Air, 300 Kh =40 W/m2'K

ME4413 Radiation HW Solution (2012/13 S2)

Intended Learning Outcomes:

Students will able to understand

(1) the concepts of thermal radiation; and

. (2) the concepts and how to design different types of fins and heat exchangers.

Problems are extracted from the Compulsory Textbook, Cengel Y.A & Ghajar A.J.,Heat and Mass Transfer-Fundamentals and Applications, 4th Edition in SI Units,McGraw Hill:

12-58

Properties: The total absorptivity of the plate is given to be 0.40.

Analysis: (a) Applying energy balance on the surface ,

G = J + qcony = J + h (Ts - TrfJ )

=4000 W/m 2 + (40 W/m 2. K)(350 - 300) K

=6000 W/m 2

(b) The total reflectivity of the plate is determined using

a + p + t = 1 ~ P = 1- a - r (for opaque surface, r = 0)

p =1- 0.40 - 0 = 0.60

(c) The emissive power of the plate is

J=E+G ref =E+pG ~ E=J-p:;

E = 4000 W/m 2 - (0.60)(6000 W/m 2) = 400 W/m 2

(d) The total emissivity of the plate is

G=~=~= 400W/m2

=0.470s, aT/ (5.67 x 10-8 W/m2

. K4)(350K)4

12-73 The absorber surface of a solar collector isexposed to solar and sky radiation. The equilibriumtemperature of the absorber surface is to bedetermined if the backside of the plate is insulated.

Analysis: The backside of the absorbing plate isinsulated (instead of being attached to watertubes), and thus

qnet =0

as G solar = 6CT(Ts4

- Ts1y) + h(Ts - Tair )

AirToo = 25°CTsky = 15°C

~ Ts = ?as = 0.87E = 0.09

Insulation

(0.87)(720W/m 2) =(0.09)(5 .67 x 10-s W/m 2 .K4)[(T

s)4 -(288K)4]+(10W/m 2 · K )(Ts -298K)

r, =356K

13-13

Analysis: We number different surfaces as

the hole located at the center of the base (1)

the base of conical enclosure (2)

the conical side surface (3)

Surfaces 1 and 2 are flat, and they have no direct view of each other.

Therefore, F I I = F22 = F12 = F21 = 0

summation rule : F I 1 + F I 2 + F 13 =1~ F 13 =1

. . nd 2 nDh d2

reciprocity rule: A,F;3 = A3F31~--(l) =--F3\~F;\ =--4 2 2Dh

2

D

13-15 The view factors between the rectangular surfaces shown in the figure are to bedetermined.

Assumptions

Analysis

The surfaces are diffuse emitters and reflectors.

We designate the different surfaces as follows :

shaded part of perpendicular surface by (1),

bottom part of perpendicular surface by (3),

shaded part of horizontal surface by (2), and

front part of horizontal surface by (4).

(a) From Fig.13-6

L2 =~ = 0.25W 4

~=~=0.25W 4

-: 2 =0.5)W 4F 23 = 0.26 and F2~(1+3) = 0.33

~=~=0.25W 4

4m

superposition rule: F2~(1+3) = F 21 + F 23~ F 21 = F2~(1+3 ) - F 23 = 0.33 - 0.26 = 0.07

reciprocity rule : Al =A2 ~AIFJ2 =A2F21~FJ2 =F21 =0.07

(c) We designate

shaded part of top surface by (1),

remaining part of top surface by (3),

remaining part of bottom surface by (4), and

shaded part of bottom surface by (2).

3m

i,"" ........ (1) ~I m(3) 1 ·····..... 1 m

.....;-- +············::·:1

From Fig.13-5,

~=~=l )D 3F(2+4) -4(1+3) =0.15

~=~=0.67D 3

and

1~<::.::.: · · ·· · · · · ·r· · · · · · · · ·· · · · · · · · · · ··(~i")· · · · · · · · · L. .

