Thermo WAH 10 -- simulating a inner-vascular cooling system

Thermo WAH10

%% Prepare environment
clear all; clc; clear figure; 

%initialize vecctors for plotting:
Tgraph=[];
Wgraph=[];

%cycle through temperature range:
for T1=273.6:340.6
	Tgraph=[Tgraph,T1];
	%mdot=9.246e-4; %kg/s, leftover from modular testing!
	%T1=0+273.6; %K , leftover from modular testing!
	P1=.2928*1000; %kPa
	fluid='R134A.fld'; %refrigerent is R134A

	%% State 1:
	h1=refpropm('H','T',T1,'P',P1,fluid);     %J/kg % enthalphy, state 1 calculation with P1 and T1
	disp(['enthalphyH1 = ',num2str(h1/1000),'kJ/kg'])

	s1=refpropm('S','T',T1,'P',P1,fluid);     % entropy
	disp(['entropyS1 =   ',num2str(s1/1000),'kJ/(kg K)'])

	%% State 2:
	s2=s1; 
	P3=.64578*1000; %kPa
	P2=P3;

	h2=refpropm('H','P',P2,'S',s2, fluid);    %J/kg % entropy
	disp(['enthalphyH2 = ',num2str(h2/1000),'kJ/kg'])

	%% State 3:
	T3=24+273.6; %K
	h3=refpropm('H','T',T3,'Q',0,fluid); %J/kg
	quality3=refpropm('Q','T',T3,'P',P2,fluid); %phase, to check that we're in the right region 
	disp(['enthalphyH3 = ',num2str(h3/1000),'kJ/kg'])

	%%State 4:
	h4=h3;

	%calculations: 
	qd14=.153*1000; %W, taken from the 2 K/hr figure published
	mdot=qd14/(h1-h4); %calculate mdot

	work=mdot*(h1-h2);   %(kg/s)*(J/kg), calculate compressor work
	Wgraph=[Wgraph,work];
end

%% plot results:
plot(Tgraph,Wgraph, 'r-*', 'LineWidth', 2.5)
title('Evaporator Intake Temperature Effect on Work Produced by Intravascular Cooling Compressor','FontSize',20)
xlabel('Evaporator Intake Temperature (K)', 'FontSize',16)
ylabel('Compressor Work (Watts)', 'FontSize',16)