In the article “Measurements of the Thermal Conductivity and Thermal Diffusivity of Polymer Melts with the Short-Hot-Wire Method”(X. Zhang, W. Hendro, et al., International Journal of Thermophysics, 2002:1077-1090), the thermal diffusivity of a liquid measured by the transient short-hot-wire method is given by

where λ is the thermal diffusivity; V and I are the voltage and current applied to the hot wire, respectively; l is the length of the wire; and A and a are quantities involving temperature whose values are estimated separately. In this article, the relative uncertainties of these quantities are given as follows: V, 0.01%; I, 0.01%; l;1%; A, 0.1%; a, 1%. ,

a. Find the relative uncertainty in λ.

b. Which would reduce the relative uncertainty more: reducing the relative uncertainty in l to 0.5% or reducing the relative uncertainties in V, I, and A each to 0?

Answer :

Step 1 of 3 :

Given,

The thermal diffusivity of a liquid by the transient short-hot-wire method is given by

=

Where, = thermal diffusivity, V and I are the Voltage and Current applied to the hot wire.

L = length of the wire, A and a are the quantities involving temperature whose values are estimated separately

Let, V=0.01% , I = 0.01%, l = 1%, A = 0.1% and a = 1%.

Step 2 of 3 :

The claim is to find the relative uncertainty inRelative uncertainty : the relative uncertainty is a unitless quantity. It is frequently expressed as a percent. In practice is unknown. So, if the bias is negligible, we estimate the relative uncertainty with .

Similarly, the relative uncertainty for is =

The given relative uncertainties are = 0.01%/100

= 0.0001

= 0.0001, = 0.01, = 0.001 and = 0.01.

ln = lnV + lnI + lnl + lnA + lna

Then, =

=

= 0.141

= 1.41%

Therefore, = 1.41%