A&nbsp;0.5−m3&nbsp;rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa.&nbsp;Calculate (a)the amount of refrigerant in the tank by mass.(b) the amount of heat transferred. Hence, show the process on a&nbsp;P−v&nbsp;diagram with respect to saturation lines.

Question

A&amp;nbsp;0.5−m3&amp;nbsp;rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa.&amp;nbsp;Calculate (a)the amount of refrigerant in the tank by mass.(b) the amount of heat transferred. Hence, show the process on a&amp;nbsp;P−v&amp;nbsp;diagram with respect to saturation lines.

Pademagk71 · Accepted Answer

For this stationary closed system, the energy balance is expressed asE∈−Eout=△Esystemtherefore&amp;nbsp;Q∈=△U=m(u2−u1)use tables (Tables&amp;nbsp;A−11&amp;nbsp;through&amp;nbsp;A−13)the properties of&amp;nbsp;R−134aP1=160Kpax1=0.4so&amp;nbsp;⇒vf=0.0007435,vg}=0.12355m3kgvf=31.06,ufg=190.31kjkgtherefore&amp;nbsp;v1=vf+x1vfgthe equation&#039;s values are changedv1=0.0007435+0.4(0.12355−0.0007435)⇒v1=0.04987m3kgso,&amp;nbsp;u1=uf+x1ufgwe replace the values in the equationu1=31.06+0.4(190.31)⇒u1=107.19kjkgthereforeP2=700Kpa(v2=v1)⇒u2=377.23kjkgthe mass of the refrigenant is determinedm=V1v1we replace the values in the equationm=0.5m30.04987m3kg⇒m=10.03kgpar (b)Q∈=m(u2−u2)we replace the values in the equationQ∈=(10.03kg)(377.23−107.19)kjkgQ∈=2708kj

Beverly Smith · Accepted Answer

V=0.5m3,R−134a,P1=160kPa,X1=0.4 P2=700kPa, Find m, Q2, show P-U DIAG. First law: Q2−W2=m(U2−U1) For a rigid tank, Wb=S12PAV=0 Q2=m(U2−U1) State 1: P1=160kPa,X1=0.4 U1=Uf+X1⋅Ufg=(31.09)+(0.4)(190.27)=107.2kjkg U1=Uf+X1(Ug−Uf) =(0.0007434)+(0.3)(0.12348−0.0007437) V1=0.04984m3kg m=VV1=(0.5m3)(0.04984m3kg)=10.03kg State 2: P2=700kPa,V2=V1=0.04984m3kg At P4at=700kPa,Vg=0.029361m3kg Since V2&amp;gt;Vg Table A-13: P=0.70mPa ММ(m3kg)U(kJkg)0.047306347.410.048497367.29 U=U1+(U2−U1V2−V1)(V−V1)

karton · Accepted Answer

Step 1 Given, V=0.5m3P1=160kPax1=40%=0.4P2=700kPa Refrigrant is r-134a From Refrigrantion tables at P = 160 KPa, we get Specific volume: vf=0.0007437m3/kg;vg=0.12348m3/kg Internal Energy: uf=31.09kJ/kg;ug=190.27kJ/kg Step 2 NSK Initial Specific Volume (v)=vf+x.(vg−vf) v=0.0007437+0.4(0.12348−0.0007437) Initial Specific Volume (v)=0.049409m3kg Volume of Tank =0.5 mass=volumeSpecific Volumem=0.50.0498409m=10.031kg Step 3 Q=m(u2−u1)u1=uf+x.ufgu1=31.09+0.4×190.27u1=107.198kJ/kg As the volume is constant, the refrigerant becomes supersaturated At P2=700kPa, and the specific volume is 0.04984 (approximately near to 0.0485) We get T=160∘C hence u2=367.29kJ/kg Q=m(u2−u1)Q=10.031(367.29−107.198)Q=2608.98kJ

A 0.5-m^{3} rigid tank contains refrigerant-134a initially at 16

Answered question

Answer & Explanation

New Questions in Other