Free convection from a cylinder.
Free convection from a cylinder.
(OP)
I am running a free Convection heat transfer experiment in which I am trying to determine the heat transfer coefficient of a long horizontal cylinder. The experiment requires that I find this coefficient by means of measuring the surface temperature and using that along with the ambient temperature and dimensions of the cylinder to determine the heat transfer coefficient. I am currently using a 120W Cartridge heater with 5/8" diameter and 6" long and varying the wattage I put into the heater.
When I calculate the heat transfer coefficient from Newtons law of cooling Q=A*h*(Ts-To) I get one value and when I use the correlations for heat transfer of a long horizontal cylinder I get a significantly lower number. I have tried this at different wattages and have come up with the same results of between 40%-55% error between the two heat transfer coefficients. Can anyone help me with what I might be doing wrong or why this error is occurring?
When I calculate the heat transfer coefficient from Newtons law of cooling Q=A*h*(Ts-To) I get one value and when I use the correlations for heat transfer of a long horizontal cylinder I get a significantly lower number. I have tried this at different wattages and have come up with the same results of between 40%-55% error between the two heat transfer coefficients. Can anyone help me with what I might be doing wrong or why this error is occurring?





RE: Free convection from a cylinder.
Note, also, a single surface temperature measurement is probably inaccurate, since the surface temperature is expected to vary depending on where on the cylinder you measure it.
TTFN
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RE: Free convection from a cylinder.
RE: Free convection from a cylinder.
RE: Free convection from a cylinder.
Patricia Lougheed
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RE: Free convection from a cylinder.
Dave
RE: Free convection from a cylinder.
Tsurface=376K
Tambient=297K
Tfilm=336K
B=.0030 (1/K)
v=19.51*10^-6 (m^2/s)
alpha=27.83*10^-6(m^2/s)
k=28.96*10^-3 (W/m*K)
emissivity=.17 ( stainless steel)
Rayleigh number =17075.7
Nu=5.49
h=9.99
Qconv=6.00793 W
Qrad=.9 W
Q=Qconv+Qrad=6.9 W
10-6.9/10=.31*100=31% error
So with the inclusion of radiation I am still 31% off from theory to measured values. I can assume that this 31% is not loss in conduction as the plate holding the cylinder only went up 4 degrees. If the loss is due to radiation the emissivity of the stainless steel would have to be around .76. Does anyone have any other reason these values do not match?
RE: Free convection from a cylinder.
http://bo
When you average your surface temperature, what is the range? does it lead you to think that the surface temperature is uniform? Would you expect some variations in radiation (T^4) with your surface temperature variations? How did you determine the .17 emissivity for stainless? I can find a lot of higher values with a 30 second internet search.
Also, how are you taking surface meassurements? With an infrared thermometer? Are you measuring through the lexan? What is the emissivity of lexan? What is the emissivity of stainless steel? What is the emissivity setting for the instrument? blah, blah, blah
RE: Free convection from a cylinder.
However, this is also part of the "correlation" process. You try to account for all the non-idealities, and what's left over is "correlated" to the actual results.
TTFN
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RE: Free convection from a cylinder.