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Steam at \(120^{\circ} \mathrm{C}\) is continuously passed through a \(50-\mathrm{cm}\) long rubber tube of inner and outer radii \(1 \cdot 0 \mathrm{~cm}\) and \(1.2 \mathrm{~cm}\). The room temperature is \(30^{\circ} \mathrm{C}\). Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber \(=0 \cdot 15 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\)

Short Answer

Expert verified
The rate of heat flow through the tube is approximately 231.41 J/s.

Step by step solution

01

Understand the Problem

We are asked to calculate the rate of heat flow through the walls of a rubber tube with given dimensions and temperature conditions. The steam inside the tube and the room air outside the tube create a temperature difference, causing heat to flow through the tube walls.
02

Identify and Convert Given Data

Given:- Inner radius, \(r_1 = 1.0\) cm = 0.01 m- Outer radius, \(r_2 = 1.2\) cm = 0.012 m- Length of the tube, \(L = 50\) cm = 0.5 m- Temperature of steam, \(T_1 = 120^{\circ} \text{C}\)- Room temperature, \(T_2 = 30^{\circ} \text{C}\)- Thermal conductivity of rubber, \(k = 0.15 \; \mathrm{J} \cdot \mathrm{s}^{-1} \cdot \mathrm{m}^{-1} \cdot ^{\circ}\mathrm{C}^{-1}\)
03

Use the Formula for Rate of Heat Flow

The rate of heat flow through a cylindrical shell can be calculated using the formula:\[Q = \frac{2 \pi k L (T_1 - T_2)}{\ln\left(\frac{r_2}{r_1}\right)}\]where \(Q\) is the rate of heat flow, \(k\) is the thermal conductivity, \(L\) is the length of the tube, \(T_1\) and \(T_2\) are the temperatures, and \(\ln\left(\frac{r_2}{r_1}\right)\) is the natural logarithm of the ratio of outer to inner radius.
04

Calculate the Logarithmic Factor

Calculate \(\ln\left(\frac{r_2}{r_1}\right)\):\[\ln\left(\frac{0.012}{0.01}\right) = \ln(1.2) \approx 0.1823\]
05

Plug Values into the Formula

Now plug in the values into the formula:\[Q = \frac{2 \pi \times 0.15 \times 0.5 \times (120 - 30)}{0.1823}\]Simplify and compute:\[Q = \frac{2 \pi \times 0.15 \times 0.5 \times 90}{0.1823} \approx 231.41 \, \text{J/s}\]
06

State the Result

The rate of heat flow through the walls of the rubber tube is approximately 231.41 J/s.

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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 material's ability to conduct heat. It is denoted by the symbol \(k\). This property tells us how easily heat can flow through a substance when there is a temperature difference. High thermal conductivity means the material is a good conductor of heat, like metals. Low thermal conductivity indicates an insulator, like rubber.

In this exercise, we deal with rubber, which has a thermal conductivity of \(0.15 \, \mathrm{J} \, \mathrm{s}^{-1} \, \mathrm{m}^{-1} \, ^{ullet}\mathrm{C}^{-1}\). This value is relatively low, demonstrating rubber's effectiveness at slowing heat transfer. It is essential for calculating the rate at which heat flows through a rubber cylinder, as we apply the thermal conductivity value in the formula for heat flow.
Cylindrical Shell Heat Flow
Heat flow in a cylindrical shell is a bit unique compared to flat surfaces. In cylinders, heat flows radially, moving from the inside to the outside or vice versa. The rate of heat flow depends on the material's thermal conductivity, the cylinder's length, the difference in temperature between the inside and outside surfaces, and a specific term related to the shell's geometry.

The formula used is:
  • \[Q = \frac{2 \pi k L (T_1 - T_2)}{\ln\left(\frac{r_2}{r_1}\right)}\]
The formula accommodates the geometrical nature by incorporating the natural logarithm of the outer and inner radii's ratio, ensuring accurate calculations. Geometrical considerations are vital since the heat path isn't just straightforward like in flat layers but curved along the cylindrical shell.
Temperature Gradient
The temperature gradient refers to how temperature changes over a particular distance in a substance. It is essentially the driving force for heat transfer. In simple terms, heat will flow from a region of higher temperature to a region of lower temperature.

