Question

A gas initially at 2.8 bar and 60oC is compressed to a final pressure of 14 bar in an isothermal internally reversible process. Determine the work and heat transfer, each in kJ per kg of gas, if the gas is (a) Refrigerant 134a, (b) air as an ideal gas. Sketch the process on p-v and T-s coordinates.

Answer 1

The work and heat transfer isothermal internally reversible process.

An isothermal procedure is a thermodynamic procedure in which the temperature of a gadget stays consistent. The transfer of heat into or out of the gadget occurs so slowly that thermal equilibrium is maintained. The melting of ice at 0 diploma is an example of isothermal technique. The reaction in a warmth pump is an example of isothermal manner. In Isothermal manner temperature is constant for the duration of the procedure and follows Boyles regulation.

Given,

Gas at initially P1= 2.8 bar

                       P2 = 14 bar

isothermal reversible process,

Compressed pressure at 60°C

(a) R = 134a

  For Refrigerant 134a R134a

At P1 = 14 bar T1 = 600C

S1 = 0.9389 KJ/Kg K

V1 = 264.64 KJ/kg

At  

P2 = 2.8 bar T2 = 600C

S2 = 1.1142 KJ/Kg K

V2 = 278.56 KJ/kg

T = 60 + 273 = 333K

Q= 2 Tds

2 Q=T ſ ds

= T(S2-S1)

= 333 ( 1.1142 - 0.9389 )

= 58.3749 KJ/Kg

Heat transfer Q =  58.3749 KJ/Kg

(b) air as an ideal gas.

Q= 2 Tds

2 Q=T ſ ds

= T(S2-S1)

 =T-R In P2 P1

 = 333 * -0.287 In 2.8 14

= 153.81 KJ/Kg

Heat transfer Q   = 153.81 KJ/Kg

W = Q - ΔV

= 28.3749 - (278.56-264.64)

= 14.4549 KJ/Kg

Work W  = 14.4549 KJ/Kg

Learn more about  isothermal  process here:-brainly.com/question/17097259

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