Solution to Quiz 2 ME3122E 2014

ME3122E Heat Transfer Quiz 2 (Convection) Date: 3 November 2014

Time: 7.30pm  –  8.15pm  8.15pm

1. An electric heating element consisting of a horizontal rod of diameter 5 mm is immersed in a tank of water at 30 C. The surface of the rod is maintained at a constant temperature of 90C. The volumetric thermal expansion coefficient for water is 0.00018 K -1. Determine the mean free convection heat transfer coefficient for the rod, and hence calculate the rate of heat transfer per metre length of the rod. (40 Marks)

Solution

Film temperature   = (90 + 30)/2 = 60°C. At T  = 60C;  = 1/0.001017 = 983.3 kg/m 3,  = 46310-6 kg/m.s,  Pr  =  = 2.97,  = k  =

0.653 W/m K,   =    = 0.00018 K -1 .

  =  =   /   =    = 0.4709 10-6 m2/s, Gr  D Pr  

 g   T s  T   D3 Pr  v

2

9.81 0.00018  90  300.005 2.97  3



0.4709 10 

6 2

From Table in Formula Sheet,

 Nu D





h





h D k 

1



q  L

0.53Gr  D Pr  4



0.53 and

m =



¼.

0.53 1.7738 105



1

4



10.8768

10.8768  0.653 / 0.005  1420.5 W/m W/m 2 K.

q  h (  DL)(T  s 

 = C  =





h (  D)(T  s

T  )





T  )  1420.5   0.005  (90  30)  1338.8 W/m. 

 1.7738 105

2. 1.2 kW of heat is generated uniformly in a thin rectangular plate of length L = 200 mm and width

 = W  =

100 mm. One surface of the plate is perfectly insulated while the other

surface is cooled by water at 30 C flowing at 1 m/s parallel to its it s length. The properties of water may by assumed to be constant at their values at 50C. (a) Obtain an expression for the average temperature difference between the plate and the water along the plate, and  show that it is given by

q s L

T  s  T   

12

13

0.680k  Re L Pr 

where all the symbols have their usual meaning. (40 Marks) (b) Hence calculate the average temperature of the plate. (20 Marks) Solution:

(a) At T  = 50C;  = 1/0.001012 = 988.14 kg/m 3,  = 544 10-6 kg/m.s,  Pr  =  = 3.54,  = k  =

0.643 W/m K.

 Re L   T  s  T    

1  L 1

988.14  1  0.2 6

544  10

 L

0 T  s  T   dx  0

 L

q s x dx

 L

 L 

1

12

13

 3.63286  105 (laminar)

 L

q s

0

h x





0.453k  Re x  Pr 

dx 

1  L

 L

q s x

0

k  Nu x



dx

q s

 L

12

13

0.453 Lk   u /    Pr   L

q s

  x

12

0

dx

12

q s L 2 3 2    x   0.453 Lk   u /    Pr   3 0.680k   u /    Pr  0 q s L  12

12

12

13

13

0.680k  Re L  Pr 

q s  1200 / 0.2  0.1  60 ,000 W/m W/m . 2

 T  s  T   

60 ,000  0.2



 29.9 C 

0.680  0.643 3.63286  10

5

1

 3.54 2

1

3

Hence the average temperature of the plate = 30 + 29.9 = 59.9 C.

13