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Chapter 1: Problem 68

Following the hot vacuum forming of a paper-pulp mixture, the product, an egg carton, is transported on a conveyor for \(18 \mathrm{~s}\) toward the entrance of a gas-fired oven where it is dried to a desired final water content. Very little water evaporates during the travel time. So, to increase the productivity of the line, it is proposed that a bank of infrared radiation heaters, which provide a uniform radiant flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), be installed over the conveyor. The carton has an exposed area of \(0.0625 \mathrm{~m}^{2}\) and a mass of \(0.220 \mathrm{~kg}, 75 \%\) of which is water after the forming process. The chief engineer of your plant will approve the purchase of the heaters if they can reduce the water content by \(10 \%\) of the total mass. Would you recommend the purchase? Assume the heat of vaporization of water is \(h_{f g}=2400 \mathrm{~kJ} / \mathrm{kg}\).

Short Answer

Expert verified
The infrared radiation heaters provide only 5.625 kJ of energy during the 18-second transport time, while the energy required to remove 10% of the water content is 52.8 kJ. Since the energy provided is significantly less than the energy required, the heaters would not be able to achieve the desired reduction in water content. Consequently, we would not recommend the purchase of the infrared radiation heaters for this purpose.

Step by step solution

01

Determine mass of water to be removed

We need to reduce the water content in the egg carton by 10% of its total mass, which is 0.220 kg. Thus, we will first find out how much mass of water to be removed. The mass of water to be removed will be: \(Mass_{removed}=0.10 \times 0.220~kg=0.022~kg\)
02

Calculate the energy required to remove the water

Knowing the mass of water to be removed, we can now find the energy needed for this purpose using the heat of vaporization of water \(h_{fg}\). The energy required to remove the water, \(E_{required}\), can be calculated as: \(E_{required} = Mass_{removed} \times h_{fg}\) Replacing the values, we get: \(E_{required} = 0.022~kg \times 2400~\frac{kJ}{kg} = 52.8~kJ\)
03

Calculate the energy provided by the heaters

Now let's find the energy provided by the infrared radiation heaters during the transport time (18 seconds). The energy provided can be calculated using the following formula: \(E_{provided}= Radiant\,flux \times \textrm{Exposed area}\times\textrm{Time}\) So, plugging in the given values, we get: \(E_{provided}= 5000 \frac{W}{m^2} \times 0.0625 m^2 \times 18s\) Note that to have compatible units, we need to convert Watts to kilojoules: \( 1 \frac{W}{m^2} = 0.001 \frac{kJ}{m^2\cdot s} \) Now the expression becomes: \(E_{provided} = 5 \frac{kJ}{m^2 \cdot s} \times 0.0625 m^2 \times 18s = 5.625~kJ\)
04

Compare the energy required and the energy provided

Now we'll compare the energy required to evaporate the needed amount of water, \(E_{required}\), to the energy provided by the infrared radiation heaters, \(E_{provided}\): \(E_{required}>E_{provided}\) Comparing the energies calculated in Step 2 and Step 3: \(52.8~kJ > 5.625~kJ\) Since the energy required to evaporate the required mass of water is significantly greater than the energy provided by the proposed heaters in the given transport time, the heaters would not be able to achieve the desired reduction in water content. Therefore, we would not recommend the purchase of the infrared radiation heaters for this purpose.

