Question

explain in the Fortran 77 code performance evaluation of evaporative cooler with corrugated fins and spherical fins...

mention it's parameter in Fortran 77 and plot graph using fortran 77

skip it.....if u can't answer it.....

Answer 1

"!2-D interpolation"

"This example uses the Interpolate2D function to return values of compressor power and mass flow rate as a function of the evaporator and condenser saturation temperatures. The compressor map data are in the Lookup table."   

P$='Power'
M$='m_dot'
P[1]=interpolate2d('compressormap', P$, T_cond,T_evap,T_cond=TC[1],T_evap=TE[1])    "[W]"
m_dot[1]=interpolate2d('compressormap', M$, T_cond,T_evap,T_cond=TC[1],T_evap=TE[1])    "[lb_m/hr]"

TE[1]=27.5 [°F]     "saturation temperature in the evaporator"
TC[1]=85 [°F]     "saturation temperature in the condensor"

"The plot window provides a plot of Power vs saturated evaporator temperature for different saturated condensing temperatures. The interpolated value of power is shown on the plot with a red square. "

$TabWidth 1 cm

C LENGTH OF CAPILLARY TUBE ID=1.63mm R-22 Tc=40 Te=5 w=0.010kg/s
C   AT THE INLET OF CAPILLARY TUBE
   IMPLICIT NONE
   Double Precision :: t1, z,y,p1, Vf1,V1, Vg1, hf1, hg1, uf1, ug1, d,
*Vel1, Re1,f1, t, p, Vf, Vg, hf, hg, uf, ug, Vel, Re, f, g, x, a,
* b,c, u, v, h,fm
   REAL:: E=2.718281828, pie=3.141592654, dia=1.63E-3, w=0.010
c   print*, "enter the value of temperature"
   t1=40
   z=2418.4/(t1+273.15)
   y=15.06-z
   p1=1000*E**y
   print*, "Pressure=", p1/1000
   Vf1=(0.777+0.002062*t1+0.00001608*t1**2)/1000
   V1=Vf1
   PRINT*, "specific volume of saturated liquid=", Vf1
   hf1= 200+1.172*t1+0.001854*t1**2
   print*, "enthalpy of saturated liquid=", hf1
   uf1=0.0002367-(t1*1.715E-6)+(t1*t1*8.869E-9)
   print*, "viscosity of saturated liquid=", uf1
   d= (w*4)/(pie*dia**2)
   Vel1=d*Vf1
   print*, "velocity=", Vel1
   Re1=(Vel1*dia)/(Vf1*uf1)
   print*, "Reynolds Number=", Re1
   f1=0.33/(Re1)**0.25
   print*, "Friction Factor=", f1

   t=39
   z=2418.4/(t+273.15)
   y=15.06-z
   p=1000*E**y
   print*, "Pressure=", p/1000
   Vf=(0.777+0.002062*t+0.00001608*t*t)/1000
   print*, "Vf=", Vf
   Vg= (-4.26+94050*(t+273.15)/p)/1000
   print*, "Vg=", Vg
   hf= 200+1.172*t+0.001854*t**2
   print*, "hf=", hf
   hg=405.5+0.3636*t-0.002273*t**2
   print*, "hg=", hg
   uf=0.0002367-(t*1.715E-6)+(t*t*8.869E-9)
   print*, "uf=", uf
   ug= (11.945E-6)+(t*50.06E-9)+(t*t*0.2560E-9)
   print*, "ug=", ug
   d= (w*4)/(pie*dia**2)
   a=0.5*(Vg-Vf)*(Vg-Vf)*d**2
   print*, "a=", a
   b=1000*(hg-hf)+Vf*(Vg-Vf)*d**2
   print*, "b=", b
   c=1000*(hf-hf1)+(0.5*d*d*Vf**2)-(0.5*Vel1**2)
   print*, "c=", c
   x=(-b+ SQRT((b*b)-(4*a*c)))/(2*a)
   print*, "dryness fraction=", x
   h=hf*(1-x)+x*hg
   print*, "enthalpy=", h
   v=vf*(1-x)+vg*x
   print*, "specific volume=", v
   u=uf*(1-x)+x*ug
   print*, "viscosity=", u
   Vel=d*v
   print*, "velocity=", Vel
   Re=(Vel*d)/(V*u)
   print*, "Reynolds Number=", Re
   f=0.33/(Re)**0.25
   print*, "friction factor=", f
  
  
   stop
   end