For a cylindrical shell, the temperature gradient helps us understand heat flow across the shell's thickness. Here, the steam inside the tube is at \(120^{\circ} \mathrm{C}\), while the surrounding room is \(30^{\circ} \mathrm{C}\). This difference creates a gradient that drives the heat from inside to outside. The larger the temperature difference, the stronger the gradient, and higher the rate of heat flow.
Natural Logarithm
Understanding the natural logarithm is crucial in dealing with cylindrical shell heat flow. It's a mathematical function denoted by \(\ln\). This function shows up in the heat flow formula for a cylindrical shape, particularly in the term \(\ln\left( \frac{r_2}{r_1} \right) \).

Natural logarithms are key when dealing with ratios of exponential growth or decay. In the context of heat flow through a cylinder, they adjust for the logarithmic increase in radius, giving a true average distance of heat transfer across the shell's thickness. Here, calculating \(\ln(1.2) \approx 0.1823\) allows us to utilize a realistic distance factor for accurate heat flow analysis.

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Most popular questions from this chapter

Water at \(50^{\circ} \mathrm{C}\) is filled in a closed cylindrical vessel of height \(10 \mathrm{~cm}\) and cross sectional area \(10 \mathrm{~cm}^{2}\). The walls of the vessel are adiabatic but the flat parts are made of \(1-\mathrm{mm} \quad\) thick \(\quad\) aluminium \(\quad\left(K=200 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{-1} \mathrm{C}^{-1}\right)\). Assume that the outside temperature is \(20^{\circ} \mathrm{C}\). The density of water is \(1000 \mathrm{~kg} \mathrm{~m}^{-5}\), and the specific heat capacity of water \(=4200 \mathrm{~J} \mathrm{k} \stackrel{-\mathrm{l}}{\mathrm{g}} \mathrm{C}^{-1}\). Estimate the time taken for the temperature to fall by \(1 \cdot 0^{\circ} \mathrm{C}\). Make any simplifying assumptions you need but specify them.

A calorimeter of negligible heat capacity contains \(100 \mathrm{cc}\) of water at \(40^{\circ} \mathrm{C}\). The water cools to \(35^{\circ} \mathrm{C}\) in 5 minutes. The water is now replaced by \(\mathrm{K}\) -oil of equal volume at \(40^{\circ} \mathrm{C}\). Find the time taken for the temperature to become \(35^{\circ} \mathrm{C}\) under similar conditions. Specific heat capacities of water and \(\quad\) K-oil are \(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\) and \(\begin{array}{llll}2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} & \text { respectively. } & \text { Density of } \mathrm{K} \text { -oil }\end{array}\) \(=800 \mathrm{~kg} \mathrm{~m}^{-3}\)

The left end of a copper rod (length \(=20 \mathrm{~cm}\), area of cross section \(\left.=0-20 \mathrm{~cm}^{2}\right)\) is maintained at \(20^{\circ} \mathrm{C}\) and the right end is maintained at \(80^{\circ} \mathrm{C}\). Neglecting any loss of heat through radiation, find (a) the temperature at a point \(11 \mathrm{~cm}\) from the left end and (b) the heat current through the rod. Thermal conductivity of copper \(=385 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\).

One end of a rod of length \(20 \mathrm{~cm}\) is inserted in a furnace at \(800 \mathrm{~K}\). The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is \(750 \mathrm{~K}\) in the steady state. The temperature of the surrounding air is \(300 \mathrm{~K}\). Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conduetivity of the rod. Stefan constant \(\sigma=6 \cdot 0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\).

A hollow metallic sphere of radius \(20 \mathrm{~cm}\) surrounds a concentric metallic sphere of radius \(5 \mathrm{~cm}\). The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at \(50^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) respectively and it is found that \(100 \mathrm{~J}\) of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

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