Heat Transfer Paper 1 Solution

HEAT TRANSFER
(HMT - I : SOLUTIONS)

1) Rth  ln r2 /r1
2KL

Rth  q
T

T  qx 2KL
ln r2 /r1

T is the loqarithmic function of radius.

2)   1

n

3) (b)

4) (d)

5) (c)

6) Pr    t  v
v e 

 
t

 Pr   ,   
t

 t

7) (b)

8) for bray body,

   f ()
b

9) (b)

10) (b)

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11) rc  K
n

 0.174
8.722

rc  0.01994m
= 19.95 mm
R = 3.25

 rc  r  thickness

= 16.7 mm

12) Ans (a) P= 13529 Kg/m3,
T = 300K,

cP = 1329.3 W/kgK

  0.1523 x 10-2 N-s/m2 ,

k = 8.54 W/mK

Pr  cp  0.0248
k

Note :

For liquid metals Pr < 01

Liquid mercury is molten metal hence is should have Pr < 0.1.

13)

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LMTD  Ti Te
ln Ti
Te

LMTD  (80)  (80)

ln 80
80

LMTD  

for LMTD Ti or Te

LMD = 80
14) Ans. 42.22 sec (range 42.0 to 42.5)

Boit Number = hLc
K

For sphere Lc = Volume  d
surfacearea 6

 Bi = hd  1000 x 0.01  0.0416  0.1
6k 6 x 40

Hence lumped heat analysis is used.

T  T  hAsT t
Ti  T
 e VCP  ets

Thermal time constant

ts  VC   16 sec
hAs

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1305000330000  t  t  42.2249 s

e16

15) i  h.A.(ti  t)eBi.F0

Bi  hLc  120 x 0.015
K 6 x 42

= 7.142x10-3

F0  L  0.045 x 
L2 (2.5x103 )2

F0  0.045 x 0.0333  239.76
(2.5x103 )2

  8.  W

16) Given , T0 = 500 K, T = 300 K
D = 5x10-3 m, K = 400 W/mK

  9000kg / m3, CP = 385J/kg K

H = 250 W/m2 K

T = T + (T0 - T ).e- / *

 dT   (T0  T )e  / * x  1 
d  0  * 

Rate of cooling at beginning is

  dT   (T0  T ) .e0
 d  0 *

 T0T  500  300
*
 .V .CP 
 h.As 
 

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 500  300
  5x103  385 
 9000x  6  x 250 
  


= 17.31 sec s

 For sphere, V  R  d 
 As 3 6 
 

(Note –ve sign Indicates cooling)

17) (d)

18) q1   (T14  T24 )
1 1 1
1 2

  (T14  T24 )  K
3
1  1 1
0.5 0.5

q2  K  K
7
1  1 1
0.25 0.25

 q2  q1 x 100
q1

K  K
7 3
 x 100
K /3

 57%

19) Total thermal resistance

 2n x 1   (n  1) x 1 + 1  1 
    

2x6x  1  0.05   (7  1)  1  0.2  1  0.6
 0.05  0.2 0.6

= 239.667

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20)

Ans. (b)

21) 0  5kg , Tci = 250L
sec
mc

Cpc = 4.2 , Cph = 15 ,

Thi = 300 , min = 1.4

NTUp = NTU)c = 5

0
(m .cp)small
C 1
0
(m .cp)big

So

 llel  1 e2 NTU
2

= 0.5

 counter  NTU  5
1 NTU 6

= 0.833

coun  1.67
 parallel

22) a  K. dT
dx

 0.51 x (600  35)
220 x 10-3

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 57.95  58

23)

U 1 1 1
L
ni  K  n0

U = 1800.84 W/m2 0C

24) Dn  4A
p

25) All the properties are taken at mean value so

Tmean  40  20  300 C
2

Entering temp. = 200C

Leaving temp. = 400C

26) (b)

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27) (d)

Qs = 1kW/m3 = 1000W/m3

T = A + Bx + Cx2, A = 9000C

B = -2000C/m, C = - 500C/m2

A = 10m2, k = 40W/mk

 .   k.A T 
  x 
Q

x0

T  B  2Cx
x

 T   B  (200)
 x x0

 .   40 x 10 x (-200) = 80000 = 80 kW
 
Q

x0

28) (d)

T = A + Bx + Cx2

At x = 0, T = A = 9000C

At x = 1, T = 900 + (-200) – 50

= 900 – 250 = 6500C

Energy balance

 .  . + . =  . 
   
Q  Q gen Q stored Q

x 0 x L

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.

80 10  Q stored  120
stored  120  90  30kW

29) Q = heat transfer by radiation

Q  A (Ts4  T4 )  =1 for black body

Q 1 x 5.67 x 10-8 x (4734 - 3734 )
A

Q  1740.52w / m2
A

30) using radiation heat transfer coefficient

Q  hrA(Ts  T )

Q  hrA(Ts-T)  1740.52=hr(200-100)
A

hr  17.4 w/m2-0c

Combined heat transfer coefficient for convection & Radiation
hc + hr = 30

hc  30 17.4  12.6w / m2 0 c Ans.

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