead of the

one in the textbook. They did not be penalized but please take note that the diagram

on textbook is Fanning friction factor.

4. When solving In() function, some students mistook an answer which is smaller than

60 as the final result. However, the answer is mathematically correct but not

physically correct.

Solution:

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

q = mcpT = 1960 Btu/hr

F131.0

0.1603.6248.7

30

1960T o

Tw, exit = 60.13

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

Take the initial guess of wall temperature = 60.6

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

83731

1076.0

6014144

48.730

1213.62

Re3

Turbulent Flow

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

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

(3) Problem 3: WWWR Chapter 22 Question (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

Grader’s Remark Q3 Grader: Vishnu Sunil Email: Vishnu.s@u.nus.edu

Comment:

Almost all the students did well in this question. Marks have been awarded to students who

have assumed co-current or counter-current flow.

Common Mistakes:

Many students have represented the overall heat transfer coefficient as “h”. The overall heat

transfer coefficient should be represented as “U”.

Question 3:

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

Solution:

Qk = mCp∆T

= 2500 lbm/hr x 0.51 Btu/lbm oF x (130-80)

= 63750 Btu/hr

Qw = mCp∆T

= ∆T = 63750 Btu/hr / (900 lbm/hr x 1.0 Btu/lbm oF)

= ∆T = 70.833 oF.

Exit temperature of water = 70.833 – 200

= 129.16 oF (3 Marks)

From Figure 22.9a,

Y = (130 – 80) / (200-80)

= 0.416

Z = (200 – 129.16) / (130 – 80)

= 1.4168

F = 0.83 (3 Marks)

Log mean temperature difference

∆TLMTD = (ΔT1 - ΔT2) / ln (ΔT1/ΔT2)

Counter-current flow

ΔT1 = 200 – 130

= 70

ΔT2 = 129.16 – 80

= 49.16

∆TLMTD = (70-49.16) / ln(70/49.16)

= 58.967 oF. (3 Marks)

Q = UA∆TLMTD

= A = 63750 / (260 Btu/hr ft2 oF x 58.967)

Heat transfer area (A) = 4.158 ft2

Corrected Heat transfer area = A / F

= 4.158/0.83

= 5.0009 ft2 (1 Mark)

Co-current flow

ΔT1 = 200 – 80

= 120

ΔT2 = 130 – 129.16

= 0.84

LMTD = (120 – 0.84) / ln(120/0.84)

= 24.0152 oF.

Q = UA∆TLMTD

= A = 63750 / (260 Btu/hr ft2 oF x 24.0152)

Heat transfer area (A) = 10.209 ft2

Corrected Heat transfer area = A / F

= 10.209 / 0.83

= 12.301 ft2