Derivation of Equation WWWR 25-10

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

Review 3

12

CN2125E/TCN2125 Heat and Mass

Transfer 2017-2018; Final

Examination

Answer all questions.

I. (1), (2).

II. (3), (4).

III. (5), (6).

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

IV. *****

V. *****

13

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

A/Prof. Praveen Linga

Open-Book Examination:

1. Textbooks and all references

2. Homework and Tutorial Solutions.

3. Certified calculators.

CN2125E/TCN2125 Heat and Mass Transfer Quiz #1 (6 March, 2018)

Venue: E5-03-19. Time: 8:30-9:30pm.

Covering Range: Week 1-5 Materials

Student Name:_________________ Matriculation Number:__________

1. 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 1400K 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 325 K.

Give your brief answer here:

kA kB kC

LA =

0.25m

LB LC =

0.1m

Ti T0 < 325 K

Air

1400K,

h = 115

W/m2-K

Air

300K,

h = 23

W/m2-K

2. You can apply the theory of unsteady heat conduction to estimate the time required to

cook a fish-ball soup. 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 3,400 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. Please construct your solution by referring to Hessler charts (see one

of the following textbooks for detailed data with eyeball interpolations, WWWR, WRF or

Wiley Custom Learning Version 2018, Appendix F3) to determine the temperature of

the sphere.

Give your brief answer here:

Bi = hV/kA=3400/0.658*(0.015/3)= 25.8357

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/(3400*0.015)= 0.0129

Referring to Figure F6, X = 0.407 = t/x12

t = 572.3 seconds.

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 30 W per meter of length. The air temperature is 22oC. 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 surface temperature is given by 306.1K. .

Give your brief answer here:

Grader’s Remarks

Average = 20.84

Medium = 19.25

Question 1:

Question 1 is about using thermal circuit-in-series model to calculate the

thickness for a layer material (kaolin) to keep the outside temperature below

325K. 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 2:

Question 2 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. Some

students read the graph wrongly and obtained a wrong value for time.

Question 3:

Question 3 is about convective heat transfer correlations for a horizontal

cylinder. Most students could list important equations but did not manage to

solve the question correctly.

1. Students need to use the proper length scale. For horizontal cylinder in

this case, the length scale is the diameter when calculating Nu and Gr.

2. While evaluating fluid properties, students need to refer to film

temperature, which is the average of surface temperature and surrounding

temperature.

3. The convective heat transfer area is the side area, in this problem.

ASSIGNMENT 2

(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

Solution:

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

Solution:

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

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)

(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

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

Remarks:

Class average: 28.88

Standard deviation: 1.35

Q1: 10 marks

Q2: 10 marks

Q3: 10 marks

Question 1:

Question 1 is an internal convective heat transfer problem. Most students followed

correct steps and solved it properly. Some mistakes found in grading are listed here.

(1) When calculating Re number, the properties of fluid should be evaluated at film

temperature.

(2) Some students forgot to multiply the density when calculating Re number. Thus they

found the fluid was laminar.

Question 2:

This question tests students’ knowledge of film condensation. Almost all students solved

it correctly.

Question 3:

Question 3 is a shell-and-tube heat exchanger problem. Most students could solve it

properly. One common mistake was found during grading: When calculating log mean

temperature, and are the temperature difference of the two fluids at respective

position, as shown in the following figure.

Figure 22.7 WWWR