Question

An ASTM B165 nickel-copper alloy tube sheathes a heating element that is used to boil water at \(1 \mathrm{~atm}\). The tube is immersed horizontally in the water. The tube diameter and length are \(5 \mathrm{~mm}\) and $15 \mathrm{~cm}$, respectively. The maximum use temperature for ASTM B165 nickelcopper alloy tube is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). Determine the highest evaporation rate of water that can be achieved by the heater without heating the tube surface above the maximum use temperature.

Short answer

Answer: The highest evaporation rate that can be achieved by the heater without heating the tube surface above its maximum use temperature is approximately 0.001668 kg/s.

Step-by-step solution

Calculate the heat transfer area of the tube

For a cylindrical tube with a given diameter and length, we can find the heat transfer area (A) using the formula: A = πDL Where D is the diameter and L is the length of the tube. The given diameter and length are 5 mm and 15 cm, respectively. To maintain consistency in units, convert the diameter into cm: D = 0.5 cm L = 15 cm A = π(0.5 cm)(15 cm) ≈ 23.56 cm²

Determine the temperature difference

The maximum use temperature for the tube is given as 260°C, and the boiling water temperature at 1 atm is 100°C. So, the temperature difference (ΔT) between the tube surface and the boiling water is: ΔT = T_max - T_water = 260°C - 100°C = 160°C

Apply the heat transfer formula related to boiling

The heat transfer formula for boiling is: q = h*A*ΔT Where q is the heat transfer rate, h is the heat transfer coefficient, A is the heat transfer area, and ΔT is the temperature difference.

Calculate the maximum heat transfer rate

In this step, we need to estimate the heat transfer coefficient (h) for the given tube material and water. According to the Nukiyama correlation for boiling heat transfer, h values for water at atmospheric pressure typically range from 5,000 to 15,000 W/m²K. As an approximation, we can select a median value of: h ≈ 10,000 W/m²K (converting the units to be consistent) Now, we can find the maximum heat transfer rate (q_max) by substituting the values in the formula: A = 23.56 cm² = 0.002356 m² (converting to m²) ΔT = 160°C = 160 K (converting temperature difference to Kelvin) q_max = (10,000 W/m²K)(0.002356 m²)(160 K) ≈ 3,769.6 W

Determine the highest evaporation rate

The heat required to evaporate a certain mass of water can be determined using the latent heat of vaporization formula: q = m_dot*L_v Where m_dot is the evaporation rate (mass flow rate) in kg/s, and L_v is the latent heat of vaporization of water at 100°C, approximately 2.26×10⁶ J/kg. To find the highest evaporation rate, we can rearrange the formula and solve for m_dot: m_dot = q / L_v Substitute the values: m_dot = (3,769.6 W) / (2.26×10⁶ J/kg) ≈ 0.001668 kg/s The highest evaporation rate that can be achieved by the heater without heating the tube surface above the maximum use temperature is approximately 0.001668 kg/s.