Question

Oxy-fuel combustion power plants use pulverized coal particles as fuel to burn in a pure oxygen environment to generate electricity. Before entering the furnace, pulverized spherical coal particles with an average diameter of \(300 \mu \mathrm{m}\), are being transported at \(2 \mathrm{~m} / \mathrm{s}\) through a \(3-\mathrm{m}\) long heated tube while suspended in hot air. The air temperature in the tube is \(900^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature of the coal particles at the exit of the heated tube, if the initial temperature of the particles is \(20^{\circ} \mathrm{C}\).

Short answer

Answer: The final temperature of the coal particles at the exit of the heated tube can be found through a few main steps: 1. Calculate the surface area of the coal particles. 2. Calculate the heat transfer rate along the tube. 3. Calculate the energy balance and find the final temperature. By combining these steps, you can use specific properties of coal, convection heat transfer equations, and the information provided to solve for the final temperature of the coal particles in the scenario.

Step-by-step solution

Calculate the surface area of the coal particles

Since the coal particles are spherical, we can use the formula for the surface area of a sphere: \(A = 4 \pi r^2\). The diameter of the coal particles is given as \(300 \mu m\), so the radius \(r\) can be calculated as: \(r = \frac{300}{2} \cdot 10^{-6} \mathrm{m}\) Now, calculate the surface area \(A\): \(A = 4 \pi (\frac{300}{2} \cdot 10^{-6})^2 \mathrm{m^2}\)

Calculate the heat transfer rate along the tube

We can use the convection heat transfer equation to calculate the rate of heat transfer along the tube: \(q = h A \Delta T\) where \(q\) is the heat transfer rate, \(h\) is the average convection heat transfer coefficient given as \(250 \mathrm{W/m^2K}\), \(A\) is the surface area of the coal particles calculated earlier, and \(\Delta T\) is the difference between the air temperature and the initial temperature of the coal particles. \(\Delta T = 900^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 880 \mathrm{K}\) Now, calculate the heat transfer rate: \(q = 250 \cdot A \cdot 880 \mathrm{W} \)

Calculate the energy balance and find the final temperature

In this step, we'll make the assumption that the internal energy of coal particles is mainly due to sensible heat and can be represented through the specific heat capacity. The equation for energy balance can be written as: \( q = m c_p \Delta T_{final} \) Where \(m\) is the mass of coal particles (we'll assume the density \(\rho\) is 1500 \( \mathrm{kg/m^3} \)), \(c_p\) is the specific heat capacity of coal which we will assume to be \(1000 \mathrm{J/kgK}\), and \(\Delta T_{final}\) is the increase in temperature from the initial value. For the mass of the coal particles, we can use the formula for the volume of a sphere and multiply it with the given density \(\rho\): \(m = \frac{4}{3} \pi r^3 \rho \) Now, we have the following equation to solve for \(\Delta T_{final}\): \(q = m c_p \Delta T_{final}\) Next, solving for \(\Delta T_{final}\): \(\Delta T_{final} = \frac{q}{m c_p}\) Now, we can find the final temperature \(T_{final}\): \(T_{final} = T_{initial} + \Delta T_{final}\) \(T_{final} = 20^{\circ} \mathrm{C} + \Delta T_{final}\) The final temperature \(T_{final}\) represents the temperature of the coal particles at the exit of the heated tube.

Key concepts

Convection Heat Transfer

Convection heat transfer plays a vital role in power plants, as it's one of the fundamental mechanisms of heat movement. It occurs when heat is transferred through a fluid (such as air or water) which is caused by the fluid's movement. In our exercise example, the hot air surrounding the coal particles transports heat to the particles' surface.

In technical terms, the rate at which heat is transferred by convection can be calculated with the formula:
\[ q = h A \Delta T \]
where:\[ q \] is the heat transfer rate, \[ h \] is the convection heat transfer coefficient (measured in Watts per square meter per Kelvin, \[ W/m^2K \]), \[ A \] is the surface area of the object was transferring heat, and \[ \Delta T \] is the temperature difference between the fluid and the object's surface. In power plants, understanding and optimizing convection heat transfer is crucial for efficient energy production.

Thermal Energy Balance

Thermal energy balance is a principle stating that energy cannot be created or destroyed in an isolated system. When applied to the context of power plants, it ensures that all heat transfers are accounted for, and the energy output matches the energy input. The coal particles in our scenario receive heat through convection, which increases their internal energy.

The formula for the energy balance involving heat transfer by convection is:
\[ q = m c_p \Delta T_{final} \]
where \[ m \] is the mass of the material (coal particles), \[ c_p \] is the specific heat capacity (amount of heat per unit mass required to raise the temperature by one degree Celsius or one Kelvin), and \[ \Delta T_{final} \] is the change in temperature of the material. This balance allows us to calculate the final temperature of coal particles after they've absorbed heat from the hot air.

Specific Heat Capacity

Specific heat capacity is a property that measures how much energy is needed to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). The specific heat capacity affects how quickly a substance heats up or cools down under the influence of an external heat source or sink.

In the power plant scenario, coal has its specific heat capacity, typically measured in \(\mathrm{J/kgK}\). With a known specific heat capacity, we can calculate how much energy, in joules, is required to bring about a certain temperature increase. The formula for the amount of energy needed can be expressed as:
\[ q = m c_p \Delta T \]
By configuring this relationship, engineers can make decisions about material selection and system design that impact the efficiency and safety of power plants.

Temperature Change Calculation

Temperature change calculation is vital to understand how much an object will heat up or cool down over time when subjected to a heat transfer process. To calculate the final temperature of an object, like the coal particles in our example, we begin with the formula:
\[ \Delta T_{final} = \frac{q}{m c_p} \]
The temperature change \( \Delta T_{final} \) is directly proportional to the heat transferred \( q \) and inversely proportional to the mass \( m \) and the specific heat capacity \( c_p \).

Therefore, the final temperature can be found using the initial temperature and adding the calculated temperature change:
\[ T_{final} = T_{initial} + \Delta T_{final} \]
For various applications in power plants, calculating an accurate temperature change is crucial for the control and optimization of processes, equipment design, and safety protocols.