Appendix 8 the way forward

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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 T h

D T T

Regime V: WWWR 21-7

**



11



12



CN2125 Heat and Mass Transfer Review #2

Chapter 19 – Convective Heat Transfer

Fundaments

Momentum diffusivity

Thermal Diffusivity

pC

k

Prandtl numberk

C p

Pr (19 – 1)

Nusselt numberk

Lh Nu (19 – 6)

Forced Convection Dimensional Analysis

Nusselt number, Pr Re,1 f Nu (19 – 7)

Stanton number

pC v

hSt

(19 – 8)

As well asPr Re

NuSt

hence, Pr Re,2 f St (19 – 9)

Natural Convection Dimensional Analysis

Grashof number2

32

T L g Gr

Pr ,3 Gr f Nu (19 – 13)

In summary,

Forced convection:

Courtesy Contribution

by ChBE Year

Representative, 2006



The convective heat transfer coefficient can be evaluated by

either Pr Re,1 f Nu

or Pr Re,2 f St

Natural convection:

The convective heat transfer coefficient can be evaluated by

Pr ,3 Gr f Nu

Exact Solution for the laminar boundary layer

2

2

y

T

y

T v

x

T v y x

(19 – 15)

2

2

y

v

y

vv

x

vv x x

y x

x

(12 – 11a)

0

y

v

x

v x x (12 – 11b)

0

S

S y x

T T

T T

v

v

v

v @ y = 0

1

S

S x

T T

T T

v

v @ y

21

Re332.0 x x x Nuk

xh (19 – 21)

Approximate integral analysis of thermal boundary layer

dyT T vdx

d

y

T

C

k t

x

y p

00

(19 – 30)

Approximate integral analysis of momentum boundary layer

dyvvvdx

d

y

T x x

y

00

(12 – 38)



dyvvvdx

d x x

o

0

y

xo

y

v

e.g. # equation 19 – 15

I. Assuming vx = vx(y) and T = T(y)

II. Substitute in momentum and energy equation

= (,x,vx)

Separation of variables

x X t ,Pr,

III

T T ht

T T k y

dy

dT k

A

qS

s

x x x x Z Nuk

hx Nu Re,,Pr,

Courtesy contribution by Yin Zhenyuan, Year 2012

Energy and Momentum Transfer Analogies

Colburn Analogy

2

Pr 32 f C St (19 – 37)

Reynold’s Analogy

2

f C St (19 – 36)



Prandtl’s Analogy

1Pr 251

2

f

f

C

C

St (19 – 57)

vonKảrmản Analogy

1Pr

6

51ln1Pr 251

2

f

f

C

C

St (19 – 58)

Energy balance of water flowing in a circular tube

x p x T C v D

q4

2

1

T T x Dhq S 2

x x p x T C v xD

q

4

2

3

0

4

2

3

T T x DhT T C v xD

q s x x x

p x

04

T T DC v

h

dx

dT s

p x

Integrating,

04

ln

D

L

C v

h

T T

T T

p xS o

S L

Chapter 20 – Convective Heat Transfer Correlations

Rayleigh Number Pr Gr Ra

2 3

2

. Pr .

p

Grashof no andtl no

C g L T Ra

k

Natural Convection

Courtesy for correction by Tan

Teoh Yuan Hao, 2012



Vertical Plate:

Churchill – Chu,

2

278169

61

Pr 429.01

387.0825.0

L L

Ra Nu (20 – 4)

For laminar range (RaL 109)

94169

41

Pr 429.01

670.068.0 L

L

Ra Nu (20 – 5)

Vertical Cylinder:

Can be evaluated using the correlations for vertical plane walls when

41

35

LGr L

D (20 – 6)

Physically, the relation serves as a limit where boundary layer thickness is small as

compared to the cylinder diameter.

Horizontal Plates:

McAdam’s Correlations:

For Hot surface facing up or cold surface facing down;75 10210 L Ra

4154.0 L L Ra Nu (20 – 7)

107 103102 L Ra 3114.0 L L Ra Nu (20 – 8)

For Hot surfaces facing down or cold suface facing up;105 10103 L Ra

4127.0 L L Ra Nu (20 – 9)

Note: film temperature, Tf , should be used for fluid property evaluation.

