79 posts
Re: PLANETARY COOLER SURFACE LOSSES
Dear Raj,
What I have given is the approximate for calculating heat balance
to be very exact for losses due to radiation and convention losses different formulas are used.
for radiation ( the standard radiation law is applied)
for convention general law is applied.
if the wind velocity in the atmosphere is above 2.75m/s another formula is used where velocity of air is taken into consideration.
it is difficult to write these formulas.
Radiation
4*(T^4-To^4)/100000000 kcal/m2.hr
Convection
80.33*(T-To)^1.33 /((T+To)/2)^0.724 kacl/m2.hr
if the velocity of wind is above 2.75m/s use for convention
28.03*V^0.805*(T-To)/(T+To)^0.351/D^0.195
where T = surface temp in degree Kelvin
To = ambient temperture in degree kelvin
V=velocity of wind in m/s
D= measuring point of wind velocity distance from kiln in meters (arial distance)
Hope this may be the exact calculation.
thanks
chari
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239 posts
Re: PLANETARY COOLER SURFACE LOSSES
Dear Chari Should I use the radn & conv. formula along with the radn factor means. Radn factor * (Radn formula+conv. formula) * surface area or Directly (rand. formula + conv. formula) * surface area which is correct ? As per FLS I think the surface area is Factor * DIAMETER * LENGTH PER COLER TUBE FACTOR = DEPENDS ON THE NUMBER OF COOLER TUBES FOR 9 COOLER TUBES FACTOR = 24.3 FOR 10 FACTOR = 26.3 FOR 11 FACTOR = 28.1 SO FOR 10 COOLER TUBES & 12.2 M LENGTH SURFACE AREA = 26.3*1.5*12.2 M2 = 481.3 M2 PER COOLER TUBE IF WE TAKE 6 POINTS ALONG THE LENGTH OF COOLER TUBE THAN THE AREA WILL BE = 481.3 / 6 = 80.1 M2 PER METER / COOLER TUBE RAJ
79 posts
Re: PLANETARY COOLER SURFACE LOSSES
Let us assume the avarage shell temperature = 300 degree C
Base temperature = 20 degree C
No of coolers = 12
Dia of cooler = 1 mts
Length of the coolers = 6 mts
Production per hour = 13 tons
Forumal -1
( Standard calculations)
Loss calculation (Kcal/kg)
Radiation loss per m2 per hour in Kcal = 4 x 10-8 x ((300+273)4 – (20+273)4) = 4017 kcal/m2.h
Convention loss per m2 per hour in Kcal =80.33 x ((300+273)-(20+273))1.333 /( ((300+273)+(20+273))/2)0.724 =1811kcal/m2.h
Total heat loss per hr per m2 = 4017+1811 = 5828 kcal/m2.h
Total surface area of coolers = 12 x pi() x 1 X 6 = 226 m2
Total heat loss per hour = 226 x 5828 = 1317128 kcal/hr
Heat loss per kg of clinker = 1317128/13000 = 101kcal/kg
Formula – 2 ( fast calculations)
By standard formula taking radiation factor of books 20.5 x 226 x (300 -20) /13000 = 99 kcal/kg
Result : The difference is only 2 kcal/kg of clinker which is negligible The factors are determined for doing the calculation fast and save time.
138 posts
Re: PLANETARY COOLER SURFACE LOSSES
Chari,
You overestimate radiation losses from the tubes.
Each tube is viewing the kiln on one side
and the ambient on the other side.
The radiation losses are:
kiln side: s (Tcooler 4 – Tkiln 4)
ambient side: s (Tcooler 4 – Tamb 4)
In these formulas, the first term is the loss by radiation from the surface, while the second term is radiation shining on the surface from the external objects. The kiln shell heats up the cooler tubes. The ambient also heats up the tubes, but very little as compared to the losses.
The radiation balance from one tube is this heat loss:
Losses [W] =
Skiln s (Tcooler 4 – Tkiln 4 ) +
Samb s (Tcooler 4 – Tamb 4)
Where Skiln is the surface on the kiln side and
Samb is the surface on the ambient side.
Assuming that the cooler and the kiln are roughly at the same temperature, then the first term vanishes:
Losses [W] =
Samb s (Tcooler 4 – Tamb 4)
Therefore, the radiation losses from the tubes are roughly 50% of what you calculated.
However, the kiln shell itself is also cooled by radiation.
Therefore, if you want to calculate the total losses from the cooler section of the kiln, then you need to take that also into account. The end result is then as explained by Dr Clark. This is the way to go for a kiln balance.
Conversly, if you are only interrested in the cooler tubes and the cooling of the clinker within these tubes, then you should take only the losses from the tubes into account. This is how to proceed for a cooler balance.