Question

A 5-m-long strip of sheet metal is being transported on a conveyor at a velocity of \(5 \mathrm{~m} / \mathrm{s}\), while the coating on the upper surface is being cured by infrared lamps. The coating on the upper surface of the metal strip has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the surrounding ambient air temperature is \(25^{\circ} \mathrm{C}\). Radiation heat transfer occurs only on the upper surface, while convection heat transfer occurs on both upper and lower surfaces of the sheet metal. If the infrared lamps supply a constant heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the sheet metal. Evaluate the properties of air at \(80^{\circ} \mathrm{C}\).

Short answer

Explain step by step how you would calculate this.

Step-by-step solution

Set up energy balance equation

Since we need to determine the surface temperature of the sheet metal, let's set up an energy balance equation. The energy entering the metal is due to the absorption of heat from infrared lamps, and the energy leaving the metal is due to combined convection and radiation from the upper and lower surfaces. The energy balance equation is: $$Q_{in} = Q_{conv,t} + Q_{rad} + Q_{conv,b}$$ where \(Q_{in}\) = Incoming heat from infrared lamps \(Q_{conv,t}\) = Convective heat transfer from the top surface. \(Q_{rad}\) = Radiative heat transfer from the top surface \(Q_{conv,b}\) = Convective heat transfer from the bottom surface.

Calculate the incoming heat

We know that the incoming heat flux is given as \(5000 \mathrm{~W} / \mathrm{m}^{2}\). To find the total heat coming into the metal, we need to multiply the heat flux with the cross-sectional area (5 m) of the sheet metal and the absorptivity (0.6). So, we have: $$Q_{in} = 5000 \mathrm{~W} / \mathrm{m}^{2} * 5\mathrm{~m} * 0.6 = 15000 \mathrm{~W}$$

Calculate the convective heat transfer from top and bottom surfaces

The convective heat transfer rate can be calculated as: $$Q_{conv} = hA(T_{s} - T_{\infty})$$ Since the convection heat transfer happens on both upper and lower surfaces, and the properties of air are to be evaluated at \(80^{\circ}C\), let's assume the film temperature \((T_s+T_{\infty})/2 = 80^{\circ}C\). So, for both surfaces, we have: $$Q_{conv,t} = 2hA(T_s - T_{\infty})$$ where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area of the sheet metal (5 m), and \(T_s\) and \(T_{\infty}\) are the surface and ambient temperatures respectively.

Calculate the radiative heat transfer from the top surface

The radiative heat transfer rate is given by the Stefan-Boltzmann law: $$Q_{rad} = \epsilon \sigma A (T_{s}^{4} - T_{\infty}^{4})$$ where \(\epsilon = 0.7\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}\)), and \(T_{\infty} = 25^{\circ}C\) is the ambient temperature.

Solve the energy balance equation

Now the energy balance equation becomes: $$15000 \mathrm{~W} = 2hA(T_s - T_{\infty}) + 0.7 * (5.67 \times 10^{-8}) * 5 (T_{s}^{4} - 298^{4})$$ Next, we will need to rearrange and solve this equation for the surface temperature, \(T_s\). This is a non-linear equation, so an iterative method like the Newton-Raphson method or a numerical solver must be used to find the value of \(T_s\). After solving this equation with the given coefficient values and air properties, the surface temperature \(T_s\) can be determined.

Key concepts

Radiation Heat Transfer

Understanding radiation heat transfer is key to solving problems involving thermal energy emission from surfaces. This type of heat transfer occurs without the need for a medium and can happen through the vacuum of space. In our exercise, radiation heat transfer plays a crucial role as the upper surface of the metal strip emits energy after being heated by infrared lamps.

Radiation heat transfer is primarily dependent on the temperature difference between a surface and its surroundings, the surface's emissivity (a measure of how well a surface emits thermal radiation), and the Stefan-Boltzmann constant. The formula to calculate the radiative heat transfer rate from a surface is expressed by the Stefan-Boltzmann law, which will be explored further in another section. Recognizing these factors allows us to accurately assess how much heat is lost due to radiation when attempting to determine the surface temperature of the sheet metal in our problem.

Convection Heat Transfer

Convection heat transfer is another method by which heat is moved from one place to another. This type of transfer is characterized by the movement of molecules within fluids (such as air or water), which carry thermal energy as they move. In the case of our sheet metal, convection occurs on both the upper and lower surfaces, where the ambient air takes away some of the heat.

The rate of convection heat transfer can be described using the formula: $$Q_{conv} = hA(T_{s} - T_{\text{\tiny\textcircled{r}}})$$where h is the heat transfer coefficient, A is the surface area, T_{s} is the surface temperature of the metal, and T_{\text{\tiny\textcircled{r}}} (typically denoted as T_{\text{\tiny\textinfinity}}) is the fluid temperature far from the surface (ambient temperature). This relationship tells us that the heat transfer rate is proportional to the surface area, the temperature difference, and the convective heat transfer coefficient, which depends on the properties of the fluid and the flow conditions.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a cornerstone in thermal physics, relating the radiative energy emitted from a black body to the fourth power of its absolute temperature. In its mathematical form, it states:$$Q_{rad} = \text{\textepsilon} \text{\textsigma} A (T_{s}^{4} - T_{\text{\tiny\textcircled{r}}}^{4})$$Here, \text{\textepsilon} represents the emissivity of the surface, \text{\textsigma} the Stefan-Boltzmann constant (5.67 \times 10^{-8} \text{W/m}^{2}\text{K}^{4}), A the area, T_{s} the absolute surface temperature (in Kelvins), and T_{\text{\tiny\textcircled{r}}} the absolute temperature of the surroundings.

Our exercise requires us to apply this law to determine how much heat is being radiated away from the metal's surface. The emissivity value given in the problem indicates that the surface does not emit radiation as efficiently as a perfect black body (which would have an emissivity of 1). This law is critical for realistic heat transfer calculations in engineering and environmental sciences.

Energy Balance Equation

The energy balance equation is a statement of the conservation of energy principle and is used to solve many heat transfer problems, including the one about the sheet metal being cured by infrared lamps. It essentially says that the energy entering a system must equal the energy leaving it plus any change in the energy stored within the system.

In steady-state conditions, where there is no change in stored energy, the energy balance equation simplifies to equate incoming and outgoing energy rates. For our exercise, the equation is:$$Q_{in} = Q_{conv,t} + Q_{rad} + Q_{conv,b}$$This indicates that the heat provided by the infrared lamps to the metal (Q_{in}) is dissipated through convection at the top (Q_{conv,t}) and bottom (Q_{conv,b}) surfaces, and through radiation from the top surface (Q_{rad}). To find the surface temperature (T_{s}), we must rearrange and solve this equation, accounting for all the given values and material properties.