~ = %= 1 I 1 ~-,"'" (2) \ JFi4 =0.082

~=!=0.33D 3

superposition rule : F(2+4)~(1 +3) = F(2+4)~1 +F(2+4)~3

symmetry rule : F(2+4)~1 =F (2+4)-43

Substituting symmetry rule gives

F. - F. - F(2+4)~(I+3) - 0.15 - 0 075( 2+4 )~ 1 - (2+4 )-43 - 2 - -2- - .

reciprocity rule : AI F I-4(2+4) =A (2+4)F(2+4HI ~(2)FI-4(2+4) =(4)(0.075)~ FI-4(2+4 ) =0.15

superposition rule: F1->(2+4) = F12 + FI4~ 0.15 = FI2 + 0.082~ F12 = 0.15 - 0.082 = 0.068

3

13-31 .

Analysis: From the Hottel 's crossed-strings method, we have

F . = L (Crossed strings) - L (Uncrossed strings)HJ 2 x (String on surface i)

For uncrossed strings, we have

L I = L2 = (w 2 + w 2 ) 1/ 2 = (w 2 + w 2 ) 1/ 2 = J2w

For crossed strings, we have

L3 =(w2 +4w2) 1/ 2 =J5w and L4 =W

Applying the Hottel 's crossed-strings method , we get F 12 as

(L3 +L4)-(L1 +L2 )F'2 = ..:.......:"--------'-'-------'----'---=..:...

2w

(J"Sw + w) - (J2w+ fiw)

2w

=0.204

The radiation heat flux between the two surfaces is

Q'2 = F\2U(T,4 - T24)

= (0.204)(5 .67 x 10-8 W/m 2 ·K 4)(7004 -3004)K4

= 2680 W/m 2

w

I'w

'I

I· w

13-40

Properties: The emissivities of surfaces are given to be E} = 0.8and E2 = 0.4.

Analysis: This geometry can be treated as a twosurface enclosure since two surfaces have identicalproperties. We consider base surface to be surface1 and other two surface to be surface 2. Then theview factor between the two becomes F12 = I . Thetemperature of the base surface is determined from

. u(11 4 - T/ )QI2 =----'---'---=---..:--

1-£\ 1 1-£2--+--+--A,el AIF12 A2e2

800W= (5.67 xI0-8

W/m2 . K 4)[(T\ )4 -(600K)4]

1-0.8 1 1-0.4----+ +----(1 m 2)(0 .8) (1 m 2)(1) (2 m 2)(0.4)

TI =630 K

Note that AI = 1m 2 and A2 = 2 m 2 .

4

b=2m

·1

13-70

D2 = 0.5 mTz = 500 K£2 = 0.4

Properties: The emissivities of surfaces are given to be £1 = 0.7, &2 = 0.4. and £3 = 0.2.

Analysis: The surface areas of the cylinders and the shield per unitlength are

A pipe,inner =Al =nD1L =n(O.l m)(l m) =0.314 m2

Apire,ouler =A2 =nD2L =n(0.5 m)(l m) =1.571 m'

A shield = A3 = nD3L =n(0.2 m)(l m) =0.628 m 2

The net rate of radiation heattransfer between the two cylinderswith a shield per unit length is

Rad iation shieldD3 = 0.2 m

. (J"(~ 4 _ T/ ) £3 = 0.2

Q I2,one shield = 1_ £ 1 1- B 1- £ 1 1- £_ _ I + _ _ + 3,1 + 3,2 + +__2

AIBI AI~3 A3B 3,I A3B 3,2 A3F3,2 A2 £ 2

(5.67xl0-8 W/m 2 ·K4)[(750K)4 -(500K)4]1-0.7 1 2 1-0.2 1 1-0.4- ---- + + + +--- -

(0.314 )(0 .7) (0.314)(1) (0.628)(0.2) (0.628)(1) (1.571)(0.4)

=725.8 W

If there was no shield (Eq. 13.40),

. Ala(~4 -r/) (0.314m 2)(5.67x10-s W/m 2 ·K4)[(750K)4 -(500K)4]

QI2,no shieid = .l.+ 1- £2 (~J = 1 1- 0.4 (0.1)

£1 £ 2 D2

0.7+~ 0.5

=2615.4 W

Then their ratio becomes

QI 2,oneshi eid = 725,8 = 0.2775Q 2615.4 W12,no shield

5

13-105

Properties: The emissivity of the thermocouple is given to be E = 0.7.