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Key Concepts

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

Infrared Radiation Heaters
Infrared radiation heaters are a type of heating technology that emit infrared light, which heats objects directly without using a medium like air or water. This makes them very efficient for certain applications, such as drying and heating surfaces rapidly.
Infrared heaters work by emitting energy in the form of infrared radiation, which is absorbed by objects and translates directly into heat. This can be particularly useful in manufacturing processes where targeted heating is needed, like the drying of materials such as egg cartons in our example.
In the exercise, we see infrared heaters being considered to dry a product by reducing the water content through evaporation. This is achieved by providing a certain amount of radiant flux, measured in watts per square meter ( ext{W/m}^2), over a specific surface area for a set period of time. In this case, the heaters offer a power density of 5000 ext{W/m}^2 over the area of the carton. The energy provided can then be used to calculate if it meets the energy required to achieve the desired water evaporation.
Heat of Vaporization
The heat of vaporization is a key concept when discussing phase changes, like turning liquid water into vapor. It represents the amount of energy required to convert one kilogram of a substance from a liquid to a gas without changing its temperature. For water, this value is quite high, indicating that water requires a significant amount of energy to vaporize.
The heat of vaporization for water, which is used in this exercise, is given as 2400 ext{kJ/kg}. This means that each kilogram of water would need 2400 kilojoules of energy to convert it from liquid to vapor. This concept is crucial when calculating the energy needed in processes that involve evaporation and drying.
In the scenario with the egg carton, to find out how much energy is necessary to achieve a 10% reduction in water weight, we multiply the mass of water to be removed by the heat of vaporization. This tells us the total energy needed for the evaporation process.
Thermal Energy Calculation
Calculating thermal energy involves determining how much energy is required or supplied during a process, such as heating, cooling, or switching phases like melting or evaporation.
In the given problem, we need to calculate the energy provided by infrared radiation heaters and compare it against the energy required to evaporate a certain mass of water. This comparison allows us to evaluate whether the suggested heating method is effective for the task at hand.
  • Firstly, determine the mass of water meant for evaporation.
  • Calculate the energy needed using the formula: \[E_{required} = ext{Mass}_{removed} \times h_{fg}\]where \(h_{fg}\) is the heat of vaporization.
  • Finally, find the energy provided by the infrared heaters using the formula: \[E_{provided} = ext{Radiant flux} \times ext{Exposed area} \times ext{Time}\]By comparing \(E_{required}\) against \(E_{provided}\), one can determine whether the heaters are sufficient for the task. In the given exercise, the heaters' energy provision falls short of the requirement, leading to the conclusion that these heaters would not meet the desired outcome.

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Most popular questions from this chapter

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Single fuel cells such as the one of Example \(1.5\) can be scaled up by arranging them into a fuel cell stack. A stack consists of multiple electrolytic membranes that are sandwiched between electrically conducting bipolar plates. Air and hydrogen are fed to each membrane through fiw channels within each bipolar plate, as shown in the sketch. With this stack arrangement, the individual fuel cells are connected in series, electrically, producing a stack voltage of \(E_{\text {stack }}=N \times E_{c}\), where \(E_{c}\) is the voltage produced across each membrane and \(N\) is the number of membranes in the stack. The electrical current is the same for each membrane. The cell voltage, \(E_{c}\), as well as the cell efficiency, increases with temperature (the air and hydrogen fed to the stack are humidified to allow operation at temperatures greater than in Example 1.5), but the membranes will fail at temperatures exceeding \(T \approx 85^{\circ} \mathrm{C}\). Consider \(L \times w\) membranes, where \(L=w=100 \mathrm{~mm}\), of thickness \(t_{m}=0.43 \mathrm{~mm}\), that each produce \(E_{c}=0.6 \mathrm{~V}\) at \(I=60 \mathrm{~A}\), and \(\dot{E}_{c g}=45 \mathrm{~W}\) of thermal energy when operating at \(T=80^{\circ} \mathrm{C}\). The external surfaces of the stack are exposed to air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and surroundings at \(T_{\text {sur }}=30^{\circ} \mathrm{C}\), with \(\varepsilon=0.88\) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the electrical power produced by a stack that is \(L_{\text {stack }}=200 \mathrm{~mm}\) long, for bipolar plate thickness in the range \(1 \mathrm{~mm}

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A cartridge electrical heater is shaped as a cylinder of length \(L=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\). Under normal operating conditions, the heater dissipates \(2 \mathrm{~kW}\) while submerged in a water flow that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, determine its surface temperature \(T_{s}\). If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at \(20^{\circ} \mathrm{C}\) but for which \(h=50\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding surface temperature? What are the consequences of such an event?
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