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Question

A sphere of radius 5 mm is initially in equilibrium at 400 °C in a furnace. It is suddenly removed from the furnace and dipped in a well-stirred water bath at 20 °C, with a convection heat transfer coefficient of 500 W/(m$^2$K). For the given range of temperatures, the thermophysical properties of the material of the sphere are: density $\rho = 3000$ kg/m$^3$, thermal conductivity $k = 10$ W/(mK), and specific heat $c = 1000$ J/(kgK). Neglecting radiation heat transfer, the time required for the centre of the sphere to cool from 400 °C to 50 °C is ________ seconds (rounded off to 2 decimal places).

Sphere Cooling Time Analysis

This problem requires calculating the transient cooling time of a sphere using the lumped system analysis.

1. Problem Setup

Given parameters are:

  • Sphere Radius: $r = 5$ mm $= 0.005$ m
  • Initial Temperature: $T_0 = 400$ ^\circ C
  • Ambient Fluid Temperature: $T_\infty = 20$ ^\circ C
  • Target Temperature: $T(t) = 50$ ^\circ C
  • Convection Coefficient: $h = 500 \, \text{W/(m}^2\text{K)}$
  • Density: $\rho = 3000 \, \text{kg/m}^3
  • Thermal Conductivity: $k = 10 \, \text{W/(mK)}
  • Specific Heat: $c = 1000 \, \text{J/(kgK)}

2. Validate Lumped System Approach

The Biot number ($Bi$) checks if internal resistance to heat conduction is small compared to external convection resistance.

  • Characteristic Length: $L_c = r/3 = 0.005 \, \text{m} / 3
  • Biot Number Calculation: $Bi = \frac{h L_c}{k}
  • $Bi = \frac{500 \, \text{W/(m}^2\text{K)} \times (0.005 \, \text{m} / 3)}{10 \, \text{W/(mK)}} \approx 0.0833

As $Bi < 0.1$, the lumped system analysis is valid, indicating the sphere's temperature is nearly uniform throughout.

3. Calculate Cooling Time

The transient temperature equation for lumped system analysis is:

$ \frac{T(t) - T_\infty}{T_0 - T_\infty} = e^{-bt} $

The cooling constant ($b$) is determined by:

$ b = \frac{h}{\rho c L_c} = \frac{hA}{\rho V c} $

  • Calculating $b$: $b = \frac{500 \, \text{W/(m}^2\text{K)}}{3000 \, \text{kg/m}^3 \times 1000 \, \text{J/(kgK)} \times (0.005 \, \text{m} / 3)} = 0.1 \, \text{s}^{-1}

Substitute known values into the temperature equation:

  • $\frac{50 \, ^\circ\text{C} - 20 \, ^\circ\text{C}}{400 \, ^\circ\text{C} - 20 \, ^\circ\text{C}} = e^{-0.1 t}
  • $\frac{30}{380} = e^{-0.1 t}
  • $e^{-0.1 t} = \frac{3}{38}

Solve for $t$ using the natural logarithm:

  • $-0.1 t = \ln\left(\frac{3}{38}\right)
  • $-0.1 t \approx -2.5381
  • $t \approx \frac{-2.5381}{-0.1} \approx 25.381 \, \text{s}

4. Conclusion

The calculated time for the sphere to cool to 50 °C is approximately 25.38 seconds.

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Important Questions from Transient Heat Transfer

  1. Consider the following statements about the thermal diffusivity of a substance:

    1. It is directly proportional to density.

    2. It is inversely proportional to specific heat.

    3. It is inversely proportional to thermal conductivity.

    4. It is directly proportional to thermal conductivity.

    Which of the given statements are INCORRECT?

  2. Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temperature is valid when

  3. The dimensionless numbers used for analyzing the transient heat conduction problems in a plane wall, cylinder, and sphere is ________.
  4. Choose the correct statement.

    The curve for unsteady state heating and cooling of bodies with Bi < 0.1 is

  5. A spherical thermocouple junction of diameter 0.706 mm is to be used for measurement of temperature of a gas stream. The convective heat transfer coefficient on bead surface is \(400~\frac{W}{{{m^2} - K}}\). The thermo physical properties of thermocouple material are \(k = 20~\frac{W}{{m - K}},\ c = 400\frac{J}{{kg - K}},\ \rho = 8500~\frac{{kg}}{{{m^3}}}\). If the thermocouple initially at 30°C is placed in a hot stream of 300°C, the time taken by the bead to reach 298°, is

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