STEM Problem Detail

Problem Statement

Prompt

Provide the solution for Convection Heat Transfer Example 6.39.

Solution Rendering...

## Heat Transfer Analysis of a Plate

### Given:

- For plate: $$C_p = 400 \text{ J/kg K}$$
- Mass: $$m = 20 \text{ kg}$$
- Initial surface temperature: $$T_s = 180^\circ\text{C}$$
- Ambient temperature: $$T_\infty = 20^\circ\text{C}$$
- Final temperature: $$T = 80^\circ\text{C}$$

### Preliminary Calculations:

$$ T_{av} = \frac{T_s + T_\infty}{2} = \frac{180 + 20}{2} = 100^\circ\text{C} = 373 \text{ K} $$
$$ \Delta T = T_s - T_\infty = 180 - 20 = 160^\circ\text{C} $$

### (a) Grashof (Gr) and Rayleigh (Ra) Numbers:

$$ Gr = \frac{L^3 g \beta \Delta T}{\nu^2} $$
$$ Gr = \frac{0.5^3 \times 9.81 \times \left( \frac{1}{273 + 100} \right) \times (180 - 20)}{(23.13 \times 10^{-6})^2} $$
$$ Gr = 982.191 \times 10^6 $$
$$ Ra = Gr \cdot Pr = 982.191 \times 10^6 \times 0.688 = 675.747 \times 10^6 > 10^5 $$

### (b) Heat Transfer Coefficient ($$h$$):

Using the correlation for Nusselt number:
$$ Nu = 0.14 \, Ra^{0.33} = 114.8 $$
$$ Nu = 0.14 \times [675.747 \times 10^6]^{0.33} $$
Since $$Nu = \frac{h \cdot L}{k}$$:
$$ 114.8 = \frac{h \times 0.5}{0.032} $$
$$ h = 7.3474 \text{ W/m}^2 \text{ K} $$

### (c) Heat Loss Rate ($$Q$$):

Heat lost from both sides:
$$ Q = 2 [h A (T_s - T_\infty)] $$
$$ Q = 2 \times [7.3474 \times (1 \times 0.5) \times (180 - 20)] $$
$$ Q = 1175.584 \text{ W} $$

### (d) Initial Rate of Cooling:

$$ \text{Rate of heat lost} = \text{Rate of cooling} $$
$$ Q = m \cdot C_p \left( \frac{dT}{dt} \right) $$
$$ 1175.584 = 20 \times 400 \times \frac{dT}{dt} $$
$$ \frac{dT}{dt} = \frac{1175.584}{20 \times 400} = 0.1469^\circ\text{C/s or } 8.816^\circ\text{C/min} $$

### (e) Time for Cooling from $$T_s$$ to $$T$$:

Instantaneous heat balance:
$$ 2 h A (T - T_\infty) = -m C_p \frac{dT}{dt} $$
$$ \int_0^t dt = \frac{-m C_p}{2 h A} \int_{T_s}^T \frac{dT}{T - T_\infty} $$
$$ t = \frac{-m C_p}{2 h A} \ln \left[ \frac{T - T_\infty}{T_s - T_\infty} \right] $$

#### 1. Time to cool to $$120^\circ\text{C}$$:

$$ t = \frac{-20 \times 400}{2 \times 7.3474 \times 0.5} \ln \left[ \frac{120 - 20}{180 - 20} \right] $$
$$ t = \frac{-20 \times 400}{2 \times 7.3474 \times 0.5} \ln \left[ \frac{100}{160} \right] = 511.74 \text{ sec.} $$

#### 2. Time to cool to $$80^\circ\text{C}$$:

$$ t_1 = \frac{-20 \times 400}{2 \times 7.3474 \times 0.5} \ln \left[ \frac{80 - 20}{180 - 20} \right] $$
$$ t_1 = 1067.94 \text{ sec.} $$

#### 3. Time required to cool from $$120^\circ\text{C}$$ to $$80^\circ\text{C}$$:

$$ \Delta t = 1067.94 - 511.74 = 556.2 \text{ sec.} $$

Loading solution...

Problem #37