Question

Liquid water enters a 10 -m-long smooth rectangular tube with $a=50 \mathrm{~mm}\( and \)b=25 \mathrm{~mm}$. The surface temperature is kept constant, and water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.25 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\).

Short answer

Answer: The approximate required tube surface temperature to heat entering water from an initial temperature of 20°C to a desired outlet temperature of 80°C is 20.8°C. Note that this value may not be accurate due to simplistic assumptions made in the calculation and may require better estimations of the heat transfer coefficient 'h' and accounting for different heat transfer mechanisms.

Step-by-step solution

Calculate the heat transfer

First, we need to calculate the heat transfer required to achieve the desired outlet temperature. The equation for heat transfer is: \(Q = m\cdot C_p \cdot \Delta T\) Given values: - \(m = 0.25 \,\text{kg/s}\) - \(C_p \approx 4.18\, \text{kJ/kg·K}\), specific heat capacity of liquid water - \(\Delta T = 80°C - 20°C = 60 \,\text{K}\) Now, we will calculate the heat transfer: \(Q = (0.25\, \text{kg/s})\times(4.18\, \text{kJ/kg·K})\times(60\, \text{K}) = 62.7\, \text{kJ/s}\) The heat transfer required to heat the water to 80°C is 62.7 kJ/s.

Calculate required tube surface temperature

To find the required surface temperature of the tube, we must use the heat transfer equation again, this time using the concept of heat flux to connect surface temperature and the required heat transfer. A common formula to describe the heat flux is Newton's law of cooling: \(q= hA\Delta T\) Where: - \(q\) is the heat flux - \(h\) is the convective heat transfer coefficient - \(A\) is the heat transfer area - \(\Delta T\) is the temperature difference between the fluid and the surface In this case, the heat transfer area can be calculated using the tube dimensions: \(A = 2(a + b)\times L = 2 (50\, \text{mm} + 25\, \text{mm}) \times 10\, \text{m} = 15\, \text{m}^2\) We can set the heat flux equal to the total heat transfer that we calculated earlier: \(q = Q = 62.7\, \text{kJ/s}\) Rearranging Newton's law of cooling to find the temperature difference between the surface and the water: \(\Delta T = \frac{q}{hA}\) We don't know the value of the convective heat transfer coefficient 'h'. It depends on the specific case such as flow conditions as well as the characteristics of the system (geometry of the tube, the fluid properties, etc.). For an accurate solution, one would need to determine it using the appropriate correlations or look up a value in the literature. For simplicity, let us assume that the value of \(h\) is approximately \(5000 \, \text{W/m2·K}\). Substituting the values, we can find the temperature difference: \(\Delta T = \frac{62.7\, \text{kJ/s}}{5000\, \text{W/m2·K}\times15\, \text{m}^2} = 0.836 \,\text{K}\)

Determine the tube surface temperature

Now we can find the needed surface temperature by adding the temperature difference to the inlet water temperature: \(T_{surface} = T_{inlet} + \Delta T = 20°C + 0.836\, \text{K} \approx 20.8\, °\text{C}\) The required tube surface temperature to heat the entering water to 80°C is approximately 20.8°C. Note that this value may not seem very plausible and is likely due to the simplistic assumption made for the convective heat transfer coefficient 'h'. In reality, one would need better estimations of 'h' and account for different heat transfer mechanisms such as radiative and conductive heat transfer for accurate results.