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Consider two walls, \(A\) and \(B\), with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is \(k_{A} / k_{B}=4\) and the ratio of the wall thicknesses is \(L_{A} / L_{B}=2\). The ratio of heat transfer rates through the walls \(\dot{Q}_{A} / \dot{Q}_{B}\) is (a) \(0.5\) (b) 1 (c) \(2 \quad(d) 4\) (e) 8

Short Answer

Expert verified
Answer: The ratio of heat transfer rates through wall A to wall B is 2.

Step by step solution

01

Recall the formula for the heat transfer rate through a wall

The formula for the rate of heat transfer through a wall is given by: \(\dot{Q} = \frac{kA\Delta T}{L}\), where \(\dot{Q}\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the surface area, \(\Delta T\) is the temperature drop across the wall, and \(L\) is the thickness of the wall.
02

Write the formula for heat transfer rates of walls A and B

Use the formula from Step 1 for wall A and wall B respectively: \(\dot{Q}_{A} = \frac{k_{A}A\Delta T}{L_{A}}\) \(\dot{Q}_{B} = \frac{k_{B}A\Delta T}{L_{B}}\)
03

Write the formula for the ratio of heat transfer rates

Divide the formula for wall A by the formula for wall B: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{\frac{k_{A}A\Delta T}{L_{A}}}{\frac{k_{B} A\Delta T}{L_{B}}}\)
04

Simplify the ratio

Cancel out the same terms and simplify: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{k_{A}L_{B}}{k_{B}L_{A}}\)
05

Substitute the given ratios

Substitute the provided ratios into the simplified equation: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{4k_{B}L_{B}}{k_{B}(2L_{B})}\)
06

Simplify the equation further

Simplify the equation by canceling out the terms: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = \frac{4}{2}\)
07

Calculate the ratio of heat transfer rates

Solve for the ratio: \(\frac{\dot{Q}_{A}}{\dot{Q}_{B}} = 2\) So, the ratio of heat transfer rates through the walls, \(\dot{Q}_{A} / \dot{Q}_{B}\) is (c) \(2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity
Thermal conductivity is a measure of how well a material can conduct heat. It is often denoted by the symbol \( k \). Materials with high thermal conductivity can transfer heat more efficiently. Imagine touching a metal surface versus a wooden one. The metal feels colder because it conducts heat away from your hand more quickly due to its higher thermal conductivity.

In the context of heat transfer through walls, thermal conductivity indicates how quickly heat can pass through the wall material from the warm side to the cold side. The greater the thermal conductivity, the faster the rate of heat transfer for a given temperature difference and wall thickness. Knowing this property is important when designing buildings and insulation because it helps determine how much heat will pass through the walls.
  • Materials with high \( k \) are more efficient at conducting heat.
  • Materials with low \( k \) are better insulators.
Wall Thickness
Wall thickness, represented by \( L \), plays a crucial role in determining the rate of heat transfer. Thickness is essentially how deep a wall is, from one side to the other.

A thicker wall will generally slow down the heat transfer because heat has to travel a longer path to get through. Think of wall thickness as a barrier that heat needs to overcome. The thicker the barrier, the more resistance there is to the flow of heat.

When comparing two walls with the same material, a thinner wall will allow more heat to pass through in a given period of time than a thicker one. This is because the heat has less distance to travel. Thus, in heat transfer equations, wall thickness is inversely proportional to the heat transfer rate:
  • Higher thickness \( \Rightarrow \) Lower heat transfer rate.
  • Lower thickness \( \Rightarrow \) Higher heat transfer rate.
Heat Transfer Rate
The heat transfer rate, denoted by \( \dot{Q} \), is the amount of heat energy transferred per unit time through a wall. It's an essential concept to understand how effective a barrier can be at controlling temperature differences.

The formula for calculating this rate through a wall is given by \( \dot{Q} = \frac{kA\Delta T}{L} \), where:\- \( k \) represents thermal conductivity,- \( A \) is the surface area of the wall,- \( \Delta T \) is the temperature difference across the wall,- \( L \) is the wall thickness.

By manipulating this formula, you can understand how changing each parameter affects the heat transfer rate. For example, increasing the thermal conductivity or the surface area or decreasing the wall thickness leads to a higher rate of heat transfer.
  • An increase in \( k, A, \) or \( \Delta T \) directly increases \( \dot{Q} \).
  • An increase in \( L \) decreases \( \dot{Q} \).
Temperature Drop
Temperature drop, symbolized as \( \Delta T \), is the difference in temperature across a wall. It represents how much cooler one side of a wall is compared to the other. This drop is the driving force for heat transfer; the greater the temperature difference, the faster the rate of heat transfer.

You can relate this to a situation where you have a hot room and a cold room separated by a wall. The higher the temperature difference between the two rooms, the more rapidly heat tries to move from the warmer side to the colder side. This movement continues until thermal equilibrium is reached or maintained through systems like heating or air conditioning.
  • Higher \( \Delta T \) results in a higher rate of heat transfer.
  • A constant \( \Delta T \) simplifies equations when comparing similar setups.

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Most popular questions from this chapter

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

What is a conduction shape factor? How is it related to the thermal resistance?

A \(5-\mathrm{mm}\)-diameter spherical ball at \(50^{\circ} \mathrm{C}\) is covered by a \(1-\mathrm{mm}\)-thick plastic insulation \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The ball is exposed to a medium at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

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