Question

two different plates A and B are in composit wall each of dimension 20cm*30cm and their thickness are 3cm and 8cm respectively.thermal conductivity are 0.25 and 0.85 in C.G.S system calculate
1-the thermal resistance of each plate
2-if the free end of plate A at temperature 150c and plate B at temperature 20c calculate the heat current through the combination
3-the temperature of the point of junction
4-the temperature gradient of each plate

Answer 1

Answer on Question #43123, Physics, Molecular Physics | Thermodynamics

Question.

Two different plates A and B are in composite wall each of dimension $20\mathrm{cm}^*30\mathrm{cm}$ and their thickness are $3\mathrm{cm}$ and $8\mathrm{cm}$ respectively. Thermal conductivity are 0.25 and 0.85 in C.G.S system.

calculate

1-the thermal resistance of each plate

2-if the free end of plate A at temperature 150c and plate B at temperature 20c calculate the heat current through the combination

3-the temperature of the point of junction

4-the temperature gradient of each plate

Given:

$a = 20cm$

$b = 30cm$

$l_{1} = 3cm$

$l_{2} = 8cm$

$\lambda_{1} = 0.25\frac{erg}{s\cdot cm\cdot{}^{\circ}C}$

$\lambda_{2} = 0.825\frac{erg}{s\cdot cm\cdot{}^{\circ}C}$

Find:

1) $R_{1} = ?R_{2} = ?$

2) $T_{1} = 150{}^{\circ}\mathrm{C}; T_{2} = 20{}^{\circ}\mathrm{C}$

$q = ?$

3) $T_{junc} = ?$

4) $\nabla T_{1} = ?\nabla T_{2} = ?$

Solution.

1) By definition thermal resistance is equal to:

$$R = \frac {l}{\lambda S}$$

$S$ is the cross-sectional area. In our case, $S = ab$ .

Therefore,

$$R _ {1} = \frac {l _ {1}}{\lambda_ {1} a b}; R _ {2} = \frac {l _ {2}}{\lambda_ {2} a b}$$

Calculate:

$$R _ {1} = \frac {3}{0 . 2 5 \cdot 2 0 \cdot 3 0} = 0. 0 2 \frac {s \cdot K}{e r g}$$ $$R _ {2} = \frac {8}{0 . 8 5 \cdot 2 0 \cdot 3 0} = 0. 0 1 5 7 \frac {s \cdot K}{e r g}$$

3) The heat current density is the amount of energy that flows through a unit area per unit time:

$$q = q _ {1} = q _ {2}$$

The Fourier equation for thermal conduction:

$$\vec {q} = - \lambda \nabla T$$

In our case,

$$q = \lambda \frac {T _ {0} - T}{l}$$ $$q _ {1} = \lambda_ {1} \frac {T _ {1} - T _ {j u n c}}{l _ {1}}; q _ {2} = \lambda_ {2} \frac {T _ {j u n c} - T _ {2}}{l _ {2}}$$ $$q _ {1} = q _ {2} \rightarrow \lambda_ {1} \frac {T _ {1} - T _ {j u n c}}{l _ {1}} = \lambda_ {2} \frac {T _ {j u n c} - T _ {2}}{l _ {2}}$$ $$T _ {j u n c} (l _ {1} + l _ {2}) = l _ {1} T _ {2} + l _ {2} T _ {1} \rightarrow T _ {j u n c} = \frac {\lambda_ {1} l _ {2} T _ {1} + \lambda_ {2} l _ {1} T _ {2}}{\lambda_ {2} l _ {1} + \lambda_ {1} l _ {2}}$$

Calculate:

$$T _ {j u n c} = \frac {0 . 2 5 \cdot 8 \cdot 1 5 0 + 0 . 8 5 \cdot 3 \cdot 2 0}{0 . 8 5 \cdot 3 + 0 . 2 5 \cdot 8} = \frac {3 5 1}{4 . 5 5} = 7 7. 1 4 {}^ {\circ} \mathrm {C}$$

2) The heat current through the combination:

$$q = q _ {1} = \lambda_ {1} \frac {T _ {1} - T _ {j u n c}}{l _ {1}} = q _ {2} = \lambda_ {2} \frac {T _ {j u n c} - T _ {2}}{l _ {2}}$$

Calculate:

$$q = q _ {1} = 0. 2 5 \frac {1 5 0 - 7 7 . 1 4}{3} = 6. 0 7 \frac {e r g}{s \cdot c m ^ {2}}$$

4) Calculate the temperature gradient of each plate:

$$\nabla T _ {1} = - \frac {T _ {1} - T _ {j u n c}}{l _ {1}}; \nabla T _ {2} = - \frac {T _ {j u n c} - T _ {2}}{l _ {2}}$$

On other hand,

$$\nabla T _ {1} = - \frac {q}{\lambda_ {1}}; \nabla T _ {2} = - \frac {q}{\lambda_ {2}}$$

Calculate:

$$\nabla T _ {1} = - \frac {1 5 0 - 7 7 . 1 4}{3} = - 2 4. 2 8 \frac {{}^{\circ} \mathrm{C}}{c m} \text{ or } \nabla T _ {1} = - \frac {6 . 0 7}{0 . 2 5} = - 2 4. 2 8 \frac {{}^{\circ} \mathrm{C}}{c m}$$ $$\nabla T _ {2} = - \frac {77.14 - 20}{8} = -7.14 \frac {{}^{\circ} \mathrm{C}}{cm} \text{ or } \nabla T _ {1} = - \frac {6.07}{0.85} = -7.14 \frac {{}^{\circ} \mathrm{C}}{cm}$$

Answer.

1)

$$R _ {1} = \frac {l _ {1}}{\lambda_ {1} a b} = 0.02 \frac {s \cdot K}{e r g}$$ $$R _ {2} = \frac {l _ {2}}{\lambda_ {2} a b} = 0.0157 \frac {s \cdot K}{e r g}$$

2)

$$q = q _ {1} = \lambda_ {1} \frac {T _ {1} - T _ {junc}}{l _ {1}} = q _ {2} = \lambda_ {2} \frac {T _ {junc} - T _ {2}}{l _ {2}} = 6.07 \frac {e r g}{s \cdot c m ^ {2}}$$

3)

$$T _ {junc} = \frac {\lambda_ {1} l _ {2} T _ {1} + \lambda_ {2} l _ {1} T _ {2}}{\lambda_ {2} l _ {1} + \lambda_ {1} l _ {2}} = 77.14{}^{\circ} \mathrm{C}$$

4)

$$\nabla T _ {1} = - \frac {T _ {1} - T _ {junc}}{l _ {1}} = - \frac {q}{\lambda_ {1}} = -24.28 \frac {{}^{\circ} \mathrm{C}}{cm}$$ $$\nabla T _ {2} = - \frac {T _ {junc} - T _ {2}}{l _ {2}} = - \frac {q}{\lambda_ {2}} = -7.14 \frac {{}^{\circ} \mathrm{C}}{cm}$$

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