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Diffstat (limited to '2510/CH17/EX17.12/Ex17_12.sce')
| -rwxr-xr-x | 2510/CH17/EX17.12/Ex17_12.sce | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/2510/CH17/EX17.12/Ex17_12.sce b/2510/CH17/EX17.12/Ex17_12.sce new file mode 100755 index 000000000..5ef496cff --- /dev/null +++ b/2510/CH17/EX17.12/Ex17_12.sce @@ -0,0 +1,32 @@ +//Variable declaration: +w = 1 //Length of tube (m) +S = 10/10**3 //Fin patch (m) +//From example 17.10: +t = 1/10**3 //Thickness of fin (m) +ro = 0.0125 //Radius of tube (m) +Af = 3.94*10**-3 //Fin surface area (m^2) +Tb = 145 //Excess temperature at the base of the fin (K) +h = 130 //Heat transfer coefficient (W/m^2.K) +Qf = 64 //Fin heat transfer rate (W) + +//Calculation: +Nf = w/S //Number of fins in tube length +wb = w-Nf*t //Unfinned base length (m) +Ab = 2*%pi*ro*wb //Unfinned base area (m^2) +At =Ab+Nf*Af //Total transfer surface area (m^2) +Qt = h*(2*%pi*ro*w*Tb) //Total heat rate without fins (W) +Qb = h*Ab*Tb //Heat flow rate from the exposed tube base (W) +Qft = Nf*Qf //Heat flow rate from all the fins (W) +Qt2 = Qb+Qft //Total heat flow rate (W) +Qm = h*At*Tb //Maximum heat transfer rate (W) +no = Qt2/Qm //Overall fin efficiency +Eo = Qt2/Qt //Overall effectiveness +Rb = 1/(h*Ab) //Thermal resistance of base (K/W) +Rf = 1/(h*Nf*Af*no) //Thermal resistance of fins (K/W) + +//Result: +printf("1. The total surface area for heat transfer is : %.3f m^2 .",At) +printf("2. The exposed tube base total heat transfer rate is : %.1f W .",Qb) +printf("Or, the exposed tube base total heat transfer rate is : %.0f Btu/h .",Qb*3.412) +printf("3. The overall efficiency of the surface is : %.1f %%",no*100) +printf("4. The overall surface effectiveness is : %.2f .",Eo) |