Let's solve the problem step by step: ### (i) Calculate the Biot number and justify lumped analysis The Biot number is given by: \[ Bi = \frac{hL_c}{k} \] For a sphere, the characteristic length \(L_c = \frac{V}{A} = \frac{r}{3}\) (where \(r\) is the radius). Here, \(D = 5\,\text{cm} = .05\,\text{m}\), so \(r = .025\,\text{m}\). \[ L_c = \frac{.025}{3} = .00833\,\text{m} \] Substitute values: \[ Bi = \frac{600 \times .00833}{45} = \frac{4.998}{45} \approx .11 \] Since \(Bi < .1\) is the strict criterion for lumped analysis, and our value is very close (.11), lumped analysis is approximately valid. For engineering purposes, this is usually acceptable. --- ### (ii) Write the transient temperature equation For lumped capacitance (lumped parameter model): \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho V C_p} t\right) \] For a sphere: - \(V = \frac{4}{3} \pi r^3\) - \(A = 4 \pi r^2\) So, \[ \frac{A}{V} = \frac{4 \pi r^2}{\frac{4}{3} \pi r^3} = \frac{3}{r} \] Therefore, \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h (3/r)}{\rho C_p} t\right) \] --- ### (iii) Time required to reach \(T = 150^\circ \mathrm{C}\) Given: - \(T_i = 400^\circ \mathrm{C}\) - \(T_\infty = 50^\circ \mathrm{C}\) - \(T(t) = 150^\circ \mathrm{C}\) - \(h = 600\,\mathrm{W/m^2K}\) - \(r = .025\,\mathrm{m}\) - \(\rho = 780\,\mathrm{kg/m^3}\) - \(C_p = 460\,\mathrm{J/kg\,K}\) Plug in the numbers: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \frac{150 - 50}{400 - 50} = \frac{100}{350} = .2857 \] Take the natural log: \[ \ln(.2857) = -1.2528 \] Set up the exponent: \[ -\frac{h (3/r)}{\rho C_p} t = -1.2528 \] \[ \frac{h (3/r)}{\rho C_p} t = 1.2528 \] Calculate the coefficient: \[ \frac{h (3/r)}{\rho C_p} = \frac{600 \times 3/.025}{780 \times 460} \] \[ = \frac{600 \times 120}{358800} \] \[ = \frac{72000}{358800} \approx .02007 \] So, \[ .02007 t = 1.2528 \] \[ t = \frac{1.2528}{.02007} \approx 62.4\, \text{seconds} \] --- ### (iv) Time constant of cooling The time constant \(\tau\) is: \[ \tau = \frac{\rho V C_p}{hA} = \frac{r}{3} \frac{\rho C_p}{h} \] Or, using the previously derived rate constant: \[ \tau = \frac{1}{\frac{h (3/r)}{\rho C_p}} = \frac{\rho C_p r}{3h} \] Plug in the values: \[ \tau = \frac{780 \times 460 \times .025}{3 \times 600} \] \[ = \frac{89700}{180} = 49.72 \,\text{seconds} \] --- ### (v) Effect if diameter is doubled If \(D\) is doubled, \(r\) is doubled. The Biot number will also double, making the lumped analysis less accurate (since Biot number will exceed .2). The time constant \(\tau\) will also double, since \(\tau \propto r\). Thus, cooling will take longer. --- ## **Final Answers** 1. The Biot number is **.11**; lumped analysis is marginally valid. 2. The transient temperature equation: \ \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h (3/r)}{\rho C_p} t\right) \] 3. Time required to reach \(T = 150^\circ C\) is **62.4 seconds**. 4. Time constant of cooling is **49.7 seconds**. 5. If diameter is doubled, the Biot number and time constant both double, so the lumped parameter model becomes less valid and cooling takes longer.