Question

A two-chambered tank is connected to a high pressure pipeline containing air. The bottom chamber of the tank contains air that is initially at 450 kPa and 300 K due to the force of an adiabatic, weighted piston (piston face area is 1.5 square meters). The piston separates the top and bottom chambers. The top chamber and bottom chamber each have an initial volume of 2.0 cubic meters. Other conditions and properties are shown in the attached figure. The valve separating the tank from the pipeline is opened and air flows from the pipeline into the top chamber section until it reaches 1100 kPa. The top and bottom tank chambers are insulated from their surroundings and each other. Compression of the bottom tank section can be assumed to be a reversible, adiabatic process. Determine the following immediately after filling (need not/may not want to solve in order given): a) Temperature of the bottom tank chamber, b) Volume of the bottom tank chamber; c) Change of tank gravitational potential energy (weight of the gas can be neglected); d) Mass added to the tank; e) Temperature of the top tank chamber; f) Entropy generation of the process. top chamber (2.0 cu.m.) (evacuated) weighted piston (1.5 sq.m. area) bottom chamber (2.0 ou.m.) air @ 450 kPa and 300 K pipeline Air 1500 kPa 300K

Answer 1

Soludions Given that - (v. 0.713 n5jka CP: 1Kikaye, RO 0.297 kJingle X: 1.4 TV, 1m² A= 1.5 m2 Viim? I P = 4500 Pa т., - зоок P=lookPa T= 30k After Valve opening pressure at the bottom of the chamber is - P = 450 +10o 1550 kPa (0) In the bottom chambeo, the compression brocess is reversible adiabatic brouss . so. = () - C (ta > Vin = 0.4 134 mi? The your oth = 300 (192) *4** T = 417. 15K ñ alte E (2) change of tank potential energy APE = Padded x nx A PE : Iloa x0.2756 1.5 LAPE : 454.74 kJ

(e mass added to the tank Die Wokp, V = V + Ax2 R: 0.297 kJI, үз\ Mass From Fin - Home - A E system mi hi - Dwy. Un is always, Conservation mass Conseovation , mis ma's. A meys ; mi Amsys so , mi hi - amisgs." Cp Ti = Cv Ti IE 1.4*340 Te = 420k So temperature of the tub tank chamber T = 420k Amsys = PA V 100x (1+0.2751x1.5) R TE 0.287 * 420 (e Dmsy. = 12.9 kg = mil