Vapor Compression Cycle

Working fluid:

• Fluid with Phase Changes

Details:

• $1\to 2$: Isentropic compression
  • $v\downarrow$ | $T\uparrow$ | $P\uparrow$
• $2\to 3$: Constant Pressure
  • $v\downarrow$ | $T\downarrow$ | $P-$
  • Phase change: $T-$ | $v\downarrow$
• $3\to 4$: adiabatic expansion (not isentropic!)
  • $v\uparrow$ | $T\downarrow$
  • Not reversible!
• $4\to 1$: Constant Pressure
  • $v\uparrow$ | $T-$ | $P-$

Assumptions:

• State 1 = saturated vapor
• State 3 = saturated liquid
• Negligible kinetic energy

T-v Diagrams:

Refrigerants:

Early days:

• Ammonia ($N{H}_3$) and sulfur dioxide ($S{O}_2$)
• Highly toxic

CFC: Chlorofluorocarbons Banned

• Ex:
  • Dichlorodifluoromethane (R-12): Original "Freon"
  • Chlorodifluoromethane (R-22)
• Lead to depletion of ozone (due to Cl)

HFC: Hydrofluorocarbons

• Ex:
  • Tetrafluoroethane (R-134a)
  • Difluoromethane (R-32)
  • Pentafluoroethane (R-125)
• No chlorine, but greenhouse gas > $C{O}_2$

Remark: $R\mathrm{\#}=CHF-90$

Example:

Conditions:

• Working fluid: R-134a
• Low temperature: $T=-20\mathrm{°}C$
• High temperature: $T=40\mathrm{°}C$

State 1:

• Saturated Vapor $\Rightarrow \{\begin{array}{l}{P}_1=P_{sat,v}(T_1)=133.7kPa\\ h_1=h_{sat,v}(T_1)=386.08kJ/kg\\ s_1=s_{sat,v}(T_1)=1.7395kJ/kgK\end{array}$

State 3:

• Saturated Liquid $\Rightarrow \{\begin{array}{l}{P}_3=P_{sat,l}(T_3)=1017kPa\\ h_3=h_{sat,l}(T_3)=256.54kJ/kg\end{array}$

Processes:

Process $2\to 3$:

• Constant Pressure $P_2=P_3=1017.0kPa$
  
  Thus $T_2\approx 47.7\mathrm{°}C$ and $h_2=428.4kJ/kg$

Process $1\to 2$:

• Isentropic $s_2=s_1=1.7395kJ/kgK$
• First Law for Constant Volume $w_{comp}=h_{T,2}-h_{T,1}\approx -42.3kJ/kg$

Process $2\to 3$ (again):

• First Law for Constant Volume $q_{out}=h_{T,3}-h_{T,2}\approx -171.9kJ/kg$

Process $3\to 4$:

• Adiabatic, no work
• First Law for Constant Volume $h_4=h_3=256.54kJ/kg$
  
  Thus: $Y_4\approx 61\mathrm{\%}$

Process $4\to 1$:

• Constant Pressure $P_4=P_1=133.7kPa$
• Phase Change $T_4=T_1=-20\mathrm{°}C$
• First law for Constant Volume $q_{in}=h_{T,1}-h_{T,4}\approx 129.6kJ/kg$

Coefficient of Performance:

Back to process $3\to 4$:

• Adiabatic
• No work

Enthalpies:

• State 3: Saturated liquid $h_3=h_l(T_3)$
• State 4: Vapor + liquid $h_4=(1-Y_4)h_l(T_4)+Y_4{h}_v(T_4)$

Entropies:

• State 3: Saturated liquid $s_4=s_l(T_3)$
• State 4: Vapor + liquid $s_4=(1-Y_4)s_l(T_4)+Y_4{s}_v(T_4)$

Recall: Gibbs Equation $du=Tds-Pdv\to ds=\frac{\xcancel{dh}}{T}-\frac{v}{T}dP>0$ since $dP\ne 0$.

Real Case:

Piping causes pressure drops and heat losses.