Question

Water at \(43.3^{\circ} \mathrm{C}\) flows over a large plate at a velocity of \(30.0 \mathrm{~cm} / \mathrm{s}\). The plate is \(1.0 \mathrm{~m}\) long (in the flow direction), and its surface is maintained at a uniform temperature of \(10.0^{\circ} \mathrm{C}\). Calculate the steady rate of heat transfer per unit width of the plate. 7-24 The forming section of a plastics plant puts out a continuous sheet of plastic that is \(1.2 \mathrm{~m}\) wide and \(2 \mathrm{~mm}\) thick at a rate of \(15 \mathrm{~m} / \mathrm{min}\). The temperature of the plastic sheet is \(90^{\circ} \mathrm{C}\) when it is exposed to the surrounding air, and the sheet is subjected to air flow at \(30^{\circ} \mathrm{C}\) at a velocity of \(3 \mathrm{~m} / \mathrm{s}\) on both sides along its surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in \(2 \mathrm{~s}\). Determine the rate of heat transfer from the plastic sheet to the air.

Short answer

Solution: Using the calculated values from the previous steps, we can find the rate of heat transfer for both cases: For the water over the plate: \(h_{water} \approx \frac{23.26 \times 30.0 \times 33.3^{\circ} \mathrm{C}}{1.0 \mathrm{~m}} = 23,228.38 \mathrm{~W/m}\) Rate of heat transfer per unit width \(= 23,228.38 \mathrm{~W/m} \times 33.3^{\circ} \mathrm{C} \times 1.0 \mathrm{~m} = 772,593.54 \mathrm{~W/m^2}\) For the air over the plastic sheet: \(h_{air} \approx \frac{23.26 \times 3.0 \times 60^{\circ} \mathrm{C}}{2 \mathrm{~s}} = 2081.4 \mathrm{~W/m}\) Total rate of heat transfer \(= (2081.4 \mathrm{~W/m} \times 60^{\circ} \mathrm{C}) \times 1.2 \mathrm{~m} = 149,860.8 \mathrm{~W/m^2}\) The rate of heat transfer per unit width of the plate for the water is 772,593.54 W/m², and for the air over the plastic sheet is 149,860.8 W/m².

Step-by-step solution

Calculate the temperature difference for water and air

First, we need to determine the temperature difference between the water flowing above the plate and the plate's surface temperature, and between the plastic sheet and the air. For the water over the plate: Temperature difference = \(43.3^{\circ} \mathrm{C} - 10.0^{\circ} \mathrm{C} = 33.3^{\circ} \mathrm{C}\) For the plastics plant: Plastic sheet temperature \(=90^{\circ} \mathrm{C}\) Air temperature \(=30^{\circ} \mathrm{C}\) Temperature difference = \(90^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 60^{\circ} \mathrm{C}\)

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient \((h)\) can be estimated for both cases (water over the plate and air over the plastic sheet) using empirical correlations related to the flow conditions, geometry, and fluid properties. For simplicity, we assume the average convective heat transfer coefficient over the length of the plate and the plastic sheet. For the water over the plate, we will use the Dittus Boelter Equation: \(h_{water} \approx \frac{23.26 \times 30.0 \times 33.3^{\circ} \mathrm{C}}{1.0 \mathrm{~m}}\) For the air over the plastic sheet, we will use the same equation: \(h_{air} \approx \frac{23.26 \times 3.0 \times 60^{\circ} \mathrm{C}}{2 \mathrm{~s}}\)

Calculate the rate of heat transfer for both cases

Now we can calculate the rate of heat transfer per unit width by multiplying the convective heat transfer coefficient with the temperature difference and the length of the plate. For the water over the plate: Rate of heat transfer per unit width \(= h_{water} \times (33.3^{\circ} \mathrm{C}) \times 1.0 \mathrm{~m}\) For the air over the plastic sheet: Total rate of heat transfer \(= (h_{air} \times 60^{\circ} \mathrm{C}) \times 1.2 \mathrm{~m}\) Use the calculated values of \(h_{water}\) and \(h_{air}\) to find the heat transfer rate for both cases.

Key concepts

Convective Heat Transfer Coefficient

The convective heat transfer coefficient, often represented as \( h \), is a crucial factor in understanding how efficiently heat is transferred between a solid surface and a fluid (like water or air) flowing over it. This coefficient depends on several factors, such as the type of fluid, the velocity of the fluid flow, and the characteristics of the surface in contact.

Some key points include:

• The higher the velocity of the fluid, the greater the convective heat transfer coefficient.
• Smooth surfaces generally have lower coefficients than rough ones due to the decreased turbulence.
• Different fluids have different thermal properties affecting \( h \).

In practical applications, engineers often use empirical correlations to estimate \( h \) for various conditions.

Dittus Boelter Equation

The Dittus Boelter Equation is a widely used empirical correlation for calculating the convective heat transfer coefficient in scenarios involving turbulent flow inside pipes. The equation is expressed as:

\[h = 0.023 imes ext{Re}^{0.8} imes ext{Pr}^{n}\]where:

• \( ext{Re} \) is the Reynolds number, indicating whether the flow is laminar or turbulent.
• \( ext{Pr} \) is the Prandtl number, describing the fluid's thermal conductivity relative to its viscosity.
• \( n \) is generally 0.3 for heating and 0.4 for cooling.

This equation is essential for evaluating how efficiently heat is transferred in industrial processes.

Temperature Difference

Understanding temperature difference is fundamental for calculating heat transfer. It refers to the difference in temperature between two surfaces or fluids. The larger the temperature difference, the more heat transfer can occur.

For instance:

• In the water over the plate scenario, the temperature difference is \( 33.3^{\circ} \mathrm{C} \).
• For the plastic sheet and air case, it is \( 60^{\circ} \mathrm{C} \).

This concept acts as the driving force for heat transfer, and greater differences often translate into higher rates of heat exchange.

Empirical Correlations

Empirical correlations are mathematical relationships derived from experimental data. They serve as essential tools for engineers to predict complex phenomena like heat transfer without exact simulations.

Some benefits include:

• They provide estimated values of variables like \( h \) under specific conditions.
• They save time and resources by avoiding detailed computational models.
• They are based on historical data, which enhances their reliability.

Using such correlations, like the Dittus Boelter Equation, allows for accurate estimations in designing more efficient heat transfer systems.