Question

A 15 -m-long polished copper tube with an outer diameter of \(5 \mathrm{~cm}\) is submerged in water at \(270 \mathrm{kPa}\). The tube is heated with hot gases flowing through it to generate steam by boiling the water. In the interest of minimizing the excess temperature, determine the surface temperature of the tube that is necessary to achieve a rate of steam generation of \(1.5 \mathrm{~kg} / \mathrm{s}\). What would be the boiling regime for the steam generation?

Short answer

Based on the given problem, the surface temperature of the copper tube necessary to achieve a steam generation rate of 1.5 kg/s is approximately 146.46°C. The boiling regime in this case is nucleate boiling.

Step-by-step solution

Calculate the heat transfer rate required to achieve the steam generation rate

To calculate the heat transfer rate, we need to know the amount of energy required to heat the water from liquid to steam. The heat transfer rate (\(Q\)) can be calculated using the following formula: \(Q = m \times L\) where \(m\) is the rate of steam generation (1.5 kg/s) and \(L\) is the latent heat of vaporization for water. For water, the latent heat of vaporization at \(270 \mathrm{kPa}\) is approximately \(2200 \mathrm{~kJ/kg}\). Therefore, the required heat transfer rate can be calculated as: \(Q = 1.5 \mathrm{~kg/s} \times 2200 \mathrm{~kJ/kg} = 3300 \mathrm{~kW}\)

Calculate the required surface temperature

To determine the necessary tube surface temperature, we can use Newton's Law of Cooling, which connects the heat transfer rate (\(Q\)), heat transfer coefficient (\(h\)), temperature difference between the tube surface (\(T_s\)), and the surrounding water (\(T_w\)) for a given surface area (\(A\)). \(Q = h \times A \times (T_s - T_w)\) We know the heat transfer rate \(Q\) from step 1. The area, \(A\) can be calculated as: \(A = \pi d L = \pi \times 0.05 \mathrm{m} \times 15 \mathrm{m} \approx 2.36 \mathrm{~m^2}\) We can rewrite the equation for surface temperature as: \(T_s = T_w + \frac{Q}{h \times A}\) We need to estimate the heat transfer coefficient (\(h\)) for the given conditions. The water temperature at \(270 \mathrm{kPa}\) saturation pressure is \(133.5^\circ C\). Since we are asked about the boiling regime, we should find the boiling heat transfer coefficient for nucleate boiling, which can be estimated as: \(h = 5.0 \times 10^3 \mathrm{~W/(m^2 K)}\) Now we can calculate the required surface temperature: \(T_s = 133.5^\circ C + \frac{3300 \mathrm{~kW}}{5.0 \times 10^3 \mathrm{~W/(m^2 K)} \times 2.36 \mathrm{~m^2}} \approx 146.46^\circ C\)

Determine the boiling regime

Now, we will determine the boiling regime in which the steam generation occurs. The boiling regimes are usually illustrated using a boiling curve, which plots the temperature difference between the surface and the liquid (\(\Delta T = T_s - T_w\)) against the heat flux (\(q = Q/A\)). Based on our calculations in step 2, the temperature difference is: \(\Delta T = 146.46 - 133.5 = 12.96^\circ C\) The heat flux can be calculated as: \(q = \frac{3300 \mathrm{~kW}}{2.36 \mathrm{~m^2}} \approx 1.4 \times 10^6 \mathrm{~W/m^2}\) Now, we can plot these values on a boiling curve and identify the boiling regime. From the boiling curve, we find that for the given temperature difference and heat flux, the boiling regime corresponds to nucleate boiling. This is the most common boiling regime encountered in industrial applications, where heat transfer is high, and the temperature difference between the heated surface and the boiling liquid is relatively low. So, the required tube surface temperature to achieve a rate of steam generation of 1.5 kg/s is approximately \(146.46^\circ C\), and the boiling regime is nucleate boiling.

Key concepts

Boiling Heat Transfer

Boiling heat transfer is a process where heat is transferred from a surface to a liquid, causing the liquid to change from a liquid phase to a vapor phase at the boiling temperature. Understanding the dynamics of boiling is crucial in thermal systems to ensure efficient and controlled heat transfer.
One common situation where boiling heat transfer is essential is in steam generation systems, like the copper tube in the exercise. Here, heat is applied to a tube submerged in water to boil the water and create steam, effectively transferring energy from the hot gases inside the tube to the water around it.
There are various boiling regimes, each characterizing a different behavior of bubbles on the heating surface. These regimes are often represented on a boiling curve, which shows the relationship between temperature difference and heat flux. The boiling regime directly influences how efficient the heat transfer process is. When analyzing a thermal system, proper identification of the boiling regime is critical, as it affects the surface conditions and heat transfer coefficient employed in calculations.

Nucleate Boiling

Nucleate boiling is one of the most effective regimes of boiling heat transfer. In nucleate boiling, bubbles form at distinct nucleation sites on the heated surface and detach into the bulk of the liquid. This process enhances the heat transfer rate due to the agitation and increased surface area for heat exchange.
During nucleate boiling, the temperature of the liquid slightly exceeds the saturation temperature, causing the liquid adjacent to the heated surface to evaporate energetically. This regime allows for high heat transfer coefficients, meaning that heat transfer occurs quickly and efficiently.
In the given problem, the boiling regime for the steam generation system was determined as nucleate boiling. The calculations of heat flux and temperature difference, using the boiling curve, confirmed this. It's crucial to correctly identify nucleate boiling as it ensures the system operates optimally, preventing overheating and maintaining energy efficiency.

• Efficiency: High heat transfer rate.
• Energy Conversion: Quick phase change helps in effective energy utilization.
• Application: Common and preferred regime in industrial applications.

Surface Temperature Calculation

Surface temperature calculation is vital for understanding and controlling heat transfer in boiling processes. The surface temperature of the copper tube in the exercise is determined using principles like Newton's Law of Cooling, which takes into account the heat transfer rate, the heat transfer coefficient, and the area over which the heat is spread.
The formula used to find the necessary surface temperature is:\[ T_s = T_w + \frac{Q}{h \times A} \]where:

• \( T_s \) is the surface temperature.
• \( T_w \) is the surrounding water temperature.
• \( Q \) is the total heat transfer rate.
• \( h \) is the heat transfer coefficient.
• \( A \) is the heat transfer area of the tube.

In the exercise, given a heat transfer rate of 3300 kW, a heat transfer coefficient of 5000 W/(m²·K), and a heat transfer area of approximately 2.36 m², the calculated surface temperature of the tube is approximately 146.46°C. Calculating this temperature ensures that the system remains within the desired boiling regime, particularly nucleate boiling, which is both effective and stable for steam generation.