A piston-cylinder device with a volume of 0.035 m^3 (refer Figure below) contain

Question

A piston-cylinder device with a volume of 0.035 m^3 (refer Figure below) contains refrigerant R134-a at 400 kPa (State 1). Initially 20% of the volume is liquid and the rest is vapor. Then heat is added to the system until the piston is just touching the stopper (State 2). At this state, all the R-134a exists 100% as vapor Heating is continued until the pressure inside the piston reaches 600 kPa (State 3). Answer the following questions: a. What is the total mass and quality of refrigerant R-134a inside the piston at initial state. b. Determine the boundary work from State 1 to State 3 (m kJ) c. Determine the amount of heat from State 1 to State 3 (in kJ) d. Show these processes on a T-v and P-v diagrams.

Explanation / Answer

Ans:- A piston cylinder device volume Refrigerator gas R-134-a = 0.035 m3  

State 1 - Pressure 400 kPa

State 2 - initial volume of liquid 20% and rest vapor . heat is added to system

State 3 - 100% vapors. Heating is continued . pressure inside the piston reaches 600 kPa.

a. A free-moving the piston cylinder can be giving as a constant pressure so such that P1 = 400 kPa P3 = 600 kPa.  in both state 1 & 3 constant pressure process, the rectangular area is P (V2 – V1). We know that P1 is 400 kPa initial, and the volume is = 0.035 m3 .

the final pressure are is pressure of 600 kPa (0.6 MPa) so V2 volume is reduce = 0.027 m3 and so as per final state has a temperature of as per condition assume 60 deg so our the superheat tables in refrigerant-134a the specific volume at this temperature and pressure is 0.037413 m3/kg as per superheat table..

So we find the volume and the specific volume of the saturated liquid, vf = 0.0006580 m3/kg at 600 kPa. We can then find the mass as follows.

m = (P1-P2) / v1 = (V1 - V2) / v1 = (0.0305 L -0.0270 L) / 0.0006580 m3/kg * m3/1000 L = 121.58 kg

b. We are find P2/P1 ratio as in the constant Pressure in process.

V1/V2 = (mRT/P2) / (mRT/P1) = P1/ P2

so our Substituting P1/P2 in change a place of V2/V1, using the equation of gas constant formula in air = 0.2970 kJ/(kg•K) from Table close boundary table and so we assume the using the state 1 & state 3 given values of m = 3.4 kg, and T = 19oC = 285.15 K, gives the following result for the work.

W = mRT log (P1/P2) = 3.4 Kg * 0.2979 kJ/ kg . K (285.15 K) log ( 400 kPa/ 600 Kpa) = -285 kJ

so The -ve sign show for the work indicates the input of 285 kJ in the state 1 to state 3.

c. So we calculate of heat in work done form. we are reduce the equation Since TVn = R, the constant R.

we are written for T1V1n or as T2V2n. so our heat work done equation

W = C/1-n ( 2V1-n - 1V1-n) = ( C2V1-n - C1V1-n) / 1-n =( P2V22V1-n - P1V11V1-n ) / 1-n = (P2V2 -P2V2) / 1-n

The above equation any substance is valid in ideal gas. The heat equation of the ideal gas equation of state1 & state3 each of PkVk product as mRTk.

W = (P2V2 - P2V2 ) /1-n = (mRT2 - mRT1) / 1-n = mR (T2-T1) /1-n

Now the gas constant for Refrigerator R-134-a in R = 0.2968 kJ/(kg•K).  so we assume mass 2 kg of refrigerator gas R134-a the temp T1 = 400 K, T2 = 460 K and n = 1.7.

Amount of heat in form of work done= mR(T2 -T1) /1-n = 2kg (0.2968 kJ/kg/K )( 460 k -400 k) / 1-1.7 = -89.0 kJ

so that -ve sing indicate the heat input in state 1 & state 3.

d. P-v diagram

T-v

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