Question

Liquid water with a specific heat of $4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\( is heated in a circular tube by an electrical heater at \)10 \mathrm{~kW}\(. The water enters the tube at \)10^{\circ} \mathrm{C}$ with a mass flow rate of \(10 \mathrm{~g} / \mathrm{s}\). The inner surface of the tube is lined with polyvinylidene chloride (PVDC) lining. According to the ASME Code for Process Piping (ASME B31.3-2014, A.323), the recommended maximum temperature for PVDC lining is \(79^{\circ} \mathrm{C}\).

Short answer

Explain your answer using the relevant calculations.

Step-by-step solution

Identify the given information

We are given the specific heat capacity of water, \(c_p = 4.18\mathrm{~kJ/kg\cdot K}\), the amount of energy added to the water by the electric heater, \(P = 10\mathrm{~kW}\), the initial temperature of the water, \(T_i = 10^{\circ}\mathrm{C}\), and the mass flow rate, \(\dot{m} = 10\mathrm{~g/s}\). The maximum recommended temperature limit for PVDC lining is \(T_{max} = 79^{\circ}\mathrm{C}\).

Convert the mass flow rate to kg/s

We need to convert the mass flow rate from g/s to kg/s. Given that \(1\mathrm{~kg} = 1000\mathrm{~g}\), we have: $$\dot{m} = 10\mathrm{~g/s} \times \frac{1\mathrm{~kg}}{1000\mathrm{~g}}$$ $$\dot{m} = 0.01\mathrm{~kg/s}$$

Calculate the total energy added by the heater

The energy added per second by the heater is the same as its power: $$E_{heater} = P\cdot t$$ Since we are looking at a rate, we can consider energy added to the water per second: $$\frac{E_{heater}}{t} = P$$

Calculate the temperature change of the water

Using the specific heat capacity formula, we can relate the energy added to the water with the temperature change: $$E_{water} = \dot{m}\cdot c_p \cdot \Delta T$$ But since the energy added to the water per second is equal to the heater's power: $$P = \dot{m}\cdot c_p\cdot \Delta T$$ Now, we need to solve for the temperature change \(\Delta T\). Rearrange the equation: $$\Delta T = \frac{P}{\dot{m}\cdot c_p}$$ Substitute the given values: $$\Delta T = \frac{10\mathrm{~kW}}{0.01\mathrm{~kg/s} \times 4.18\mathrm{~kJ/kg\cdot K}}$$

Calculate the final temperature

Since the specific heat capacity has units of \(\mathrm{kJ}\), convert the heater's power to \(\mathrm{kJ/s}\): $$P = 10\mathrm{~kW} \times \frac{1000\mathrm{~J}}{1\mathrm{~kJ}} = 10000\mathrm{~J/s} = 10\mathrm{~kJ/s}$$ Now, substitute the converted power and calculate \(\Delta T\): $$\Delta T = \frac{10\mathrm{~kJ/s}}{0.01\mathrm{~kg/s} \times 4.18\mathrm{~kJ/kg\cdot K}} \approx 239.23\mathrm{~K}$$ Next, we find the final temperature by adding the initial temperature and the temperature change: $$T_f = T_i + \Delta T \approx 10^{\circ}\mathrm{C} + 239.23\mathrm{~K}$$

Check if the final temperature is within the recommended limit

The final temperature of the water is: $$T_f \approx 10^{\circ}\mathrm{C} + 239.23\mathrm{~K} = 249.23^{\circ}\mathrm{C}$$ Since the final temperature, \(249.23^{\circ}\mathrm{C}\), is higher than the maximum recommended temperature for PVDC lining, \(79^{\circ}\mathrm{C}\), the temperature in the tube is not safe for the PVDC lining.