Horizontal Cylinders

Churchill and Chu,

2

278169

61

Pr 559.01

387.060.0

D

D

Ra Nu (20 – 10)



for125 1010

D Ra

Morgan,n

D D RaC Nu (20 – 11)

values of C and n can be found in table 20.2

Spheres

Yuge correlation:

4143.02 D D Ra Nu (20 – 12)

for 1Pr and 5101 D Ra

Rectangular Enclosures

Refer to W3R page 316 – 318.

Forced Convection for Laminar Flow

Peclet number Re Pr Pe

.Pr .Re noandtl

p

no ynold

avg

k

C Dv

Pe

Graetz number Pe x

DGz

4

Laminar flow (Courtesy contribution by Yin Zhenyuan, Year 2012)

Turbulent flow



Di ttus and Boelter :n

D D Nu Pr Re023.0 8.0 (20 – 26)

where,

1. n = 0.4 if the fluid is being heated, n = 0.3 if the fluid is being cooled.2. all fluid prooertiese are evaluated at the arithmetic – mean bulk temperature.

3. the value of ReD should be > 104.4. Pr is in the range 0.7 < Pr < 100; and5. L/D > 60.

Colburn: 322.0 Pr Re023.0 DSt (20 – 27)

where,1. ReD and Pr are evaluated at film temperature, and St is evaluated at bulk

temperature.2. ReD, Pr and L/D should have values within the following limits:

a. 410Re D

b. 160Pr 7.0

c. 60 D L

Sieder and Tate: 14.0

322.0

Pr Re023.0

w

b DSt

(20 – 28)

Where,1. All fluid properties except μw+ are evaluated at bulk temperature.

2.410Re D

3. 17000Pr 7.0

4. 60 D L

Points to note:

I. Dittus and Boelter, and Colburn equations are more frequently used for fluids

with Pr within the specified range.II. Dittus and Boelteris easier to use as compared to Colburn equation as the fluid

properties are evaluated at bulk temperature for the former.

III. Sieder and Tate is used to account for fluids with high Pr numbers.

Forced Convection for External Flow

Flow parallel to plane surfaces



Laminar range 510Re

3121 Pr Re332.0 x x Nu (19 – 25)

and mean Nusselt number,3121

Pr Re664.0 L L Nu (19 – 26)

Turbulent Range 6103Re

With turbulent flow in a boundary layer, application of Colburn analogy yields

2Pr 3

2 f C St (19 – 37)

coupled with equation (13 – 31),3154 Pr Re0288.0 x x Nu (20 – 31)

and mean Nusselt number (Courtesy contribution by Yin Zhenyuan, Ng Charmaine, Year 2012)

(20 – 32)

Fluid properties should be evaluated at film temperature for the above correlations.

Cylinders in Cross – flow

McAdams

1. Plotted empirical data obtained from flow of air normal to single cylinders.2. Plot is reproduced in Fig. 20.9 ( pg 329 W3R)

3. Values of NuD are for Pr 1.

4. for other fluids,31

Pr (correction factor) should be employed:

ie. 31Pr

figureactual Nu Nu

The widely used correlation31Pr Ren

D B Nu

1. B and n are functions of Reynold’s number.

2. values of constants are provided in Table 20.3.3. Film temperature is appropriate for evaluation of physical properties.

Churchill and Bernstein

^(1/3)



5485

4132

3121

282000

Re1

Pr 4.01

Pr Re62.03.0

D D D Nu (20 – 34)

correlating equation for which 2.0Pr Re D

Single Spheres

Whitaker’s Correlation

41

4.03221 Pr Re06.0Re4.02

s

D D D Nu

(20 – 35)

380Pr 71.0

4106.7Re5.3

2.30.1

S

All properties are evaluated at T, except for μs, which is evaluated at Ts.

Case of falling liquid droplets modeled as spheres.

Ranz and Marshall 3121 Pr Re6.02 D D Nu (20 – 36)

Chapter 21 – Boiling and Condensation

Boiling

Rohsenow correlati on Equation form of Addom’s pool boiling data (fig. 21.2 pg 343 W3R)

3

7.1

21

Pr

L fg sf

sat s pLv L fg L

hC

T T C g h

A

q

(21 – 5)

C pL – heat capacity of the liquid.

