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Diffstat (limited to '2510/CH19/EX19.11/Ex19_11.sce')
-rwxr-xr-x | 2510/CH19/EX19.11/Ex19_11.sce | 29 |
1 files changed, 29 insertions, 0 deletions
diff --git a/2510/CH19/EX19.11/Ex19_11.sce b/2510/CH19/EX19.11/Ex19_11.sce new file mode 100755 index 000000000..2dc08e48c --- /dev/null +++ b/2510/CH19/EX19.11/Ex19_11.sce @@ -0,0 +1,29 @@ +//Variable declaration: +h1 = 800 //Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2. F) +h2 = 40 //Heat transfer coefficient for oil outside coil (Btu/h.ft^2. F) +h3 = 40 //Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2. F) +h4 = 2 //Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2. F) +k1 = 0.039 //Thermal conductivity of insulation layer (Btu/h.ft. F) +l1 = 2/12 //Thickness of insulation layer (ft) +D = 10 //Diameter of tank (ft) +H = 30 //Height of tank (ft) +k2 = 224 //Thermal conductivity of copper tube (Btu/h.ft. F) +l2 = (3/4)/12 //Thickness of insulation layer (ft) +T1 = 120 //Temperature of tank ( F) +T2 = 5 //Outdoor temperature ( F) + +//Calculation: +Uo1 = 1/(1/h3+(l1/k1)+1/h4) //Overall heat transfer coefficient for tank (Btu/h.ft^2. F) +At = %pi*(D+2*l1)*H //Surface area of tank (ft^2) +Q = Uo1*At*(T1-T2) //Heat transfer rate lost from the tank (Btu/h) +//From table 6.3: +l2 = 0.049/12 //Thickness of coil (ft) +A = 0.1963 //Area of 18 guage, 3/4-inch copper tube (ft^2/ft) +Uo2 = 1/(1/h2+(l2/k2)+1/h1) //Overall heat transfer coefficient for coil (Btu/h.ft^2. F) +//From steam tables: +Tst = 240 //Temperature for 10 psia (24.7 psia) steam ( F) +Ac = Q/(Uo2*(Tst-T1)) //Area of tube (ft^2) +L = Ac/A //Lengt of tube (ft) + +//Result: +printf("The length ofcopper tubing required is : %.1f ft",L) |