A group of engineers wanted to calculate the heat transfer coeffi
A group of engineers wanted to calculate the heat transfer coefficient of air in free convection (room air, T∞ = 25°C). They decided to use lumped heat capacitance method for calculation and different shapes of aluminium for comparison. Initially, they took an aluminium ball of 5 mm in diameter and heated it to 300°C. They plotted variation of aluminium ball temperature (measured using a thermocouple embedded inside) with time. The plot is given below.
The properties of aluminium are as follows: ρ = 2800 kg/m3; k = 100 W/m K; Cp = 0.8 kJ/kg K;
A. The heat transfer coefficient of the air is 46.67 W/m<sup style="">2</sup> K
B. In the same environment, the maximum diameter that engineers can go is 1.28 cm
C. The time constant of the thermocouple is 40 sec.
D. The assumption of lumped capacitance method by the engineers is not valid.
Please scroll down to see the correct answer and solution guide.
Right Answer is:
SOLUTION
Concept:
Lumped capacitance method in transient heat conduction.
When a body of initial temperature Ti is suddenly placed in a fluid whose temperature is T∞, then the temperature distribution of the body over time is given by
\(\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{t}{\tau }}};\;\;\;\;\;\tau \left( {time\;constant} \right) = \frac{{\rho VC}}{{hA}}\)
For lumped capacitance method to be valid, the Biot number should be less than 0.1,
\(Bi = \frac{{h{L_c}}}{k} < 0.1\)
Calculation:
Given:
Ti = 300°C; T∞ = 25°C; d = 5 mm = 0.005 m; ρ = 2800 kg/m3; k = 100 W/m K;
Cp = 0.8 kJ/kg-K ⇒ 800 J/kg-K.
From the plot,
\(\frac{{126.167 - 25}}{{300 - 25}} = {e^{ - \frac{{40}}{\tau }}} \Rightarrow \tau = 40\) (Option 3)
We have
\(\tau = \frac{{\rho VC}}{{hA}} \Rightarrow 40 = \frac{{2800 \times 0.005 \times 800}}{{h \times 6}} \Rightarrow h = 46.67W/{m^2}\;K\) (Option 1)
Now,
\(Bi = \frac{{46.67 \times 5 \times {{10}^{ - 3}}}}{{6 \times 100}} = 3.8 \times {10^{ - 4}} < 0.1\)
Hence, lumped capacitance method assumption is valid (Option 4 is wrong)
Maximum diameter they can go is given by Bimax
\(0.1 > \frac{{46.67 \times D}}{{6 \times 100}} \Rightarrow D < 1.28\)
They can go up to 1.28 m diameter. (Option 2 is wrong)