Bromley

413 4.0

62.0

sat svo

pv fg v Lvv

T T D

T C h g k h

(21 – 7)

Do – outside diameter of tube.



Condensation

F ilm Condensation: The Nusselt model

Thickness of condensation film

41

8

3

4

w sat pL fg v Ll

w sat

T T C h g

xT T k

(21 – 18)

Objected aligned vertically (parallel to direction of g)

41

3

8

3

943.0

w sat

w sat pL fg v L L

T T L

T T C h gk

h

(21 – 19)

Vertical alignment use length scale ‘L’, length of plate/cylinder.

Object aligned horizontally ( perpendicular to direction of g)

e.g. horizontal cylinder.

41

3

8

3

725.0

w sat

w sat pL fg v L L

T T D

T T C hk g

h

(21 – 25)

Correlation between vertical and horizontal cylinders

4141

3.1725.0943.0

L D

L D

hh

horizontal

vertical (20 – 26)

For the case of equal heat transfer coefficients i.e. 1horizontal

vertical

h

h

Relation between D and L simplifies to



86.2 L

D (21 – 27)

Film Condensation: Banks of n no of horizontal tubes

41

3

8

3

725.0

w sat

w sat pL fg v L L

T T nD

T T C hk g

h

(21 – 28)

Chapter 22 – Heat Transfer Equipment

SinglePass heat exchangers

1. Parallel flow or co-current flow

2. Counter flow or counter current flow.

3. Cross flow – where two fluids flow perpendicularly to each other.

Types of heat exchangers

1. Double pipe heat exchangers – common single pass configuration

2. Cross flow heat exchangers

3. Shell and tube Heat exchangers

Heat transfer calculations

2

1

21

lnT

T

T T AU q (22 – 9)

Where:

o

2

1

21

ln T

T

T T - called the logarithmic – mean temperature difference ( lmT )

o 111 C H T T T and 222 C H T T T

o U – Overall heat transfer coefficient.

o A – Contact area for heat transfer process.

Can be simplified to:

lmT AU q (22 – 10)



Correction – Factor for Shell and Tube heat transfer

Parameters to be evaluated

1.

in sin s

int out t

T T

T T

Y

and,

2.

int out t

out sin s

s

t

shell

p

tube

p

T T

T T

C

C

C m

C m

Z

Where t – tube side fluids and s – shell side fluids

3. Look up for correction factor Z Y f F , on the ordinate of the plots with the

corresponding Y and Z parameters previously evaluated in steps 1 and 2.

4. calculate heat transfer with modified form of 22 – 10.

lmT F AU q (22 – 14)

Chapter 23 – Mass Transfer Fundamentals

Gas Mass Diffusivity 21

3

232

32

*3

2

A A

AA M

N

P

T D

(24 – 30)

Hirchfelder Equation

D AB

B A

AB P

M M T

D

2

21

23 11001858.0

(24 – 33)

AB - Lennard – Jones parameter (“collision diameter”) in angstrom (Å).

AB - “collision integral” for molecular diffusion, evaluated using Appendix Table K.1

(pg 742 W3R)

Temperature – Pressure effect in Gas phase

Modif ied H irchfelder Equation

2

1

23

1

2

2

1)1,1()2,2(

T D

T D

P T AB P T ABT

T

P

P D D

(24 – 41)

Full er correlation



23131

21

75.13 1110

B A

B A

AB

P

M M T

D

(24 – 42)

To determine the υ terms, look up Table 24.3 (pg 435 W

3

R)

Gas mixtures

nn

mixture D y D y D y

D

1313212

1'''

1

(24 – 29)

where

n y' is the mole fraction of component n in the gas mixture evaluated on a

component 1 free basis, i.e.

n y y y

y y

32

22'

mixture D 1 - mass diffusivity for component 1 in the gas mixture.

n D 1 - is the mass diffusivity for the binary pair, component 1 diffusing through

component n.

Liquid Mass Diffusivity

Stokes – Einstein equation

B

ABr

T D

6 (24 – 50)

Wil ke and Chang equation

6.0

218104.7

A

B B B AB

V

M

T

D

(24 – 52)

Note:

μB – viscosity of solution in centipoises.

