Question

2. (10 points) Assume that you have rigid container with a volume of 0.75 m? The container is initially evacuated (empty). Unfortunately, a tiny hole develops in the side of the container, and air from the surroundings leaks in. The surrounding air is at 100 kPa and 25°C. The air leaks in until the pressure in the container is the same as the pressure of the surrounding air. The process occurs slowly enough that heat transfer between the container and the surroundings keeps the temperature of the air inside the container constant at 25°C. Find the total amount of heat transfer, in kJ. Is the heat transfer going into the container or out of the container?

Answer 1

Unsteady flow Energy Equation. means the propesties varies with respect to time, ie The mass flow rate at the entry and exit Control volume are different. Unsteady problem solve by using two conservation mass_conservation Energy conservation ie Let us assume me is the mass flow rate at the entry of the Control Volume mass flow rate at the exit of cv. me ท, Initial mass inside the cv Final mass inside the cr. i 2 e -- mi 10 me

Applying mass concessation Punciple dm d mi dt d me dt at Applying Energy Conservation mini_+ K€; + PE; + @) - (me be (me be + KEętlę+W DEC dt d dt CV

25°C Control Volume 100 kPa Volume 0.75 m² Air 100 kPa Evacuated 25°c Initial condition. Final Condition Applying mass conservation principle, ami - ame= dm dt dt - As no may exit the CV, dm ami - mi at Applying energy conservation principle + 2 de d dt (mi hi + WE + PC: 978) - g (mohet pr. 736.94 & mihi) + a leglecting KE and PE terms, du do 0 mi dhi thidmi dr dt + inki + 9 det As dm mi dt du d dm hi+Q = de -U, = (m - m ) hi + 9 U₂ - U, = m₂ hi + 6

Using ided gas equ, 100 x 10² x 0.75 m PV RT (273+25) m₂ = U₂ = m hi to CT = M₂ CP T + 287 x G=0.710 kJ/kg (p = 1.005 kilugh 0.8 769 kg. 0.718x (273+25) = 0.8769 x 1.005x(273 +25) + 0 8 = - 48.6587 kJ theat rejowed to As heat supplied to system is tive and theat the system is negative in Heat is going out of 48.6587 KJ the container and is equal to