How to plot smooth curve

12 vues (au cours des 30 derniers jours)
muhammad nur
muhammad nur le 13 Déc 2020
Commenté : muhammad nur le 14 Déc 2020
Y=[0:25]; %lenght (m)%
L= 25; %lenght (m)%
r1=0.043; %inner radius of pipe (m)%
r2=0.045; %outer radius of pipe (m)%
rg=0.090; %radius surrounding of ground (m)%
D=0.180; %diameter pipe (m)%
AsurL=pi*D*L; %surface area (m2)%
AsurY=pi*D*Y; %area (m2)%
Tair=34; %air temperature inlet (°C)%
Tg=24; %temperature ground (°C)%
C=1007; %specific heat air (J/kg.K)%
h=37.56; %convection of heat transfer coefficient %
Rc=1./(AsurL*h); %convection resistance%
ln=@log;
Kp=0.19; %PVC pipe Thermal conductivity (W/m.K)%
Kg=0.655; %Ground thermal conductivity (W/m.K)%
Rp=(ln((r2)/r1)./(2*pi*L*Kp)); %pipe resistance%
Rg=(ln((rg)/(r2))./(2*pi*L*Kg)); %ground resistance%
Rt=Rg+Rp+Rc; %total resistance%
U=1./Rt;
m=0.21; %mass flow rate (kg/s)%
Ty1=Tg+((Tair-Tg).*exp((-U.*AsurY)./(m.*C)));
T=Tair-Ty1;
Q=m*C*T
m=0.11; %mass flow rate (kg/s)%
Ty1=Tg+((Tair-Tg).*exp((-U.*AsurY)./(m.*C)));
T=Tair-Ty1;
Q1=m*C*T
m=0.02; %mass flow rate (kg/s)%
Ty1=Tg+((Tair-Tg).*exp((-U.*AsurY)./(m.*C)));
T=Tair-Ty1;
Q3=m*C*T
m=0.05; %mass flow rate (kg/s)%
Ty1=Tg+((Tair-Tg).*exp((-U.*AsurY)./(m.*C)));
T=Tair-Ty1;
Q4=m*C*T
plot(Y,Q,Y,Q1,Y,Q3,Y,Q4)
axis([0 25 0 2500]);
ylabel('heat transfer,Q');
xlabel('pipe length');
legend('m=0.21');Y=[0:25]; %lenght (m)%

Réponse acceptée

Mathieu NOE
Mathieu NOE le 13 Déc 2020
hello
simply increase the x axis points density
change the first two lines so you have now 10 more points in your curves - looks smoother
L= 25; %lenght (m)%
Y=linspace(0,L,250); %lenght (m)%
  1 commentaire
muhammad nur
muhammad nur le 14 Déc 2020
its work, thank you

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur 2-D and 3-D Plots dans Help Center et File Exchange

Produits


Version

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by