B – “association” parameter for solvent B (Table in pg 440 W3R).

Hayduk – Laudie equation For evaluating infinite dilution diffusion coefficient of nonelectrolytes in water .

589.014.151026.13 A B AB V D (24 -53)

Note:

μB – viscosity of water in centipoises



Pore Diffusivity

Knudsen Di ff usion

Knudsen number

pore

nd

K

If 1n K , then

A

pore

A

pore pore

KA M

T d

M

NT d u

d K 4850

8

33

Effective diffusivity

For straight, cylindrical pores aligned parallel to one another,

KA AB Ae D D D

111

For random pores with porosity (volume – void fraction: ε)

2' Ae Ae D D

Differential Equations of Mass Transfer Problems

For component A (in molar units)

Mass transfer Di ff erential Equation

0

A

A

A Rt

C N

Fick’s 1st Law

termconvection Bulk

B A A A AB A N N y ycD N

For component A(in mass units)



Mass transfer Di ff erential Equation

0

A

A A r

t n

Fick’s 1st Law

termconvection Bulk

B A A A AB A N N D N

Pointers for Mass transfer

i. How to use either general differential equations or “mass” balance for adifferential control volume to solve real problems?

Please refer to examples 1 (pg 467 W3R) and example 3 (pg 470 – 472 W3R).ii. On extension to “steady state: solutions for these differential equations.

iii. The scalar equations for different coordinate systems:

a. Cartesian coordinates2 2 2

2 2 2

A A A A AB A

C C C C D R

t x y z

b. Cylindrical coordinates2 2

2 2 2

1 1 A A A A AB A

C C C C D r R

t r r r r z

or

2 2 2

2 2 2 2

1 1 A A A A A AB A

C C C C C D R

t r r r r z

c. Spherical coordinates2

2

2 2 2 2 2

1 1 1sin

sin sin

A A A A AB A

C C C C D r R

t r r r r r



Derivation of Equation WWWR 25-10



CN2125 Heat and Mass Transfer Quiz #1 (March 4, 2016)

Venue: MPSH1A. Time: 12:40-1:40pm.

Covering Range: Week 1-5 Materials

Student Name:_________________ Matriculation Number:__________

1. A furnace wall consisting of 0.35m of fire clay brick (thermal conductivity k A = 1.13W/m-K), 0.21m of kaolin (thermal conductivity k B = 1.45 W/m-K), and a 0.21m outer

layer of masonry brick (thermal conductivity k C = 0.66 W/m-K) is exposed to furnace gas

at 1340K with air at 300K adjacent to the outside wall. The inside and outsideconvective heat transfer coefficients are 135 and 27 W/m2-K, respectively. Determine the

heat loss per square meter of wall and temperature of the outside wall surface (T0) under

these conditions.

Give your brief answer here:

k A k B k C

LA =

0.35m

LB =

0.21m

LC =

0.21m

Ti T0

Air

1340 K,h1 = 135

W/m2-K

Air

300 K,

h2 = 27

W/m2-K

Note: 4 pages total. Continued on the back of this page.



Assume 1 m2 area:

= 1ℎ

= 1135

× 1 = 0.0074

= 1

ℎ

= 1

27

× 1

= 0.037

=

= 0.35 1.13

× 1 = 0.31

= = 0.21

1.45 × 1

= 0.14

=

= 0.21 0.66

× 1 = 0.32

= ∆

Σ= , ,

= 1340 300

0.81 /= 1284

′′ = = 1284 /

= ℎ ( ,)

= , ℎ

= 300 1284 27

× 1 = 347.6



2. (a) Define the following dimensionless groups and provide short (one-sentence)

description about their physical meanings: (i) Bi, (ii) Nu.

(b) Air at 18 oC is blown against a pane of glass 5.0 cm thick. If the glass is initially at -1oC and has frost on the outside, estimate the length of time required for the frost to begin

to melt. The following properties of glass may be used: k = 0.78 W/mK, = 2720 kg/m3,

c p = 840 J/kgK. The following expression for semi-infinite wall might be useful in yourcalculation,

t

xerf

T T

T T

o s

s

2

You can ignore the temperature difference between the air and frost and the respective

contacting glass surfaces.


