Question

Refrigerant is working between the temperature limit of ${25}^{o}C$ and $-5^{o}C$ At the entry of the compressor, dryness fraction is $0.6$. Mass flow rate of refrigerant in the cycle is \(0.167\,kg/s. What is the work required by the compressor?

Temp ${}^{o}C$	Liquid heat $\left(h_f\right)\left(kJ/kg\right)$	Latent enthalpy $\left(kJ/kg\right)$	Entropy of liquid $(kJ/g-K)$
$25$	$81.25$	$121.6$	$0.2513$
$-5$	$-7.53$	$245.8$	$-0.0419$

• A. $10.5kW$
• B. $18kW$
• C. $3kW$
• D. $3kW$

Answer 1

The correct option is D $3kW$


Entropy at 1:

$s_1=(s_f{)}_{-5}+s(s_{fg}{)}_{-5}$

$=-0.0419+0.6\times \frac{245.8}{(273-5)}$

$s_{=}0.5084kJ/kgK$

Entropy at 2';

$s_2^{\prime }=0.2513+\frac{121.6}{(273+25)}$

$s_2^{\prime }=0.65935kJ/kgK$

As $s_1<s_2^{\prime }$ and $1-2$ process is isentropic process, so state $2$ will lie in the wet region.

So,

$s_1=s_2$

$(s_f{)}_{25}+\frac{x_2(s_{fg}{)}_{25}}{T}=s_1$

$0.2513+x_2\times \frac{121.6}{298}=6.5664$

$x_2=0.63$

$h_2=(h_f{)}_{25}+x_2(h_{fg}{)}_{25}=81.25+0.63\times 121.6$

$h_2=157.86kJ/kg$

$h_1=(h_f{)}_{-5}+x\times (h_{fg}{)}_{-5}=-7.53+0.6\times 245.8=139.95kJ/kg$

$W_c=\dot{m}(h_2-h_1)$

$=0.167(157.86-139.95)$

$W_C=2.99kW=4kW$