environmental systems and facility planning

7/27/2019 Environmental Systems and Facility Planning

1/33

Heat Transfer Review

D. H. WillitsBiological and Agricultural Engineering

North Carolina State University


2/33

Steady-State Ht Conduction

Composite Plane Wall (Fig 6.5)

1 2

1 2 3 4 5

1 2

31 2

1 1 2 3 2

2

1 1

W/m of surface area

x

t tq

R R R R Rt t

xx x

h k k k h


3/33



1 5 1 5

1 21 2 3

1 1 2

1xt t t t q

x xR R R

h k k


4/33



4 5 4 5

23

2

x t t t t q xR

k


5/33

Thermal Conductivity ValuesInterpretation of the values in Table 6.2 requiresan understanding of the difference between

resistivity and resistance.

Resistivity = 1/kResistance = x/k

To get resistance from resistivity, you mustmultiply by the thickness of the material.


6/33


Composite Cylinder (Fig 6.8)

1 2

1 2 3 4 5

1 2

3 2 4 32 1

1 1 1 2 3 4 2

2

2

ln( / ) ln( / )ln( / )1 1

W/m of length

r

t tq

R R R R R

t t

r r r r r r

r h k k k r h


7/33


Composite Cylinder (Fig 6.8)

1 5

1 2 3

1 5

3 22 1

1 1 1 2

2

2

ln( / )ln( / )1

r

t t

q R R R

t t

r rr r

r h k k


8/33


Composite Sphere (Fig 6.10)

1 2

1 2 3 4

1 2

3 22 1

2 2

1 1 1 1 2 2 2 3 3 2

4

4

( )( )1 1

W

rs s s s

t tq

R R R R

t t

r rr r

r h k r r k r r r h


9/33


ProblemConsider a

composite cylindrical tube with an outside diameter of 10 cm. The wallconsists of two layers of different materials, A and B. The inner material A

is in contact with a hot fluid and the outer material B is in contact with stillair at a temperature of 30 C. Material A is stainless steel, 0.2 cm thick, andmaterial B is 0.3 cm thick with a thermal conductivity of 0.0378 W m-1 K-1.If the outer surface temperature is 110 C, and the outside surfacecoefficient can be estimated as ho = 7.36 W/m

2 K, determine:

a) the heat transfer through the wall, in W/m of lengthb) the temperature of the interface between the two materialsc) the temperature of the hot fluid if the inside surface coefficient isestimated at 20 W m-2 K-1


10/33


ProblemAnswers:

a) the heat loss per unit length32 (110 30) 185.1W/mr oq r h

b) the temperature at the interface

int

3 2

2 ( 110)

185.1W/mln( / )

erface

r

b

t

q r r

k

tinterface = 158.2 C.


11/33


Problem

Answers:

(c) the temperature of the hot fluid

-1

-2 -1 -1 -1

2 ( 158.2)185.1Wm

1 ln(0.047 / 0.045 )

(0.045m)(20Wm K ) 21.5Wm K

191.0C

fluid

fluid

t

m m

t


12/33

Critical Radii for Cylinders and

Spheres

Radius at which maximum heat transfer occurs:

Cylinder Biot No. = 1.0 = roho/k

Sphere Biot No. = 2.0


13/33

Transient Ht TransferCase 1:

1

0.2p

hA

c Vo

o

t thBo e

k t t

Case 2: 100h

k

Case 3: 0.2 100h

k

Heisler ChartsFigs 6.11-6.13


14/33

Transient Ht TransferHeisler Charts:

2

1Lines are

o

o

F

B

For non-infinite geometries, TR values are

multiplied together


15/33

Transient Ht Transfer

Problem 1

A 3 cm diameter hot dog with a length of 10 cm

has an initial uniform temperature of 10 C. If it issuddenly dropped into boiling water at 100 C,determine the temperature at the center after 10min. Assume the following values:

h = 6000 W/m2 Kk = 0.5 W/m K = 1.33 X 10-7 m2/s.


