Question

A $R-12$ refrigerant reciprocating compressor operates between the condensing temperature of ${30}^{\circ }C$ and evaporator temperature of $-{20}^{\circ }C$. The clearance volume ratio of the compressor is $0.03$. Specific heat ratio of the vapour is $1.15$ and the specific volume at the suction is $0.1089m^3/kg$. Other properties at various states are given in the figure. To realize $2$ tons of refrigeration , the actual volume displacement rate considering the effect of clearance is

• A. $6.35\times {10}^{-3}{m}^3/s$
• B. $63.5\times {10}^{-3}{m}^3/s$
• C. $635\times {10}^{-3}{m}^3/s$
• D. $4.88\times {10}^{-3}{m}^3/s$
• E. No option matches

Answer 1

The correct option is E No option matches

Given data:

$C=0.03$

$\gamma =1.15$

$v_1=0.1089m^3/kg$


Capacity : $Q=2tons=2\times 3.5kW=7kW$

$P_1=1.50bar;P_2=7.45bar$

$h_1=176kJ/kg;h_2=207kJ/kg$

$h_3=h_4=65kJ/kg$

Refrigerating effect, $RE=h_1-h_4=176-65=111kJ/kg$

Mass flow rate,
$m=\frac{\text{Refrigerating capacity(Q)}}{\text{Refrigerating effect(RE)}}$

$=\frac{7}{111}=0.063kg/s$

We know that volumetric efficiency of the compressor,

${\eta }_v=1+C-C{r}_p^{1/\gamma }$

$=1+C-C{\left(\frac{P_2}{P_1}\right)}^{1/\gamma }$

$=1+0.03-0.03{\left(\frac{7.45}{1.50}\right)}^{1/1.15}$

$=0.9091$

also ${\eta }_v=\frac{\text{Actual suction volume}}{\text{Displacement volume}}$

${\eta }_v=\frac{m{v}_1}{V_s}$

$0.9091=\frac{0.063\times 0.1089}{V_s}=\frac{0.00686}{V_s}$

or $V_S=0.00754m^3/s$

$=7.54\times {10}^{-3}{m}^3/s$