Question

Superheated steam at an average temperature \(200^{\circ} \mathrm{C}\) is transported through a steel pipe \(\left(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{o}=8.0 \mathrm{~cm}\right.\), \(D_{i}=6.0 \mathrm{~cm}\), and \(L=20.0 \mathrm{~m}\) ). The pipe is insulated with a 4-cm thick layer of gypsum plaster \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is \(10^{\circ} \mathrm{C}\). The steam and the air heat transfer coefficients are estimated to be 800 and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) the daily rate of heat transfer from the superheated steam, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation.

Short answer

(b) What is the temperature on the outside surface of the gypsum plaster insulation? (a) The daily rate of heat transfer is 1171.6 kWh/day. (b) The temperature on the outside surface of the gypsum plaster insulation is 162°C.

Step-by-step solution

Calculate the Thermal Resistance of Steel Pipe

Find the thermal resistance of the steel pipe using the formula and given values: \(R_{\text{steel}} = \frac{\ln\left(\frac{8\,\text{cm}}{6\,\text{cm}}\right)}{2 \pi (50\, \mathrm{W/m \cdot K})(20\,\text{m})}\) \(R_{\text{steel}} = \frac{\ln\left(\frac{4}{3}\right)}{2 \pi (50\, \mathrm{W/m \cdot K})(20\,\text{m})} = 9.74 \times 10^{-5}\, \mathrm{K} / \mathrm{W}\)

Calculate the Thermal Resistance of Gypsum Plaster Insulation

Similarly, find the thermal resistance of the gypsum plaster insulation using the formula and given values: \(R_{\text{gypsum}} = \frac{\ln\left(\frac{12\,\text{cm}}{8\,\text{cm}}\right)}{2 \pi (0.5\, \mathrm{W/m \cdot K})(20\,\text{m})}\) \(R_{\text{gypsum}} = \frac{\ln\left(\frac{3}{2}\right)}{2 \pi (0.5\, \mathrm{W/m \cdot K})(20\,\text{m})} = 0.00379\, \mathrm{K} / \mathrm{W}\)

Calculate Temperature Difference

Calculate the temperature difference between steam and air required for heat transfer: \(\Delta T = (200^{\circ}\mathrm{C}) - (10^{\circ}\mathrm{C}) = 190\,\text{K}\)

Calculate Overall Thermal Resistance

Now, we will find the combined thermal resistance of both layers: \(R_{\text{total}} = R_{\text{steel}} + R_{\text{gypsum}} = 9.74 \times 10^{-5}+ 0.00379 \,\mathrm{K}/\mathrm{W} = 0.00389\, \mathrm{K} / \mathrm{W}\)

Calculate Heat Transfer Rate

To determine the rate of heat transfer, we will use the formula: \(q = \frac{\Delta T}{R_{\text{total}}}\) Plug in the values: \(q = \frac{190\, \mathrm{K}}{0.00389\, \mathrm{K} / \mathrm{W}} = 48817\, \mathrm{W}\)

(a) Daily Rate of Heat Transfer

Now, to find the daily rate of heat transfer, convert the rate from watts to kilowatt-hours and then multiply by the number of hours in a day (24 hours): \(\text{Heat Transfer Rate} = \frac{48817\, \mathrm{W}}{1000\, \mathrm{W/kW}} \times 24\, \mathrm{hr/day} = 1171.6\, \mathrm{kWh/day}\)

Calculate Temperature of Outer Surface

Next, we will find the temperature on the outer surface of the gypsum plaster insulation. We know that \(q = U A \Delta T\). We can find \(U\) by using the formula: \(U = \frac{1}{R_{\text{total}}}\) Now, we can find the temperature difference between outside and inside of the insulation: \(\Delta T_{\text{gypsum}} = \frac{q}{U A} = \frac{48817\, \mathrm{W}}{\frac{1}{0.00389\, \mathrm{K/W}} \times 2\pi \left(\frac{12\,\text{cm}}{100\,\text{cm/m}}\right) 20\,\text{m}} = 38\,\mathrm{K}\)

(b) Temperature on the Outside Surface of Gypsum Plaster Insulation

Now, using the temperature difference, we can find the temperature of the outer surface of the gypsum plaster insulation: \(T_{\text{outer}} = T_{\text{steam}} - \Delta T_{\text{gypsum}} = 200^{\circ}\mathrm{C} - 38\,\mathrm{K} = 162^{\circ}\mathrm{C}\) So, \((a)\) the daily rate of heat transfer from the superheated steam is 1171.6 kWh/day, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation is 162°C.

Key concepts

Thermal Resistance

Thermal resistance plays a critical role in understanding how well materials can resist the flow of heat. It's like thermal insulation, which is crucial when you want to maintain a certain temperature within a space or object.
It is denoted by the symbol \( R \), and it quantifies how easily heat can pass through a material. The formula for thermal resistance in a cylindrical object like a pipe is:

• \( R = \frac{\ln\left(\frac{D_{o}}{D_{i}}\right)}{2 \pi k L} \)

- Here, \( D_{o} \) and \( D_{i} \) refer to the outer and inner diameters of the pipe.- \( k \) is the thermal conductivity of the material, and \( L \) is the length of the pipe.
Calculating thermal resistance helps in determining how well a material can block heat. In our case, it was necessary to calculate the thermal resistance of the steel pipe and the gypsum plaster to find the total resistance to heat flow.
By finding the thermal resistance of each layer, you can more accurately predict the heat transfer rates and temperatures, which are crucial for efficient thermal management.

Superheated Steam

Superheated steam is steam that is heated beyond its boiling point, without any liquid involved.
This type of steam can hold more energy, which makes it especially useful in industrial applications. Let's consider some important aspects about superheated steam:

• It doesn't condense easily because it's "dry."
• It carries more energy compared to the saturated steam.
• It expands better and is often used in turbines for electricity generation.

The concept of superheated steam is important in thermal management applications because it helps in efficiently transferring heat.
In the given exercise, superheated steam at 200°C is being transported, which requires insulation (like gypsum plaster) to minimize heat loss and maintain the high energy content of the steam.
This makes superheated steam ideal for situations where maximum energy transfer and efficiency are desired.

Gypsum Plaster Insulation

Gypsum plaster insulation is used for its thermal insulating properties, often applied as an additional layer around objects to reduce heat loss.
In our exercise, gypsum plaster surrounds the steel pipe to ensure that the superheated steam maintains its high temperature. Here are some reasons why gypsum plaster is chosen for insulation:

• It has a relatively low thermal conductivity \( (k = 0.5 \, \text{W/m} \cdot \text{K}) \), meaning it is effective in reducing heat flow.
• It's easy to apply and provides a smooth finish, ideal for creating insulated layers.

When calculating the thermal resistance of gypsum plaster, the same formula used for the steel pipe can be applied, but the lower thermal conductivity indicates a higher thermal resistance.
The use of gypsum plaster plays a vital role in this scenario by helping maintain consistent temperature levels around the steam pipe, ultimately conserving energy and reducing the need for additional heating.