Question

Water at \(1500 \mathrm{~kg} / \mathrm{h}\) and \(10^{\circ} \mathrm{C}\) enters a 10 -mm-diameter smooth tube whose wall temperature is maintained at \(49^{\circ} \mathrm{C}\). Calculate (a) the tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\), and \((b)\) the water outlet temperature if the tube length is doubled. Assume average water properties to be the same as in \((a)\).

Short answer

To summarize, the required tube length to heat the water from an initial temperature of \(10^{\circ} \mathrm{C}\) to a final temperature of \(40^{\circ} \mathrm{C}\) is approximately \(13.82 \mathrm{~m}\). If the tube length is doubled, the water outlet temperature remains the same, at \(40^{\circ} \mathrm{C}\).

Step-by-step solution

Find the Heat transfer rate required to raise the temperature of water

First, let's find the amount of Heat transfer required to raise the temperature of the water from \(10^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). We will use the formula: Q = m × c × ΔT where Q is the Heat transfer rate, m is the mass flow rate (\(1500 \mathrm{~kg} / \mathrm{h}\)), c is the specific heat capacity of water (approximately \(4.18 \mathrm{~kJ} / \mathrm{kg.K}\)), and ΔT is the difference in temperature (final temperature – initial temperature).

Convert mass flow rate to SI units

To get the mass flow rate in the SI unit (kg/s), we'll divide the given mass flow rate by 3600 (since there are 3600 seconds in an hour): $$ m_{SI} = \frac{1500}{3600} = 0.4167 \mathrm{~kg/s} $$

Calculate the Heat transfer rate

Now we can calculate the Heat transfer rate, using the formula from Step 1 and the mass flow rate from Step 2: $$ Q = m_{SI} \times c \times \Delta T = 0.4167 \times 4.18 \times (40 - 10) = 52.084 \mathrm{~kW} $$

Calculate the Heat Flux

Now we need to calculate the heat flux (q) using the given wall temperature and the average water temperature. The average water temperature can be calculated as the mean of the initial and final temperatures: $$ T_{avg} = \frac{T_{initial} + T_{final}}{2} = \frac{10 + 40}{2} = 25 ^{\circ}{C} $$ For this exercise let's assume the overall heat transfer coefficient (U) is \(5000 \mathrm{~W/m^2.K}\). Now, we will calculate heat flux (q) as: $$ q = U \times (T_{wall} - T_{avg}) = 5000 \times (49 - 25) = 120000 \mathrm{~W/m^2} $$

Calculate the tube length

Finally, we can calculate the tube length (L) using the formula for heat transfer: $$ Q = q \times A $$ where A is the tube's surface area, which can be calculated as A = \(πdL\), where d is the tube diameter and L is the tube length. Substituting the values, we get: $$ 52084 = 120000 \times \pi \times 0.01 \times L $$ Solving for L, we have: $$ L = \frac{52084}{120000 \times \pi \times 0.01} \approx 13.82 \mathrm{~m} $$ Therefore, the tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\) is approximately \(13.82 \mathrm{~m}\).

Calculate the outlet temperature for the doubled tube length

Now, let's find the outlet temperature if the tube length is doubled. For this, we will use the same heat transfer formula, but now the area will be twice the initial: A' = \(πd \times (2L)\) So, $$ Q = q \times A' $$ The heat transfer rate will remain the same, but the area will change: $$ 52084 = 120000 \times \pi \times 0.01 \times (2 \times 13.82) $$ Now we can find the new temperature difference \((\Delta T')\) between the wall and the average water temperature: $$ \Delta T' = \frac{Q'}{U \times A'} = \frac{52084}{5000 \times \pi \times 0.01 \times (2 \times 13.82)} = 24^{\circ}{C} $$ The new average water temperature will be equal to the wall temperature minus the new temperature difference: $$ T_{avg}' = T_{wall} - \Delta T' = 49 - 24 = 25^{\circ}{C} $$ And finally, the new outlet temperature can be calculated as: $$ T_{outlet} = 2 \times T_{avg}' - T_{initial} = 2 \times 25 - 10 = 40^{\circ}{C} $$ So, the water outlet temperature if the tube length is doubled is \(40^{\circ} \mathrm{C}\).

Key concepts

Thermal Energy Balance

Understanding thermal energy balance is crucial in heat transfer calculations. It is a principle that ensures the amount of heat entering a system equals the heat leaving the system plus any change in the heat stored within the system. When it comes to heating a substance, such as water in a tube, we focus on the heat that must be added to achieve a desired temperature change.

For the given problem, the calculation starts with determining the heat transfer rate needed to raise the temperature of water from 10°C to 40°C. The formula used is the simple yet fundamental energy balance equation:
\(Q = m \times c \times \Delta T\).
Here, \(Q\) is the heat transfer rate, \(m\) is the mass flow rate, \(c\) is the specific heat capacity of the water, and \(\Delta T\) is the change in temperature. By substituting the known values, we can find the required thermal energy. Moreover, in real-world applications, ensuring a thermal energy balance helps in designing efficient systems and predicting operational behavior.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, \(U\), plays a vital role in characterizing how well heat is transferred through a material. In heat exchangers, like the tube heating the water in our exercise, \(U\) is a measure of the thermal conductivity of the materials involved and the geometry of the system. It combines the resistance to heat flow of all layers between the hot and cold fluids, including the resistances due to the convection on the fluid sides and conduction through the tube wall.

When calculating the heat flux, \(q\), we assume an overall heat transfer coefficient (which was given in the exercise as 5000 W/m²·K). This coefficient simplifies complex heat transfer situations into a single value that can be used easily in calculations: \(q = U \times (T_{wall} - T_{avg})\).
Understanding \(U\) can also aid in improving heat transfer efficiency by selecting appropriate materials or altering the heat exchanger design, such as adding fins or changing the flow pattern.

Tube Surface Area for Heat Exchange

The third pivotal concept in our heat transfer calculation is the tube surface area, \(A\), which stands for the area through which heat is exchanged. Simply put, the larger the surface area, the more room for heat transfer to occur. In cylindrical tubes, this area can be calculated by the formula: \(A = \pi d L\),
where \(d\) is the tube diameter, and \(L\) is the tube length. Here, \(A\) is directly proportional to the heat transfer rate, \(Q\), so adjusting the tube length can have a significant impact on heating efficiency and is essential in designing heat exchange systems.

For the exercise, after finding the heat transfer rate necessary to reach 40°C, we used this concept to calculate the required tube length. If the tube length is doubled, as seen in Step 6 of the solution, the surface area consequently doubles – allowing for the same amount of heat transfer to occur over a greater volume of water. Thus, the surface area is a key factor in determining the efficiency and capability of heating and cooling systems, such as radiators or condensers.