Question

The calculated value of the thermal conductivity of the carbon nano tube was found as: KCN = 3113 W/m-K, however, the theoretical value of the thermal conductivity of the wire is actually: K = 4500 W/m-K and the island separation is 5 μm (this is the actual spacing between the two islands). The difference between the measured and theoretical values is due to the contact resistance between the nano tube and the islands in the experiment.
Required:
a. Calculate the thermal contact resistance (Rtd) that exists between the carbon nano tube and the top surfaces of the heated and sensing islands.
b. Using the value of thermal contact resistance calculated in part A, calculate the fraction of the total resistance between the heated and sensing islands that is due to the thermal contact resistances for island separation distance of 5, 10, 15, and 20 μm.

Answer 1

Answer:

a) 1,607,973.9  K/W

b)

i)  0.3082 = 30.82%

ii)  0.1821 = 18.21%

iii)  0.1293 = 12.93%

iv)  0.1002 = 10.02%

Explanation:

Value of thermal conductivity ( calculated value ) KCN  = 3113 W/m-k

Thermal conductivity ( theoretical value ) K = 4500 W/m-k

Island separation = 5 μm

a) Determine the thermal contact resistance

Resistance due to contact between carbon nano tube and top surfaces can be determined using the relation below

( I / A*K ) + 2Rc  =  ( l / A*KCN ) ------- ( 1 )

where ; I = 5 * 10^-6 m

A = π * ( 14 * 10^-9 )^2 m^2  = 153.93 * 10^-18 , K = 4500 , KCN = 3113

input  values into equation 1 above

hence Rc = 1,607,973.9  K/W

b) Determine fraction of total resistance between heated and sensing

fraction of total resistance ; f1 = $\frac{2 Rc}{I/KA + 2Rc}$

where : Rc = 1607973.9,  K = 4500, A =  153.93 * 10^-18 ,  

i) for I = 5 * 10^-6 m  

fraction = 0.3082 = 30.82%

ii) for I = 10 * 10^-6 m

fraction = 0.1821 = 18.21%

iii) for I = 15 * 10^-6 m

fraction = 0.1293 = 12.93%

iv) for  I = 20*10^-6

fraction = 0.1002 = 10.02%