$Q_{a}=70 kW,$ c=4 L=$1.2\times D, N=1200rpm$
$T_{E}=T_{sat_{Pe}}=-8^{\circ}, T_{E}=T_{sat_{Pc}}=42^{\circ} Cpl=1.24\frac{kj}{kgK}$
Cpv=0.74 $\frac{kJ}{kgk}$
h1=$h g_g^\circ+Cpv(Tsup-Tsat)$
=402.3+0.74(2-8)
h1=409.7kJ/kg
h2= $h \ g_{42^{\circ}}+CPv(Tsup-Tsat)$
h2=416.8+0.74(T2-42)...(1)
For T2,
S1=S2
$S \ g _{-g^{\circ}}+CPv\times ln\frac{Tsp}{Tsat}=s \ g_{42^{\circ}}+Cpv \times ln\frac{Tsup}{Tsat}$
1.764+$0.74\times$ ln$\frac{275}{265}=1.697+0.74\times ln\frac{T2}{315}$
Assuming compression to be isentropic
T2=357.86 K=$84.86 °C$
Put in (1),
h2=448.52 $\frac{kJ}{kg}$
h2=448.52 $\frac{kJ}{kg}$
h3=$h_(f_42℃)+Cpl(Tsub-Tsat)$
=252.4+1.24(37-42)
h3=h4=246.2 kJ/kg
We know, ideal COP =$ \frac{T_E}{(T_C-T_E )}$= 5.3
COP of given system = $\frac{R.E.}{Wc}$=$\frac{(h1^{'}-h4)}{(h2-h1)}$
=402.3 $\frac{kJ}{kg}$
COP=4.02
Also, $Q ̇_a=m ̇_R×(h1^{'}-h4)$
mass flow rate,$m ̇_R=0.45 kg/sec$
Also,
$m ̇_R×v1$=theoretical piston displacement in m3 per min×$\frac{η_vol}{60}$
v1=$v_(g_{-8}℃)×\frac{Tsup}{Tsat}=0.064×\frac{275}{265}=0.066 \frac{m3}{kg}$
Therefore, theoretical piston displacement in $m^{3}/min=2.24 m^{3}/min$
We know, theoretical piston displacement in m3 per min=$c.\frac{\pi}{4} D^2$ LN
2.24=$4×\frac{\pi}{4}×D^{2}×1.2 D×1200$
D= 0.079 m = 7.9 cm
L=9.5 cm