Question

A sphere of radius \( 5 \) mm is initially in equilibrium at \( 400^{\circ}\text{C} \) inside a furnace. It is suddenly taken out and dropped into a well stirred water bath held at \( 20^{\circ}\text{C} \), with a convection heat transfer coefficient of \( 500 \) W/(m²K). Over this range of temperatures the sphere material has density \( \rho = 3000 \) kg/m³, thermal conductivity \( k = 10 \) W/(mK), and specific heat \( c = 1000 \) J/(kgK). Neglecting radiation, find the time needed for the centre of the sphere to cool from \( 400^{\circ}\text{C} \) to \( 50^{\circ}\text{C} \). Give the answer in seconds, rounded off to 2 decimal places.

Hint:

Check the Biot number first to confirm the sphere can be treated as lumped, then use the exponential cooling formula.

Answer 1

Correct Answer: 25.39

Step 1: Check whether lumped capacitance can be used.
For a sphere the characteristic length is \( L_c = V/A = r/3 \), so \( L_c = 5/3 = 1.667 \) mm \( = 0.001667 \) m.
The Biot number is \( Bi = hL_c/k = (500)(0.001667)/10 = 0.0833 \), which is below \( 0.1 \), so we can treat the whole sphere as one lump at a uniform temperature.

Step 2: Write the lumped cooling equation.
For lumped cooling by convection, \( \dfrac{\theta}{\theta_i} = \dfrac{T-T_\infty}{T_i-T_\infty} = e^{-t/\tau} \), where the time constant is \( \tau = \dfrac{\rho V c}{hA} = \dfrac{\rho c L_c}{h} \).

Step 3: Compute the time constant.
\( \tau = \dfrac{(3000)(1000)(0.001667)}{500} = \dfrac{5000}{500} = 10 \) s.
A small \( \tau \) means the sphere responds fast, so \( \tau=10 \) s tells us this is a quick cool down.

Step 4: Insert the given temperatures.
Here \( T_i = 400^{\circ}\text{C} \), \( T_\infty = 20^{\circ}\text{C} \), \( T = 50^{\circ}\text{C} \), so \( \theta_i = 380 \) and \( \theta = 30 \).
\( \dfrac{380}{30} = e^{t/10} \) gives \( t = 10\ln(380/30) = 10\ln(12.667) \).

Step 5: Solve for the time.
\( \ln(12.667) = 2.539 \), so \( t = 10 \times 2.539 = 25.39 \) s.

Final Answer:
The centre of the sphere needs this much time to cool from \( 400^{\circ}\text{C} \) to \( 50^{\circ}\text{C} \). \[ \boxed{t \approx 25.39 \text{ s}} \]