Part 2: Detailed Analysis Questions (70 Points total) 1. (25 pts) Exergy LLC produces commercially available tube-in-tube heat exchangers. Consider their model #AS-00448, with specifications and flow conditions listed below: ●

● ● Overall length: L = 6 m, all 316 stainless steel construction, surface roughness of 0.5 µm Inner tube: ID; = 7.1 mm, OD¡ = 9.5 mm Outer tube: ID. = 16.7 mm, OD. = 19.1 mm -1 Cold water enters the inner tube at: V₁ = 10 L min-¹ and Ti,in = 20°C Counterflow warm water enters the outer tube at: V = 30 L min¹¹ and To,in = 50°C/na. (6 pts) Calculate the frictional pressure drop for the tube-side and annulus-side water flows. (Hint: pressure drop curves are available from the manufacturer) b. (9 pts) Calculate the overall UA for this heat exchanger at the listed operating conditions. c. (5 pts) Find the effectiveness of this heat exchanger for the listed operating conditions. Could the heat transfer rate be significantly improved by increasing the heat exchanger length? d. (5 pts) Often, there is a “limiting" resistance in a heat transfer system that is more significant than others. Is there a specific heat transfer stage in this heat exchanger that contributes a "limiting" resistance? If so, suggest a strategy to improve the overall performance./n2. (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. TEMPERATURE CURRE THERMOSTAT UPPER ELEMENT LOWER HEATING Handyman ANODE ROD COLD WATER PRESSURE UNIVE DRAIN VALVE/na. (10 pts) Assume that the water in the heater tank is stationary and at temperature To = 50°C. 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 (Ts) 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 ε = 0.7. Would the heater element likely survive being powered in this case?/n3. (25 pts) Consider a flat plate solar collector for water heating installed on a roof. The collector has a single layer of glass (glazing) above the absorber to retain heat. ● ● Glass cover Insulation Air space Absorber plate The geometry and operating conditions for the solar collector are summarized as follows The face area of the collector assembly is H = 1 m × W= 1 m. The assembly is inclined = 30° from the horizontal. The air gap between the absorber plate and the glass cover is L = 15 mm thick. The absorber plate is at a uniform temperature TA = 60°℃ and has emissivity &a=0.9. The glass cover plate is thin, and has an emissivity of &G = 0.5. The ambient air is blowing over the surface of the absorber at U = 2 m s¹ with T% = 10°C. a. (12 pts) Assume, for now, the glass cover is at uniform temperature TG = 30°C. Calculate the natural convection and radiation heat transfer rates from the absorber plate to the glass cover./nb. (13 pts) Solve for the steady state temperature of the glass cover plate and the total heat loss rate through the front face of the solar collector. To do this, perform the following steps: Calculate the external forced convection and radiation heat transfer rate from the glass cover plate to the ambient (assuming TG = 30°C) Set up a steady state energy balance for the glass plate: Qconv,A→G + Qrad,A→G = Qconv,G→∞ + Qrad,G→∞ Adjust TG until the above energy balance equation is satisfied. Make sure to update the radiation heat transfer rate and natural convection heat coefficient at each iteration. o If using EES, you can just "Update Guesses" (CTRL-G), add the energy balance equation, and comment out the guess for Ts.