Page 363 - Defrosting for Air Source Heat Pump

Appendices 359

kmw1=mf(j-1,i-1)+0.0192; % the water comes from 1st circuit, kg/s
kmw2=kmw1+mf(j-1,i)+0.0192; % the water left from this 2nd
circuit, kg/s
ksmrw=sfrost(17,i); % the sum of water retained on the
coil, kg
kTw1=Tw(j-1,i); % °C
kTri=Tri(j,i); % °C
2
kRr=Rr(j,i); % (K m )/W
kMr=Mr(j,i); % kg/s
khri=hri(j,i); % kJ/kg
% all the input parameters in the function listed here
x0=[0.0042 0.0042 0.335 1200 0.001]; % mf=x(1), mr=x(2),
Tw=x(3); qr=x(4); Tro=x(5) the values of debugging;
options=optimset(’display’,’off’,’MaxIter’,100000,
’MaxFunEvals’,20000);% number
[A,fval,exit]=fsolve(@(x)mystage32(x,kmw1,kmw2,ksmrw,kTw1,
i,kTri,kRr,kMr,khri),x0,options);
mf(j,i)=A(1); % melted water, kg/s; after this stage, mf is 0 kg/s
mrw(j,i)=A(2); % retained water, kg/s
Tw(j,i)=A(3); % retained water temperature, °C
qr(j,i)=A(4); % energy used in defrosting from refrigerant, W
Tro(j,i)=A(5); % the temperature of tube surface at exit of
each circuit, °C
A
x00=real(A);
fval
exit
qm(j,i)=334000.*mf(j,i); % W
sfrost(j,i)=5.*sum(mf(:,i)); % after this stage, sfrost(j,i)
=0.350, kg
qair(j,i)=1.4748.*Tw(j,i).^(4/3).*2.6852*2.5*0.50*((sfrost
(j-1,i))./0.323).^1.5;
%W
s_qair(j,i)=sum(qair(:,i))*5; % W
2
hair(j,i)=1.4748.*Tw(j,i).^(1/3); % W/(K˙m )
smvaw(j,i)=5.*sum(mvaw(:,i)); % kg
2
hd(j,i)=0; % W/(K m )
qvap(j,i)=mvaw(j,i)*2443*1000; % W
s_qvap(j,i)=sum(qvap(:,i))*5; % W
watertray(j,i)=kmw2; % kg/s
swatertray(j,i)=sum(watertray(:,i)); % kg
hro(j,i)=44518+1170.36*Tro(j,i)+1.68674*Tro(j,i)^2+5.2703/
2
1000*Tro(j,i)^3; % W/(m °C)
qr2(j,i)=kMr*(khri-hro(j,i)); % W
s_qr2(j,i)=sum(qr2(:,i))*5; % W
% here is the end of stage 3 for Circuit 2: frost melting with water
flow to down circuit
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