(a)

(i)k

A/hVBi

The Biot number is the ratio of conductive (internal) resistance to heat transfer to theconvective (external) resistance to heat transfer.

(ii) K

hL Nu

The Nusselt number is the ratio of conductive thermal resistance to convective thermal

resistance of the fluid.

(b) For glass:

s/m1041.3

8402720

78.0

c

k 27

p

Apply semi-infinite wall expression:

t

xerf

T T

T T

o s

s

2

18 0

0.94718 12

xerf

t

By interpolation:

37.1t2

x

x = 0.05 m , t = 977s.

Note: 4 pages total. Continued on the back of this page.



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 thecable at a rate of 30 W per meter of length. The air temperature is 30oC. 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 transfercoefficient (or Nusselt number) for a horizontal cylinder can be estimated by the

Morgan correlation ( n

D D RaC Nu ). The values of constants C and n are given by,

C n

102 < D Ra < 104 0.850 0.188

104 < D Ra < 107 0.480 0.250

107 < D Ra < 1012 0.125 0.333

The properties of air may be estimated based on interpolation of the following data:

T (K) K (W/m.K) Pr gβρ2 /μ2

300 2.62 ×10-2 0.708 1.33 ×108

320 2.78 ×10-2

0.703 9.94 ×107

Recommended initial guess for this problem is given by Tsurface = 314.2 K.


Initial guess: , = 314.2

Film temperature: = ,+ = . +

= 308.6

By interpolation: = 2 . 6 9 × 1 0− / , = 0.706,

=1.19×10 Δ = , = 314.2 303 = 11.2

=

Δ =1.19×10 ×11.2×0.25 =2.08×10

= = 2.08 × 10 ×0.706=1.47×10 Morgan correlations: C=0.125; n=0.333

=0.125×. = 30.45

= ℎ

ℎ =

=

30.45×2.69×10−

0.25 = 3.28

Assume the length of the cable is L meter, convective area =

= ℎ (, )

30 × = 3.14 × 0.25 × × 3.28

× (, 303)

, = 314.7 ≈ ,314.2

Converged! Thus initial guess is valid. = 314.2



Grader’s comments by Shen Ye ([email protected]):

Quiz #1 (Mar. 4, 2016)

Class average: 28.5

Standard Dv: 2.49

Q1: 10 marksQ2: 10 marks

Q3: 10 marks

Question 1:Question 1 tests basic concept of heat conduction and convection. It can also be solved

using the concept of thermal resistance. Most students answered the question correctly.

Only one mistake I would like to highlight.

Comment mistake:In this problem, convective resistance cannot be ignored. When calculating the heat loss

across the entire system, total resistance should including two convective resistances and

three conductive resistances. = ∆ = ,−,++++

Question 2:Question 2 tests the definition and physical meanings of two dimensionless numbers, Bi

and Nu. Also, an unsteady-state heat conduction to a semi-infinite wall problem isintroduced to determine the time required to melt the frost. Interpolation skill of error

function chart is also tested. Most students solved this problem correctly. Followings are

some typical mistakes.

Common mistakes:1. In question 2 (a), some students only showed the definition of Bi and Nu and did not

give the physical meanings of these two dimensionless groups. The physical meanings of

these two numbers are the ratios of conductive resistance to convective resistance. You

cannot change their orders.2. In question 2 (b), some students did not understand the meaning of T, Ts and T0 in the

equation

0=erf

2√ . T0 means the initial temperature of the frost, which is -1 ℃. Ts is the

temperature of heat surface (air), which is 18℃. T is the targeted temperature of the frost, which

is 0 ℃ in this case.3. Some students could not interpolate the values correctly. For interpolation of two groups of

numbers, take this question as an example (erf 1.3 = 0.9340 and erf 1.4 = 0.9523). If our

target is to find Φ where erf Φ=0.947, as 0.9340<0.947<0.9523, Φ should be between1.3 and 1.4. Some students got interpolation results beyond the range of the two selected

ends, which could not be right. This mistake was also found in Question 3.

Question 3:Question 3 is a nature convection problem for horizontal cylinder. Trial and error method

is needed to solve this iteration problem using the given initial guess. Also, to get the

thermal parameters of a certain temperature, interpolations are required. Most students

followed the right procedures for this kind of questions and got the converged answer in



the first trail. However, students could not get the correct answers because of the

following mistakes.

