Q60P Figure shows the cross section o... [FREE SOLUTION]

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Chapter 18: Q60P (page 544)

Figure shows the cross section of a wall made of three layers. The layer thicknesses are $L_{1}, L_{2} = 0 .700 L_{1}, and L_{3} = 0 .350 L_{1}$ . The thermal conductivities are, $k_{1}, k_{2} = 0 .900 k_{1}$ and $k_{3} = 0 .800 k_{1}$ . The temperatures at the left and right sides of the wall are $T_{H} = 30 .0 掳颁$ and $T_{C} = 鈭� 15 .0 掳颁$ , respectively. Thermal conduction is steady. (a) What is the temperature difference $螖 T_{2}$ across layer 2 (between the left and right sides of the layer)? If $k_{2}$ were, instead, equal to $1 .1 k_{1}$ , (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously, and (c) what would be the value of $螖 T_{2}$ ?

Short Answer

Expert verified

Temperature difference $螖 T_{2}$ across layer 2is . $15.8 掳颁$
If, $k_{2} = 1.1 k_{1}$ then the rate at which energy is conducted through the wall is greater.
Temperature difference localid="1662433196049" $螖 T_{2}$ whenlocalid="1662433188116" $k_{2} = 1.1 k_{1}$ islocalid="1662433192284" $13.8 掳颁.$

Step by step solution

The given data

Thickness $L_{2} = 0.700 L_{1}$
Thickness $L_{3} = 0.350 L_{1}$
Thermal conductivity $k_{2} = 0.900 k_{1}$
Thermal conductivity $k_{3} = 0.800 k_{1}$
The temperature oftheleft side of wall $T_{H} = 30.0 掳颁or 303 K$
The temperature of the left side of wall $T_{c} = 鈭� 15 掳颁or 258 鈥块$

Understanding the concept of Fourier's law of thermal conduction

According to Fourier's Law, the amount of time it takes for heat to go through a material is proportional to the negative gradient of the temperature as well as the cross-sectional area that is perpendicular to the gradient. We will use the equation for the conduction rate through the slab under the assumption that the transfer is a steady-state process, i.e., the temperature everywhere in the slab and the rate of energy transfer do not change with time. In a steady-state process, the conduction rate through the two materials must be constant.

Formula:

The rate at which the sphere emits thermal radiation, $P_{c o n d} = \frac{Q}{t} = \frac{k A (T_{H} 鈭� T_{C})}{L}$ 鈥�(颈)

(a) Calculation of the temperature difference

Considering the rate of heat transfer through each layer to be the same, the rate of heat transfer across the entire wallis equal to the rate across layer $2 (P_{2})$ . Using equation (i) and cancelling out the common factor of area, we obtain

$\begin{matrix} \frac{T_{H} 鈭� T_{C}}{\frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}} + \frac{L_{3}}{k_{3}}} = \frac{螖 T}{\frac{L_{2}}{k_{2}}} \\ \frac{30 掳颁鈭� (鈭� 15 掳颁)}{2.21} = \frac{螖 T_{2}}{\frac{7}{9}} \\ \frac{45 掳颁}{2.21} = \frac{螖 T_{2}}{\frac{7}{9}} \\ 螖 T_{2} = 15.8 掳颁 \end{matrix}$

Hence, the value of temperature difference is $15.8 掳颁$

(b) Calculation of the rate of energy of conduction

If $k_{2} = 1.1 k_{1}$ , then the rate at which energy is conducted through the wall is greater because we are multiplying it by conductivity $k_{2.}$ Hence the conductivity of $k_{2.}$ is going to increase.

(c) Calculation of the temperature difference if k2=1.1k1

Repeat the above calculation of part (a) with new value for $k_{2}$ .We get the temperature difference as:

$\begin{matrix} \frac{45}{1 + \frac{7}{11} + \frac{35}{80}} = \frac{螖 T_{2}}{\frac{7}{11}} \\ 螖 T_{2} = 13.8 掳颁 \end{matrix}$

This is less than the result calculated in (a), which shows that the temperature gradient across 1 and 3 is greater than in part (a).The larger temperature gradient leads to large conductive heat currents.

Hence, the required temperature difference is $13.8 掳颁$