Question

Liquid water flows in a thin-walled circular tube, where the pumping power required to overcome the turbulent flow pressure loss in the tube is $100 \mathrm{~W}\(. The water enters the tube at \)10^{\circ} \mathrm{C}$, where it is heated at a rate of \(3.6 \mathrm{~kW}\). The average convection heat transfer coefficient for the internal flow is $120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The tube is \)3 \mathrm{~m}$ long and has an inner diameter of \(12.5 \mathrm{~mm}\). The tube surface is maintained at a constant temperature. At the tube exit, an ethylene propylene diene (EPDM) rubber o-ring is attached to the tube's outer surface. The maximum temperature permitted for the o-ring is \(150^{\circ} \mathrm{C}\) (ASME Boiler and Pressure Vessel Code, BPVC. IV-2015, HG-360). Is the EPDM o-ring suitable for this operation? Evaluate the fluid properties at \(10^{\circ} \mathrm{C}\). Is this an appropriate temperature at which to evaluate the fluid properties?

Short answer

Answer: Yes, the EPDM o-ring is suitable for this operation since the calculated temperature at the tube exit (138.72°C) is less than the maximum allowable temperature for the o-ring (150°C).

Step-by-step solution

Calculate the thermal energy supplied to the water

The rate at which heat is supplied to the water is given as 3.6 kW. Convert this to watts: Thermal energy supplied = 3.6 kW × 1000 = 3600 W

Calculate the water flow rate

The power required to overcome the turbulent flow pressure loss is given as 100 W. The water flow rate can be calculated using the relation: \(\dot{m} = \frac{P}{C_p(T_f - T_i)}\) where \(\dot{m}\) is the mass flow rate, \(P\) is the pumping power, \(C_p\) is the specific heat of water (approximately 4186 J/kg·K at 10°C), \(T_f\) is the final temperature of the water, and \(T_i\) is the initial temperature of the water (10°C). We will solve for the mass flow rate: \(\dot{m} = \frac{100}{4186(T_f - 10)}\)

Determine the heat transfer rate from convection

The heat transfer rate due to convection can be calculated as: \(Q = hA_s\Delta T\) where \(Q\) is the heat transfer rate, \(h\) is the average convection heat transfer coefficient (given as 120 W/m²·K), \(A_s\) is the surface area of the tube, and \(\Delta T\) is the temperature difference between the tube surface and the water. The surface area of the tube can be calculated using the formula: \(A_s = 2\pi rL\) where \(r\) is the radius of the tube, and \(L\) is its length. Convert the diameter to meters and find the radius: \(r = \frac{12.5}{2}\times10^{-3} \mathrm{m}\) The length of the tube is given as 3 m, so the surface area of the tube is: \(A_s = 2\pi\cdot (6.25\times10^{-3}\mathrm{m})\cdot 3 \mathrm{m}\approx 0.235 \mathrm{m}^2\) Now we can calculate the heat transfer rate: \(Q = 120\ \mathrm{W/m}^2\cdot \mathrm{K\:} \times0.235 \: \mathrm{m}^2\times\Delta T\)

Calculate the temperature of the water at the tube exit

We know that the thermal energy supplied to the water is equal to the heat transfer rate: Thermal energy supplied = Heat transfer rate \(3600 \:/\: \mathrm{W} = 120\ \mathrm{W/m}^2\cdot \mathrm{K\:} \times0.235\: \mathrm{m}^2\times\Delta T\) Solving for \(\Delta T\) gives \(\Delta T = \frac{3600}{120\times0.235}= 128.72 \:/\: \mathrm{K}\) The temperature of the water at the tube exit is: \(T_f =T_i +\Delta T =10+128.72 = 138.72 \:/\: ^\circ \mathrm{C}\)

Compare the temperature at the tube exit to the maximum allowable temperature for the o-ring

The maximum allowable temperature for the EPDM o-ring is given as 150°C. Since the calculated temperature at the tube exit (138.72°C) is less than the maximum allowable temperature for the o-ring (150°C), the EPDM o-ring is suitable for this operation. As for evaluating fluid properties at 10°C, this temperature is suitable since it is the initial temperature of the water and the fluid properties are not expected to vary significantly within the range of temperatures experienced in this operation.