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alex0919
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Homework Statement
A block of steel dimension is 230mm*230mm*40mm=2,116,000cubic millimeters
0.00785 gram/cubic millimeters is the given steel density
Using 180 degree celsius oil (density given is 0.9) to heat up the steel from 25 degree celsius to 110 degree celsius in a 10mm diameter and 428.5mm long channel with oil pressure at 10 bar from the inlet. How long will it take to heat up the steel to the target temperature? What's the oil flow rate?
Then
Using 15 degree celsius water (density given is 1) to cool down the steel from 110 degree celsius to 60 degree celsius in a 8mm diameter and 489.5mm long channel with water pressure at 10 bar from the inlet. How long will it take to cool down the steel tempeerature to 60 degree celsius? What's the water flow rate?
Known S.H.C. of steel is 0.112 Kcal/Kg/per celsius degree
oil is 0.5 Kcal/Kg/per celsius degree
Water is 1 Kcal/Kg/Per celsius degree
Homework Equations
1atm = 14.7psi =101,325Pa
1kg/cm^2 =14.2psi= 97,878.57Pa
2kg/cm^2 =195,757Pa
10kg/cm^2 =978,785Pa
u^2 = 978,785/1,000
u^2/gc = P/d
gc=1
u = 31.29 m/s (Water with density of 1)
u = 32.98 m/s (Oil with density of 0.9)
The Attempt at a Solution
Oil flow rate=32.98m/sec*3.14*0.004m*0.004m=0.001657cubic meters/sec=1.66 Liter/sec.
Water flow rate=31.29m/sec*3.14*0.005m*0.005m=0.002456cubic meters/sec=2.46Liter/sec.
Is my calculation on the velocity right? It looks to me there is something wrong compare with the result we got in a practice similar to this.
(110-25)*0.112*16.61Kg(steel weight)=158.1272Kcal
(180-110)*0.5=140Kcal
158.1272/140=1.12948Kg of required Oil /0.9(Oil density)=1.2550 Liter
1.2550Liter/Oil flow rate at 1.66L/Sec=0.76 sec.
I am not quite sure if my calculation is right because practically, I don't think we can achieve the target in that short period of time. I think there must me something wrong on my calculation. Can anybody point out my mistake? Thank You!
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