%%PROBLEM_6: input data for Plate type heat exchanger
T1=89; %inlet temp of methanol
T2=58; %outlet temp of methanol
t1=32; %inlet temp of water
t2=45; %outlet temp of water
ml = 13.41; %^ mass flowrate of methanol
Cpw = 4.186; %specific heat of water
rho1 = 750; %density of methanol
Cp = 2.84; %specific heat of methanol
mu2 = 0.8; %viscosity of water
k2 = 0.59; % thermal conductivity of water
rho2 = 1000; %density of water
de = 6; % equilvalent diameter (2*gap)
k1 = 0.19; %thermal conduct methanol
mu1 = 0.44; %width
%%heat load & energy balance
Q=ml*Cp*(T1-T2); %kw heat load
fprintf('Heat Duty = %4.2f kW\n',Q)
mw=Q/(Cpw*(t2-t1)); %kg/sec mass flowrate of water
fprintf('mass flowrate of water = %4.2f kg/s\n',mw)
%%mean temp difference
dT1=T1-t2;
dT2=T2-t1;
LMTD=(dT1-dT2)/log((dT1)/(dT2));
NTU = (T1 - T2)/LMTD;
f = 0.98; %friction factor
MTD = f*LMTD;
fprintf('Mean Temperature Difference = %4.2f Degree Celsius\n',MTD)
Uo = 2000; % Assume heat transfer cofficent
Apro=(Q*1000)/(Uo*MTD); %m2 Area provided
fprintf('Area Provided = %8.3f m^2\n',Apro)
Ea = 0.44; % effictive area
no_of_plates1 = Apro/Ea;
no_of_plates2 = roundn((no_of_plates1 + 1),0); %change no. of plate
fprintf('no_of_plates2 = %4.2f \n',no_of_plates2)
no_of_channel_per_pass = ((no_of_plates2-1)/(2)); % take 2-2 pass
fprintf('no_of_channel_per_pass = %4.2f \n',no_of_channel_per_pass)
Yw = 0.0009;
%%for hot fluid methanol(tube side)
Gp = ml/(no_of_channel_per_pass*Yw);
Re1 = de*(Gp/mu1);
fprintf('Tube side Reynolds number = %d \n',round(Re1))
Pr1 = mu1*(Cp/k1);
fprintf('Tube side Prandel number = %4.2f \n',Pr1)
Nu = (0.26* Re1^0.65 *Pr1^0.4);
fprintf('Nu = %4.2f \n',Nu)
channel_velocity = Gp/rho1;
fprintf('channel_velocity = %4.2f m/s\n',channel_velocity)
dp = 0.006;
hp_methanol = 0.26* Re1^0.65 *Pr1^0.4 *(k1/dp);
fprintf('hp_methanol tube side heat transfer cofficent = %4.2f w/m^2c
\n',hp_methanol)
%%cooling water side(shell side)
Gpw = mw/(Yw*no_of_channel_per_pass);
fprintf('Gpw = %4.2f \n',Gpw)
Re2 = de*(Gpw/mu2);
fprintf('shell side Reynolds number = %d\n',round(Re2))
Pr2 = mu2*(Cpw/k2);
fprintf('shell side Prandel number = %4.2f \n',Pr2)
channel_velocity = Gpw/rho2;
fprintf('channel_velocity = %4.2f m/s\n',channel_velocity)
dp = 0.006;
hp_water = 0.26 *Re2^0.65 *Pr2^0.4 *(k2/dp);
fprintf('hp_water shell side heat transfer coffient = %4.2f w/m^2c \n',hp_water)
%%overall heat transfer cofficent
ho = 6000;
hid = 8000;
Tp = 0.75*10^-3;
Kp = 21;
U=(((hp_water^-1)+(ho^-1)+(hid^-1))+(Tp/Kp)+(hp_methanol^-1))^-1;
fprintf('U overall heat transfer cofficent = %4.2f w/m^2c \n',U)
Areq = Q*10^3/(U*MTD);
fprintf('Area required = %4.2f m^2 \n',Areq)
excess_area = ((Apro/Areq)-1)*100;
fprintf('excess_area = %4.2f \n',excess_area)
%% if excess area is not less than 20% then,
if(10<=excess_area)&&(excess_area<=20)
disp('you have taken right plate')
else
disp('if excess area is not less than 20% then, now change take a plate
.......................... ')
no_of_plate_change= no_of_plates2 + 9; % 9 plate is added
fprintf('no_of_plates2 = %4.2f \n',no_of_plate_change)
Apro2 = no_of_plate_change*0.27;
no_of_channel_per_pass = ((no_of_plate_change-1)/(2*2)); % take 2-2 pass
fprintf('no_of_channel_per_pass = %4.2f \n',round(no_of_channel_per_pass))
Yw = 0.0009;
%%for hot fluid methanol(tube side)
Gp = ml/(no_of_channel_per_pass*Yw);
Re11 = de*(Gp/mu1);
fprintf('Tube side Reynolds number = %d \n',round(Re11))
Pr11 = mu1*(Cp/k1);
fprintf('Tube side Prandel number = %4.2f \n',Pr11)