Common mistakes:1. Fluid properties should be evaluated at the film temperature (average of air and surface

temperature) when using all the equations instead of using the surface temperature or air

temperature.2. The heat transfer area for this horizontal cylinder should be the side area, which is = instead of the top and bottom surface.

3. In the equation = , L represents for the characteristic length. For a horizontal

cylinder, it is the diameter , d.4. Interpolation mistakes as I mentioned in the previous question.



CN2125 Heat and Mass Transfer

2015-2016; 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



F eT

D

LSt Exp

T T

T T

L

so

s L

9.231240300

4

72000175.0

2

nd

iteration,

at which 510304.0 N/m

5

25 1036.3

/10304.0

)/25.12)(12/1(Re

s ft

s ft ft DV x

f = 0.00349, St = 0.001745

Therefore, the exit temperature, TL = 231.9oF is valid assumption.

(b) Column analogy,

Assume TL = 200 F

F T avg

1302

60200, F T f

2152

300130 at which 51032.0 N/m

5

5 1019.3

1032.0

)25.12)(12/1(Re

f = 0.00357

0012.0)79.1(10785.1Pr 2

32

332

f C

St

F eT

D

LSt Exp

T T

T T

L

so

s L

8.198240300

4

7200012.0

(a) T bulk = 145.95 F

at F 95.145 = 61.23 lb/ft3

Cp = 1 Btu/lb F

F T

F T

f

avg

9.2222

30095.145

95.1452

9.23160

F F eT L

9.2317.231240300 720001745.0



T T F lb

Btu ft

lb ft T Cpmq

6.245123.61min/

48.7

303

3

(a) F T 9.171609.231

s Btu Btuq /6.703min/6.218,426.2459.171

(b) T bulk = 129.4 F

at F 4.129 = 61.77 lb/ft3

Cp = 0.999Btu/lb F

T T F lb

Btu ft

lb ft T Cpmq

5.247999.077.61min/

48.7

303

3

F T 8.138608.198

s Btu Btuq /6.572min/353,345.2478.138

(2) WWWR # 20.38

A valve on a hot-water line is opened just enough to allow a flow of 0.06 fps. The water

is maintained at 180ºF, and the inside wall of the 1/2-in. schedule-r0 water line is at 80ºF.

What is the total heat loss through 5 ft of water line under these conditions? What is theexit water temperature?

ANS

for ½ in schedule-40 water line

Assume bulk temperature = 150 F , TL = 120 F

at F 150 = 61.3 lbm/ft3 b =sec.

1029.0 3

ft

lb

, Pr = 2.72

L = 5 ft,

s ft

lbm s ft ft lbV G

2

3 68.3)/06.0)(/3.61(

0.622 D in

F ft hr

Btuk

.383.0



6571029.0

12

622.068.3

Re3

GD (Laminar)

14.0

33.0)Pr (Re86.1

s

b

L

D Nu

s at F 80 =3

310578.0 ft

lb

F hrft

Btu Nu

D

k h

2

14.033.0

33578.0

29.0

)5(12

622.072.2657

12622.0

)86.1(383.0

F ft s

Btu

2

3

.1017.9

96.03600

12622.0

54

106.03.61

334ln

D

L

VCp

h

T T

T T

so

sl

F F eT T eT T so s L

120118)80180(80)( 96.096.0

The exit water temperature is 118 F .

at F 118 =3

6.61 ft

lb Cp=

F lb

Btu

999.0

Total heat loss,

s BtuT CpVAT Cpmq /48.0)118180(999.06.6112

622.04/06.0

2

(3) WWWR # 21.5

Four immersion heaters in the shape of cylinders (made of brass) 15 cm long and 2 cm in

diameter are immersed in a water bath at 1 atm total pressure. Each heater is rated at 500W. If the heaters operate at rated capacity, estimate the temperature of the heater surface.

What is the convective heat-transfer coefficient in this case?

