5-190 A steel cylinder initially contains nitrogen gas at 200 kPa and 25 degree

Question

5-190 A steel cylinder initially contains nitrogen gas at 200 kPa and 25 degree at Celsius. The cylinder is in is connected to a supply line containing nitrogen at 800 kPa and 25 degree at Celsius. A valve is opended, allowing nitrogen to flow into the cylinder until the pressure reaches 800 kPa. The cylinder internal volume is 0.1 m^3, its mass (cylinder only) is 50kg, and its specific heat is 0.43 kJ/kg K. Find the final mass of nitrogen in the tank and the final temperature of the nitrogen assuming constant specific heats at room temperature for nitrogen, and assuming (a) no heat transfer from the nitrogen to the tank, and (b) rapid heat transfer between the nitrogen and tank such that the cylinder and nitrogen remain in thermal equilibrium during the process. (with negligible heat transfer from the cylinder to its external surroundings.)
Answer: (a) 380K, 0.733 kg, (b) 301K, 0.927 kg
Note: For the Solutions I will rate lifesaver (*****) Please give fully detail explanation along with the steps.

Explanation / Answer

this is an unsteady process. and in such kind of process we always need to consider 2 equations

MASS BALANCE :

min - m out = m

m = m2 - m1

ENERGY BALANCE :
Ein - E out = E  

Now here the energy goes only in form of internal energy, no work or heat addition is there

m2.u2 - m1.u1 = m.h

m2.Cv.T2 - m1.Cv.T1 = m.h

m2.Cv.T2 - m1.Cv.T1 = (m2- m1).Cp.T

here T = constant temp = 25 C

Cv for nitrogen = 0.743 KJ/Kg.K and Cp = 1.039 KJ/Kg.K

taking Nitogen as ideal gas :

P1.V1 = m1.R.T, R for nitrogen = 0.2968 KJ/Kg.K

200 x 0.1 = m1 x 0.2968 x (25+273)

m1 = 0.2261 kg

also P2.V2 = m2.R.T2

800 x 0.1 = m2 x 0.2967 x T2

m2 = 278.7 / T2

Putting this value in equation:

( 278.7/T2 - 0.2261) x 1.039 x 298 = 278.7/T2 x 0.743 x T2 - 0.2261x 0.743 x 298

T2 = 380 K

m2 = 278.7/380 = 0.733 KG

b) for heat transfer, equation becomes :

(m2- m1).Cp.T =  m2.Cv.T2 - m1.Cv.T1 + Q

(m2- m1).Cp.T =  m2.Cv.T2 - m1.Cv.T1 + (m.C.T)tank

( 278.7/T2 - 0.2261) x 1.039 x 298 = 278.7/T2 x 0.743 x T2 - 0.2261x 0.743 x 298 + 50 x 0.43 x (T2 - 298)

T2 = 301 K

m2 = 278.7 / 301 = 0.927 kg

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