Nitrogen N2 is compressed in a pistoncylinder assembly from

Nitrogen (N_2) is compressed in a piston-cylinder assembly from p_1=.7 bar, T_1=280 K to p_2=11 bar. The initial volume is 0.262 m^3. The process is described by pV^1.25=constant. Assuming ideal gas behavior and neglecting kinetic and potential energy changes, determine the work and heat transfer for the process, each in kJ using (a) constant specific heats evaluated at 300K, and (b) data from table A-23. Compare results and discuss. (tables found here https://eclass.upatras.gr/modules/document/file.php/MECH1206/CENGEL_%20PROPERTY%20Tables%206%20NEW-SIappendix1.pdf )

5.4 Nitrogen (N2) is compressed in a piston-cylinder assembly from pi0.7 bar. T 280 K to h-11 bar. Te initial volume is 0.262 m, The process is described by pV125 = constant. Assuming ideal gas behavior and neglecting kinetic and potential energy changes, determine the work and heat transfer for the process, each in kJ using (a) constant specific heats evaluated at 300K, and (b) data from Table A-23. Compare the results and discuss. 3 pts. for given/find/assumptions/identifying the system and energy transfers 2 pts. for correctly plotting the process on a p-v diagram 3 pts. for correctly identifying governing equations 4 pts. for correctly computing all intermediate values 4 pts. for correct calculations for heat transfer and work

Solution

N₂ compression in a piston-cylinder assembly

Given, P₁ = 0.7 bar, T₁ = 280 K, V₁ = 0.262 m³ ; P₂ = 11 bar, T₂ = ?, V₂ = ?

Process follows : P₁V₁^1.25 = P₂V₂^1.25   i.e. V₂ = 0.0289 m³.

Use ideal gas equation : P₁V₁/T₁ = P₂V₂/T₂   i.e T₂ = 485.75 K

(a) From 1st law of Thermodynamics for closed system, q₁₂ - w₁₂ = e₂ - e₁

= u₂ - u₁(Neglecting kinetic and potential energy)

= c_v(T₂ - T₁)

[Constant property cv = 0.743 kJ/kg-K from Table A-2 at 300K ]

= 0.743 x (485.75 - 280)

= 152.87 kJ/kg

w₁₂ = Work Transfer = (P₂V₂ - P₁V₁)/(1 - k) = {(11 x 0.0289 - 0.7 x 0.262) x 100}/(1 - 1.25) = -53.8 kJ/kg.( Work done on the system)

So, q₁₂ = Heat Transfer = 152.87 - 53.8 = 99.07 kJ/kg (Heat added to the system).

(b) Data from table A-18 :

u₁ = 5813 kJ/kmol = 207.61 kJ/kg [ 1 kmol of N₂ = 28 kg of N₂]

u₂ = 10103.5 kJ/kmol = 360.84 kJ/kg [ 1 kmol of N₂ = 28 kg of N₂]

w₁₂ will be same as k=1.25 (given). so, w₁₂ = -53.8 kJ/kg.

Heat transfer, q₁₂ = 153.23 - 53.8 = 99.43 kJ/kg.

The above results are close to each other as the temperature range ( 280 K - 486 K), c_v changes slightly. [See table A-2].