A closed rigid, tank with a volume of 2.0 m^3, contains a two-phase mixture of R

Question

A closed rigid, tank with a volume of 2.0 m^3, contains a two-phase mixture of Refrigerant R-134a with a temperature of 4 degree C. A paddle wheel stirs the mixture as the tank process is underway in the amount of 5,000 kJ. As this occurs, the heat is transferred to the tank over the same time interval, until the contents of the tank are saturated vapor at a temperature of 44 degree C. Kinetic and potential energy effects negligible. For the Refrigerant R-134a, determine (a) the amount of energy transfer by heat transfer in kJ, and (b) the final pressure of the Refrigerant R-134a at the end of the process in bars.

Explanation / Answer

transfer of energy is through heat only when it is due to temperature difference. In this case energy is transfered by stirring R-134a.and heat transfer simultaneously. so this transfer of energy is due to work done on the system and the heat transfer.

A) apply first law of thermodynamics-

delta Q= delta U+ delta W ....... eq(1)

work is done on the system so delta W will be negative

initial state of system is 2m3 volume, acutally mass is not mentioned in the problem. which is required to calculate of dryness fraction.

as you have asked value of heat transfer also in KJ which is only possible when you provide mass.

i shall give you equation if you put value of mass you will get answer

specific volume at initial condition=2/M=v

v=vf+xvfg

at 4oc vf= .0007801m3/kg ,vfg=0.0592199m3/kg

you will get x.

so internal energy can also be calculated at 4oc

U1=Uf+xUfg

Uf= 55.08kj/kg ,Ufg= 174.19 kj/kg

so u will get U1 at 4oc

from property table U2 at 44oc for saturated vapour

U2=249.96kj/kg

take U2-U1=delta U

so put value in eq(1)

delta Q= M*delta U-5000

you will get heat transfer in kj

B) we ca use R-134a property table to get final pressure

as the system is saturated vapour at 44oc.

from propety table pressure is 11.299 bar.

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