Analysis: Assuming the area of the shield to be very close to the sensor of thethermometer, the radiation heat transfer from the sensor is determined from

Using the concept ofEqs. 13.42 & 13.43

Q. = (j(r..

4 - r/ ) =(5.67x10-sW/m2.K4)[(490K)4-(320K)4]=2095W/

2

rad, fromsensor (1 J (1 J (1) ( 1) . m- -1+ 2--1 - -1+ 2--151 5 2 0.7 0.15

Then the actual temperature of the gas canbe determined from a heat transfer balance tobe

. .q conv.to sensor =qconv.from sensor

h(Tf -Tth)=257.9W/m2

120W/m 2 »ccr, -490)=209.5W/m 2

~Tf =492K

ThermocoupleAir, Tf Tlh = 490 K

- 1;,=0.7~:w=320 1;2=0.15-

Without the shield the temperature of the gas would be4 4

T - T ctha(Tth - Tw )f - th + h

=490K+ (0.7)(5.67 xI0-s W/m 2 ·K4)[(490K)4 -(320K)4]

120 W/m 2 . oC=506K

6

13-114

Properties: The emissivity of the bottom surface is 0.90.

Analysis: We consider the top surface to be surface 1,the base surface to be surface 2, and the side surface tobe surface 3. This system is a three-surface enclosure.The view factor from the base to the top surface of thecube is from Fig, 13-5 FI2 =0,2 . The view factor fromthe base or the top to the side surfaces is determined byapplying the summation rule to be

FII + FI2 + FI3 =1~ FI3 =1- FI2 =1-0.2 =0,8

since the base surface is flat and thus F,} =0 . Other view factors are

3m

T1 = 700 K£ = ?

T2 = 950 K£2 = 0,90

F21 = F12 = 0.20, F23 =F13 =0.80,

Surface 1:

Surface 2:

Surface 3:

-8 2 4 4 1- GI [ ](5,67 xl0 W/m.K )(700K) =.1, +-- O.20(.I j-.I2)+O,80(.I, -.13 )GI

4 1-G?[ ]rrT2 =.12 +--- F21(.I2 -.II)+F23(.I2 -.13 )G2

(5.67 xl0-8 W/m 2.K 4)(950K)4 =.12 + 1-0.90[0 ,20(.12 -.1,)+0.80(.12 -.13

) ]0,90

4aT3 = J 3

(5.67x10-8 W/m 2 .K 4)(450K)4 =.13

We now apply Eq. 9-34 to surface 2

Q2 =A2[F21(J 2 -.I})+F23(.I2 -.13)]= (9m 2)[0 .20(.I2 -.11)+0.80(.12 -.13 ) ]

Solving the above four equations, we find

G! =0.44, .II =11,736W/m 2, .12 =41 ,985W/m 2

, J 3 =2325W/m 2

The rate of heat transfer between the bottom and the top surface is

Al =A2 =(3m)2 =9m 2

, 2 2Q21 = A2F2\ (.12 - .II) =(9 m )(0 ,20)(41 ,985 - 11,736)W1m =54.4 kW

The rate of heat transfer between the bottom and the side surface is

A3 =4A\ =4(9m 2)=36m 2

. 2 2Q23 =A2F23 (J2 -.I3)=(9m )(O,8)(41 ,985-2325)W/m =285.6 kW

TLC/20 12/1 3s2

7