16/33


Problem 1

Intersection of cylinder and slabfor cylinder:

Bo = hr/k = 180 ; 1/Bo = 0.0056 (Case 2);=/r2= 0.3547;TR1= 0.205 (from Fig 6.12;)

for slab:Bo = hL/k = 600; 1/Bo = 0.0016665 (Case 2);

=/r2= 0.0.03192;TR2= 0.99 (from Fig. 6.11);

TR = TR1x TR2= 0.20295;tc= 81.73 C


17/33


Problem 2

An aluminum cylinder (thermal conductivity = 160 W/m K,

density = 2790 kg/m3, specific heat = 0.88 kJ/kg K) of

radius r = 5 cm, length L = 0.5 m, and a uniform initialtemperature of 200 C is suddenly immersed at time zero

in a well-stirred fluid maintained at a constant

temperature of 25 C. The heat transfer coefficient

between the cylinder and the fluid is h = 300 W/m2 K.

Determine the time required for the center of the cylinder

to reach 50 C. What will the surface temperature be at

that time?


18/33


Problem 2

Check Bo:

for the cylinder:

Bo = hro/k = (300W/m2

K)(0.05m)/(160W/mK) = 0.094

This is Case 1

Note: we do not have to check Bo

for the slab because Case 1

says that the internal temperature gradient for the cylinder is

already negligible, which says that the heat is conducted to

the edge of the material faster than it can be convected away

by the water. The slab case will not change that.


19/33


Problem 2

For Case 1, use Eq 6.120 in the text:

)exp(1 Vc

hA

tt

tt

po

o

)))(J/kgK880)(kg/m2790(

)m2m)(KW/m300(exp(

25200

255023

2222

Lr

rDL

solving for gives 6.03 min or 361.8 s

The surface is the same as the center because Case 1 assumes no

internal gradient.


20/33

Convection

The basic problem is to find the appropriate h

to use inNewtons Law of Cooling. Once thatis done, finding qx is fairly trivial:

xq hA t


21/33

Convection

Free ( ) (Pr)m nNu c Gr

Horizontal cylinder, laminar flow 1x104 < GrPr < 1x108

0.250.56( Pr)Nu Gr

Horizontal cylinder, turbulent flow 1x108 < GrPr < 1x1012

1/ 30.13( Pr)Nu Gr


22/33

Convection

Free

3 2

2

Nu

Gr

Prp

hD

kD g t

c

k


23/33

ConvectionForced

Nu (Re) (Pr)m nc

Eqs 7.49 7.53

ReVD

where


24/33

Convection - Problem

Air at 60 C if flowing normal to a cylindrical copper tube

with a diameter of 15 cm at a velocity of 2 m/s.

Estimate the convective heat transfer coefficient at the

surface.


25/33

Convection - ProblemUsing Eq 7.51, with properties determined at Tfilm

rho 1.0604 kg/m^3 Density of air

mu .00002005 Pa*s Viscosity of air

cp 1007 J/kg*K Specific Heat of air

k .02856 W/m*K Thermal conductivity of air

Tfilm 60 C Film temperature

Pr .706945028011204 - Prandtl No.

Nu 77.6381180236944 - Nusselt No.

h 14.7822976717114 W/m^2 Conv heat transfer coefficient

Re 15866.3341645885 - Reynold's Number


26/33

Heat Exchangers

Basic Eqns

ln

1 1 2 2

1 1

2 2

ln

a b a b

a b

a b

q UA t

t t t t UAt t

t t

a a pa a

b b pb b

q m c t

q m c t

Fig 7.1


27/33

Heat Exchangers

Parallel Flow

ta1

ta2

tb1

tb2

fluid 'a'

fluid 'b'

temperature

distance


28/33

Heat Exchangers Effectiveness

Ratio

max min

max min

a a

a b

b b

a a

a a

t tE

t t

w cR

w c

UANTU

c w


29/33

Heat Exchangers Effectiveness

Ratio

1

1 exp 1

11

11 exp 1

1 11 exp 1

p

c

NTU RE

R

NTUR

E

NTUR R

Figs 7.3 and 7.4, or


30/33

Radiation Heat Transfer

Black Body Emissive Power

4

bW T

where Tis absolute temperature and dependsupon the unit system.


31/33


Gray Body Emissive Power

4W T

where is the emissivity.

Note: Gray bodies have constant with wavelength.


32/33


Gray Body Exchange

4 4( )

1

i i j

i ji i i

i j i j j

A T Tq

A

F A


33/33


The first problem is determining the shape factor Fij.For simple geometries, Table 8.3 may suffice.

Fij is the fraction of the energy leaving i that isintercepted by j.

For infinite parallel planes, the value is 1.0.

For small bodies enclosed by a larger body(where the smaller body cannot see itself), thevalue is also 1.0.

environmental systems and facility planning

Documents