
#1
Apr1812, 08:20 AM

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I have an anaerobic digester with a volume of 2500 m^3 and internal temperature is 60°C. But, average outer temperature is 20°C. We can neglect the insulations, accept c= 1kcal/kg, and assume that the surface area of the reactor is 300 m^2. I can calculate the heat loss from e=m.c.deltaT..
Is there someone who can tell me how long does it take to balance the outer and internal temperatures? Or, the time for dropping the internal temperature to 58°? (I guess watt= joule/second does not work in such case) Thanks in advance.. Gorkem 



#2
Apr1812, 10:48 PM

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What exactly is c? The units you give it imply energy density, but then your e= formula suggests power density.
You can't ignore the insulation due to the contents, even if you ignore that from the container. If it had no insulation value it would always track external temperature. So there's not enough information here. The shape of the container also matters. There will be a nonuniform distribution of temperatures. If we assume a cylinder radius R (infinite length) with uniform power density P (power/unit area) and conductivity K (power per unit length per temperature gradient), for an annulus at a radius r: Temperature = θ(r) Heat flow F(r) = K.2.π.r.∂θ/∂r In steady state, this must match the total power generated inside the annulus: F(r) = P.π.r^2 So ∂θ/∂r = P.r/2K θ(r) = core temp  (P/4K).r^2 



#3
Apr1912, 03:35 AM

P: 3

I just want to calculate the cooling time very roughly. So i did not give any information on insulation, etc. We can assume that the temperature distributed uniformly and c is heat capacity.
For a 2500 m^3 reactor with 60°C internal solid waste temperature, the energy amount which waste need to loose until the temperature drops to 58°C is equal to (c: 1 kcal/kg, d: 1kg/L): e=m.c.ΔT = 2.500.000 kg* 1* (6058)= 5.000.000 kcal Of course, there will be temperature gradient from outer to inner (or vice versa) part of the reactor, which is determined by the radius (it also defines the total surface area). But in this case we can ignore the temperature gradient, cause the tank is complete mix. If the height of the tank is ≈ 10 m, we can accept the surface area as 300 m^2. I sincerely appreciated that your calculation the temperature of the waste at any point in the reactor, but i need to shift to the determination of the time it will take.. i just reached an equation: m.c.dT/dt=h.A.ΔT=Q here h is the heat transfer coeff. do you think this works? Thanks 



#4
Apr1912, 05:04 PM

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my digester
You simply cannot treat the temperature as uniform AND ignore external insulation.
If you do both then all heat generated is lost instantly. If the internal heat transfer coefficient is significant but the external insulation is not then the internal heat gradient must be considered. I did assume that the digester was continuing to generate heat, but it looks like you're assuming it isn't. OTOH you want cooling time, so we're not lookng at steady state. That makes it tough. I can write down the diffusion equation but I don't know if I can solve it. A quick search of the web was fruitless. Simplest might be to model it. Might think about it some more. 



#5
Apr2012, 03:48 AM

P: 3

Thank you haruspex..




#6
Apr2212, 11:06 PM

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Here goes...
Taking the digester to be a uniform solid sphere radius R, surface temp fixed at 0 (we only care about relative temperatures), some initial internal profile. At time t, radius r, temperature = T(r, t). Diffusion equation in polar is δ/δr(r^2.δT/δr) = k.r^2.δT/δt some constant k (based on conductivity and specific heat). This has solutions of the form sin(a.r).exp(λt)/r where λ = (a^2)/k A general solution will be a linear combination of these. The T(R, t) = 0 condition means a = n.π/R for some integer n (making sin(a.R) = 0). So T(r, t) = Ʃc[itex]_{n}[/itex].sin(n.π.r/R).exp(λ[itex]_{n}[/itex]t)/r where λ[itex]_{n}[/itex] = (n.π/R)[itex]^{2}[/itex]/k The final step is to find the sequence c[itex]_{n}[/itex] s.t. T(r, 0) matches the initial profile (standard Fourier analysis). If you are interested in the core temperature at time t, i.e. at r=0, you need to take the limit of sin(a.r)/r as r tends to 0. This is simply a, giving: T(0, t) = Ʃc[itex]_{n}[/itex].(n.π/R).exp(λ[itex]_{n}[/itex]t) But you can't avoid having to supply the c[itex]_{n}[/itex]. Over time, the dominant term will become the one with the smallest λ[itex]_{n}[/itex], i.e. the n=1 term: T(0, t) ~ c[itex]_{1}[/itex].(π/R).exp(λ[itex]_{1}[/itex]t) for large t HTH 


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