ANS

Using SI Unit



222 01.0)02.0)(2(4

)15.0)(02.0(4

2 m D DL A

22 000,50

01.0

500

m

W

m

W

A

q

ASSUME NUCLEATE BOILING

3

7.1

2/1

Pr

)()(

L fg sf

sat s Lvl fg l

hC

T T Cp g h

A

q

Cf = 0.006 ( water-brass )

At 373.15 K ,

34.958

m

kg L ,

K kg

kJ Cp L

.211.4 ,

sm

N L 2

610278

Pr = 1.72 , m N /101.56 3

3/5955.0/1 mkg v g v

K T

K kg

kJ

kg kJ

m N

mkg smkg kJ

sm

N

mkW T

Cp

hC

g h

AqT

L

L fg sf

v L fg L

69.4

.211.4

)72.1)(/2257)(006.0(

/101.56

/)5955.04.958(/81.9/225710278

/50

Pr

)(

/

3/1

2/1

3

32

2

6

2

7.1

3/1

2/1

K T etemperatur surface

T T T

s

sat s

84.37715.37369.4,

K m

kW h

hT

T h A

q

266.10

50000



Using English Unit

22 14

2 ft D DL A

hr ft

Btu

A

q2

15849

ASSUME NUCLEATE BOILING

3

7.1

2/1

Pr

)()(

L fg sf

sat s Lvl fg l

hC

T T Cp g h

A

q

Cf = 0.006 ( water-brass )

At 212 F

,

39.59

ft

lbm L ,

F lb

BtuCp

m

L.

01.1 , s ft

lbm L 2

310195.0

Pr = 1.72 , ft lb f /1097.6 3

3/0372.0 ft lbmv

F T

CphC

g h

AqT L

L fg sf

v L fg L

13.9

Pr )(

/

7.1

3/1

2/1

F T etemperatur surface

T T T

s

sat s

13.22121213.9,

F h ft

Btuh

hT

T h A

q

29.1735

15849



(4) WWWR 22.12

A shell-and-tube exchanger having one shell pass and eight tube passes is to heatkerosene from 80 to 130ºF. The kerosene enters at a rate of 2500 lbm/h. Water entering at

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

coefficient is 260 Btu/h ft

2

ºF. Determine the required heat-transfer area.

ANS

200 water

80 130 kerosene

wwwk k T CpmT Cpm

hr lbmmk /2500 , hr lbmmw /900

F ft hr BtuU

2./260 , T bulk at 105 F , F lbm BtuCpk

/51.0

F T w

8.70

)1(900

)51.0)(80130(2500

F T out w

1298.70200, , F T lm

9.58

4970ln

4970

)( lmT F UAq

lmT U

q AF

, F = 0.8

16.4)9.58)(260(

)50)(51.0(2500

lmT U

q AF

A = 4.16/0.8 = 5.2 ft2

416.0120

50

80200

80130

,,

,,

int in s

int out t

T T

T T Y

4.15071

80130129200

,,

,,

int out t

out sin s

T T

T T Z



Grader’s comments by Yao Zhiyi, Office E5-01-01 / Phone: 90841445.

E-mail: [email protected]

(1) Problem 19.25:

* This problem has to be solved by a trial & error method. Do not stop iterating until your

result satisfies the convergence criterion (Temperature difference less than 3oF in

successive iterations).

* should be evaluated at film temperature ( = + ) to calculate Re =

and should be evaluated at bulk mean temperature ( = + ) to calculate

= ∆

(2) Problem 20.38:

* All fluid properties except μ should be evaluated at the bulk mean temperature to

calculate = 1.86

.

* = ∆=ℎ∆.

∆T in equation = ∆ is defined as ( ). This term means the temperature

difference between inlet and outlet of the fluid stream.

∆T in equation = ℎ∆ is defined as ∆T. This will give an exact solution of the

heat transfer rate. In a simplified solution ∆ can be approximated as ∆ =

mean. This will give approximate solution of the problem.

(3) Problem 21.5:

* There are two equations for calculation of surface tension of water in textbook WWWR pg 334 and WRF pg 363. One is in SI system and another is in English system. You may

use either SI or British (American) unit and both equations yields different values for

surface tension with the same units.

(4) Problem 22.12:

*) Some students made mistakes in calculation of ∆T. Please refer to P341, WWWR

5th Edition to learn how to calculate ∆T.

mailto:[email protected]




Appendix 8 the way forward

Documents

Transcript of Appendix 8 the way forward