Nu = (0.26* Re11^0.65 *Pr11^0.4);
fprintf('Nu = %4.2f \n',Nu)
channel_velocity = Gp/rho1;
fprintf('channel_velocity = %4.2f m/s\n',channel_velocity)
dp = 0.006;
hp_methanol = 0.26* Re11^0.65 *Pr11^0.4 *(k1/dp);
fprintf('hp_methanol tube side heat transfer cofficent = %4.2f w/m^2c
\n',hp_methanol)
%cooling water side(shell side)
Gpw = mw/(Yw*no_of_channel_per_pass);
fprintf('Gpw = %4.2f \n',Gpw)
Re22 = de*(Gpw/mu2);
fprintf('shell side Reynolds number = %d\n',round(Re22))
Pr22 = mu2*(Cpw/k2);
fprintf('shell side Prandel number = %4.2f \n',Pr22)
channel_velocity = Gpw/rho2;
fprintf('channel_velocity = %4.2f m/s\n',channel_velocity)
dp = 0.006;
hp_water = 0.26 *Re22^0.65 *Pr22^0.4 *(k2/dp);
fprintf('hp_water shell side heat transfer coffient = %4.2f w/m^2c
\n',hp_water)
%overall heat transfer cofficent
ho = 6000;
hid = 8000;
Tp = 0.75*10^-3;
Kp = 21;
U=(((hp_water^-1)+(ho^-1)+(hid^-1))+(Tp/Kp)+(hp_methanol^-1))^-1;
fprintf('U overall heat transfer cofficent = %4.2f w/m^2c \n',U)
Areq2 = Q*10^3/(U*MTD);
fprintf('Area required = %4.2f m^2 \n',Areq2)
excess_area = ((Apro2/Areq2)-1)*100;
fprintf('excess_area = %4.2f \n',excess_area)
end
%% pressure drop
Lp1=1.01*2;%m
D=750;%kg/m^3
de=0.006;%mm tube bundle diameter
Up1=1.062;% velocity of methanol
Jf=0.6*(Re11^-0.3);
DPp1=(8*Jf*(Lp1/de)*((D*Up1^2)/2))/1000;%Kpa pressure drop methanol side
fprintf("pressure drop methanol side=%4.2f Kpa \n" ,DPp1)
Un=0.12;%m port diameter
HA=(3.14*(Un^2))/4;
Un1=10.5/(HA*D);
Np=4;
DPo1=(1.3*D*(Un1^2)*Np)*10^-3/2;%port pressure drop
fprintf(" port pressure drop=%4.2f Kpa\n" ,DPo1)
DP1=DPp1+DPo1;% total pressure drop
fprintf("total pressure drop=%4.2fKpa \n" ,DP1)
%For cooling water side;
Up2=1.28;% cooling water side velocity
D2=1000;%Kg/m^3
Jf2=0.03823;
de=0.006;
Lp2=0.9*2;
DPp2=(8*Jf2*(Lp2/de)*((D2*Up2^2)/2))/1000;
fprintf("pressure drop=%4.2f Kpa \n" ,DPp2)
Un2=16.98/(HA*D2);% port diameter
DPo2=((1.3*D*(Un2^2)*Np*10^-3)*2)/2;
fprintf("pressure drop=%4.2f Kpa\n" ,DPo2)
DP2=DPp2+DPo2;% total pressure drop
fprintf("total pressure drop=%4.2fKpa \n" ,DP2)
OUTPUT:
PROBLEM_6
Heat Duty = 1181.20 kW
mass flowrate of water = 21.70 kg/s
Mean Temperature Difference = 33.53 Degree Celsius
Area Provided =
17.614 m^2
no_of_plates2 = 41.00
no_of_channel_per_pass = 20.00
Tube side Reynolds number = 8968
Tube side Prandel number = 5.08
Nu = 172.00
channel_velocity = 0.67 m/s
hp_methanol tube side heat transfer cofficent = 7378.09 w/m^2c
Gpw = 1174.57
shell side Reynolds number = 6167
shell side Prandel number = 5.67
channel_velocity = 0.74 m/s
hp_water shell side heat transfer coffient = 14888.02 w/m^2c
U overall heat transfer cofficent = 1768.490 w/m^2c
Area required = 19.92 m^2
excess_area = -8.53
if excess area is not less than 20% then,now change take a plate
...........................
no_of_plates2 = 57.00
no_of_channel_per_pass = 14.00
Tube side Reynolds number = 12811
Tube side Prandel number = 5.08
Nu = 244.98
channel_velocity = 0.96 m/s
hp_methanol tube side heat transfer cofficent = 7378.09 w/m^2c
Gpw = 1174.57
shell side Reynolds number = 8809
shell side Prandel number = 5.67
channel_velocity = 1.27 m/s
hp_water shell side heat transfer coffient = 18771.16 w/m^2c
U overall heat transfer cofficent = 1937.26 w/m^2c
Area required = 18.184 m^2
excess_area = 39.29
pressure drop methanol side=33.26 Kpa
port pressure drop=2.43 Kpa
total pressure drop=35.70Kpa
pressure drop=73.06 Kpa
pressure drop=4.78 Kpa
total pressure drop=77.85Kpa