2. (20 pts) An electric heating element in a residential hot-water heater can be approximated as a D = 10 mm diameter horizontal cylinder with length L = 0.5 m. When powered, the element produces Q = 1200 W of heat. T H CUTOFF THEAT HEATING LOWE TH L Hantar A GUTE PE ww a. (10 pts) Assume that the water in the heater tank is stationary and at temperature T% = 50C. Find the heat transfer coefficient from the surface of the heater element to the water and the surface temperature of the heater element. Solution tips: First, guess a plausible surface temperature (T.) for the heater element, and use this to calculate the natural convection heat transfer coefficient If you are using EES, you can use the built-in function for the volumetric expansion coefficient: volexpcoef (Water, T=T_film, P=Po#) Next, calculate the convection heat transfer rate to the water (Qconv). If you solving this problem by hand, adjust your guess for Ts until Qconv=Q. You will need to update your calculation for the heat transfer coefficient each iteration. If you are solving this with a program like EES, you can just "Update Guesses" (CTRL-G), add the equation Qconv=Q, and comment out the guess for Ts. b. (10 pts) Now, consider what would happen if the heater element was powered if the tank was dry (i.e., full of air). Solve for the heater element surface temperature in this case. You will need to also consider radiation heat transfer from the element in this case. Assume a surface emissivity of = 0.7. Would the heater element likely survive being powered in this case?