Question

Cooling water available at \(10^{\circ} \mathrm{C}\) is used to condense steam at \(30^{\circ} \mathrm{C}\) in the condenser of a power plant at a rate of \(0.15 \mathrm{~kg} / \mathrm{s}\) by circulating the cooling water through a bank of 5 -m-long \(1.2-\mathrm{cm}\)-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of \(4 \mathrm{~m} / \mathrm{s}\) and leaves at a temperature of \(24^{\circ} \mathrm{C}\). The tubes are nearly isothermal at \(30^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient between the water, the tubes, and the number of tubes needed to achieve the indicated heat transfer rate in the condenser.

Short answer

Answer: The average heat transfer coefficient (h) is approximately 1,558 W/(m^2K), and two tubes are needed to achieve the heat transfer rate in the condenser.

Step-by-step solution

Calculate the mass flow rate of the cooling water

First, we must find the mass flow rate of the cooling water, using the given flow rate: $$ m = 0.15~\text{kg/s} $$

Calculate the temperature difference

We will find the temperature difference between the inlet and outlet temperature of the cooling water: $$ \Delta T = T_\text{out} - T_\text{in} = 24^\circ\text{C} - 10^\circ\text{C} = 14^\circ\text{C} $$

Determine the specific heat capacity of water

We must know the specific heat capacity of water to calculate the heat transfer rate. For water, the specific heat capacity is approximately: $$ c_p = 4.18~\text{kJ/(kg·K)} $$

Calculate the heat transfer rate from the cooling water

Using the mass flow rate, specific heat capacity, and temperature difference, we can calculate the heat transfer rate from the cooling water: $$ Q = m\cdot c_p\cdot \Delta T = 0.15~\text{kg/s} \times 4.18~\text{kJ/(kg·K)} \times 14~\text{K} $$ $$ Q = 8.781~\text{kJ/s} $$

Calculate the convective heat transfer coefficient

Now we can use the convective heat transfer formula to find the average heat transfer coefficient, h, considering the tube surface area A: $$ Q = h \cdot A \cdot (T_\text{tubes} - T_\text{in}) $$ Rearranging to solve for h: $$ h = \frac{Q}{A \cdot (T_\text{tubes} - T_\text{in})} $$ We will first find the surface area of a single tube. The surface area of a cylinder can be found with the formula: $$ A = 2\pi r_\text{external} L $$ Where r_external is the internal radius of the tube and L is the length of the tube. We're given the following values: Internal diameter: 1.2 cm, so the internal radius is 0.6 cm or 0.006 m Length of tubes: 5 m Therefore, the surface area of a single tube is: $$ A = 2\pi (0.006~\text{m}) (5~\text{m}) \approx 0.1885~\text{m}^2 $$ Now, we can find the average heat transfer coefficient, h: $$ h = \frac{Q}{A \cdot (T_\text{tubes} - T_\text{in})} $$ $$ = \frac{8,781~\text{W}}{0.1885~\text{m}^2\times (30^\circ\text{C} - 10^\circ\text{C})} $$ $$ h \approx 1,558~\text{W/(m}^2\text{K)} $$

Calculate the number of tubes required to achieve the heat transfer rate

Finally, we can find the number of tubes needed as: $$ Q = n \cdot h \cdot A \cdot (T_\text{tubes} - T_\text{in}) $$ Rearranging for the number of tubes, n: $$ n = \frac{Q}{h\cdot A \cdot (T_\text{tubes} - T_\text{in})} $$ Plugging in the known values: $$ n = \frac{8,781~\text{W}}{1,558~\text{W/(m}^2\text{K)}\times 0.1885~\text{m}^2\times{20}^\circ\text{C}} $$ $$ n \approx 1.2695\text{ (tubes)} $$ As you cannot use a fraction of a tube, we will round up to 2 tubes. So, two tubes are needed to achieve the heat transfer rate in the condenser.

Key concepts

Cooling Water Mass Flow Rate

The cooling water mass flow rate is a critical parameter in many industrial processes, including the operation of a condenser in a power plant. It refers to the amount of cooling water that flows through the system per unit of time, typically measured in kilograms per second (kg/s).

In our example, the mass flow rate is given as 0.15 kg/s. This value is crucial because it directly influences the amount of thermal energy that can be transported away from the steam being condensed. To improve understanding of this concept, consider that a higher mass flow rate means more water passes through the condenser per second, thus potentially increasing the system's ability to remove heat.

To illustrate, if you have a garden hose supplying water to cool down a hot surface, the rate at which you pour the water affects how quickly the surface cools. Similarly, in an industrial setting, managing the cooling water mass flow rate is important for maintaining optimal operating conditions and ensuring efficient heat transfer.

Convective Heat Transfer

Convective heat transfer is the process of heat transfer between a solid surface and a fluid (liquid or gas) that is moving. This phenomenon explains how heat is carried away from the surface of the tubes in our condenser by the flow of cooling water.

When discussing convective heat transfer, the heat transfer coefficient is a fundamental concept. It quantifies the efficiency of heat transfer per unit area and per unit temperature difference between the solid surface and the fluid. A higher coefficient indicates that heat can be transferred more effectively.

Application in the Condenser

In the power plant condenser scenario, the heat transfer coefficient helps us understand how effectively the copper tubes can transfer heat from the steam to the cooling water. By calculating this coefficient, we can design a system that maximizes heat transfer and efficiency. Convective heat transfer is also influenced by factors such as the fluid's velocity, its physical properties, and the nature of the fluid flow (whether it is turbulent or laminar).

Specific Heat Capacity

Specific heat capacity, symbolized as c_p, is a property of a material that measures the amount of energy required to raise the temperature of a unit mass by one degree in temperature. It is typically measured in joules per kilogram per Kelvin (J/kg·K) or kilojoules per kilogram per Kelvin (kJ/kg·K).

In our exercise, the specific heat capacity of water is used to determine how much energy is needed to raise the temperature of the cooling water as it absorbs heat from the steam. With a specific heat capacity of 4.18 kJ/(kg·K), water is an excellent medium for transferring heat due to its relatively high capacity to store thermal energy.

Why Is It Important?

Understanding the specific heat capacity of the cooling water is vital because it impacts how much heat the water can carry away from the condenser. It's like knowing how much heat you can absorb with a sponge; the higher the specific heat capacity, the more heat the water can 'soak up' before it becomes too hot.

Thermal Energy Balance

Thermal energy balance is used to analyze heat exchange systems by applying the principle of conservation of energy. It states that the energy lost by the system must be equal to the energy gained by the surrounding environment.

In the context of our power plant's condenser, the thermal energy balance involves ensuring that the heat removed from the steam equals the heat absorbed by the cooling water. The calculation of the condenser's heat transfer rate, which is the rate at which heat is transferred from steam to cooling water, is an application of thermal energy balance.

Practical Implications

In practice, engineers use the thermal energy balance to size equipment and design systems that achieve the desired heat transfer. For example, knowing the energy balance allows us to calculate the number of tubes required in a condenser to handle the expected heat load. It ensures that the system operates within its capacity and that the cooling water can effectively remove the desired amount of heat from the steam.