I can't solve 2 problems for an assignment ans was wondering is someone could help please.
1) An insulated rigid tank is initially evacuated. A valve is opened, and air from the atmosphere at 100kPa and 20C enters the tank. Once the pressure inside the tank reaches 100kPa the valve is closed. Determine the final temperature of the air in the tank. Assume K = 1.4 (K=Cp/Cv) Hint: The answer is not 20C
2) A piston cylinder device with a spring attached initially contains 0.02m^3 of R-134A at 50C. A valve connected to the cylinder is opened and 20kg of R-134A enters the device from a supply line at 800kPa and 80C. The spring is compressed and the piston moves up. The process stops when the pressure inside the cylinder reaches 1MPa, at a temperature of 100C and a volume of 0.6m^3.
a) The quality (x) of the R-134A in state 1?
B) Work done in this process (kJ)?
c) Amount of heat transferred to/from device (kJ)?
If anyone can help me that would be great. Thanks!
You must be wondering how bullet b) suddenly transmutated to a cool emotion. I learnt the hard way how to permanently fix the problem. Sorry, I an a bit rusted in physics. Can't help.
Enthalpy
18th January 2011 - 04:35 PM
(Problem 1) The part of the atmosphere that doesn't enter the tank gives a work of 1*RT to each mole entering. Flow losses convert it to heat. Admitting that the air entering the tanks receives no other heat from the rest of the world, this RT is the variation of its internal heat, which equals 2.5*RT for the perfect diatomic gas having Cp/Cv=1.4, so its temperature becomes (273+20)*(1+1/2.5) Kelvin.
Please check...
Sithdarth
18th January 2011 - 09:24 PM
QUOTE
(Problem 1) The part of the atmosphere that doesn't enter the tank gives a work of 1*RT to each mole entering. Flow losses convert it to heat. Admitting that the air entering the tanks receives no other heat from the rest of the world, this RT is the variation of its internal heat, which equals 2.5*RT for the perfect diatomic gas having Cp/Cv=1.4, so its temperature becomes (273+20)*(1+1/2.5) Kelvin.
Please check...
This is wrong. What is occurring is an adiabatic free expansion. Thus no work is done getting the gas into the chamber. At best if it is an non-ideal gas work is done against the interatomic forces causing a net decrease in temperature. However, the question doesn't seem to give enough information. If the gas is not ideal what approximation are we supposed to use? If it is the Van der Waals approximation then what are the coefficients for air? Without these things any calculation is impossible. If we are to assume that the gas is idea there would be no temperature change so if that hint was part of the problem I assume the gas is non-ideal. In which case somewhere in your notes or you book you can probably find the Van der Waals approximation with the constants you need to work this out.
Work done in this process (kJ)?