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# Estimating Heat-Up Time For Product Tank

tanks heating start-up time coils steam hot water calculations

8 replies to this topic
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### #1 Frankdj

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Posted 20 March 2017 - 10:27 AM

After calculating the KJ required to heat the steel of a tank, say Q1,  plus the KJ to heat up the product, say Q2,  and knowing the rate of loss of heat once up to target temperature, say P3 ( and estimating that during the start-up that the rate of loss of heat is 50% of the full rate,  due to the temperature gradient during that phase ).   We can use (Q1+Q2) / (Latent heat of the steam)   = part of the mass of steam required, enabling one to attempt a first pass estimate of the time for the start-up, and then apply that time to convert P3 to KJ also,  to then add to Q1  and Q2 for an approx total KJ .......  it would be nice if it were simple and just a function of the steam flow rate.  But it is compounded by a heat balance equation,   and the unknown loss of heat of the ejected steam / condensate mix  plus unknowns such as the product ( bulk liquid ) changing properties during the heat up, eg  viscosity, density, thermal conductivity, etc.   It seems that one has to resort to empirical data for estimating a start-up duration.  Any  advice or comments are welcome.  There are no specific bulk liquids in question,  This is a general issue, in particular for heating coils inside tanks,  with steam or even hot water as the heat source. Thank you.

Regards  Frank

### #2 Frankdj

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Posted 20 March 2017 - 06:15 PM

Please note, in the above case Natural Convection is envisaged for the product, not forced circulation.
Thanks
Frank

### #3 sgkim

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Posted 21 March 2017 - 04:33 AM   Best Answer

Hi Frank,

To calculate the heating up time easily, you'd rather simplify the system by assuming the following system parameters be constants:

steam temperature (ts), overall heat transfer coefficient(U), surface convection loss coefficient,(h), specific heats of product and steel(Cp & Cps), ta=air temperature,

m = mass of product,   ms=mass of steel, Ac=surface area of coil (or jacket), Av=surface area of vessel

Heating duty can be supplied by the heat transfer in the system deducting the heat loss to atmosphere due to natural convection on the surface Av.

Heat balance:  (m*Cp + ms*Cps)*dt/dθ  =  U*Ac*(ts-t) - h*Av*(t-ta)  .........(1)  <= the only variables are time, θ, and product temperature, t.

You can calculate the time required θ by integrating the equation (1) from initial temperature t1 to final temperature t2 :

θ =(integrate from 0 to θ) ∫dθ = (integrate from t1 to t2) ∫ {(m*Cp + ms*Cps) dt / [U*Ac*(ts-t) - h*Av*(t-ta)] } .........(2)

You can neglect the heat loss term, h*Av*(t-ta), if the surface of the vessel is well insulated to minimize the heat loss.

~Stefano

Edited by sgkim, 21 March 2017 - 04:37 AM.

### #4 Frankdj

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Posted 21 March 2017 - 12:26 PM

Hi Stefano,

Your response is fantastic.  Problem solved .   Thank you very much.

Kind regards,

Frank

### #5 sgkim

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Posted 23 March 2017 - 02:01 AM

Frank,

The total steam requirement for the whole heating sequence and the instant steam requirement at time t are not yet answered by me.  Evaluate these data from the equations given in my previous posting.

~Stefano

### #6 Frankdj

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Posted 28 March 2017 - 01:46 PM

Stefano,

I have attached a file which shows my working on the heat  equation derivation for the start-up time to heat the liquid in a tank from T1 to T2.   Unfortunately the final part of the equation involves a Ln denominator element which tends to  yield the problem of trying to find the Ln of zero.  Which is nonsense.  I wonder if you would kindly have a look at it and see if you can spot the mistake.  Thank you.  Kind regards,

Frank

### #7 sgkim

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Posted 06 April 2017 - 04:59 AM

Frank,

In your equation dT/dt =0 is equivalent to "no heat transfer in the system"  i.e. dQ/dt=0.   "ln zero" means that "no additional time is required if T1 is equal to T2".

~Stefano

### #8 Frankdj

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Posted 09 April 2017 - 06:00 PM

Stefano,

If T1 = product import temperature, and T2 =  target storage temperature of product.  Then let T3 = the steady state temperature of the product as  time approaches infinity.   On a graph of Product Temperature versus time,  I imagined the gradient at the steady state temperature would be dT/dt = (T3-T3)/infinity  = 0.  Which is when "heat input" ="heat losses".

My only interest in T3 was to check that  T3 > T2, because if this were not so,  then T2 would not be achievable.

My main  interest is the validity of the finally derived equation on page 4 of my attachment ( see above ).   It seems to be academically correct,  but the answers it produces are at times erratic and in general seem  contrary to expected reality.  The denominator of the "ln" portion often tends to "ln" zero,  which gives rise to nonsense and instability in the time calculated.  I hope you can spot my mistake.  The derivation of the final formula on page 4 is independent of T3.

Thanks

Frank

### #9 sgkim

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Posted 21 April 2017 - 02:34 AM

Frank,

Based on the integrated equation, a temperature vs. time profile can be drawn.  You can find that the steady state temperature shall be reached at time is infinite.

The case that  K1-K2T = 0  is a singularity point and a special case in actual situation.  If T is equal to K1/K2,  the temperature of the product does not depend on the time, but determined only from weighted balance temperature of Uc*Ac and Us*As.  The temperature of the product shall be

Weighted temperature,  T = (Uc*Ac*Th + Us*Ac*Ta)/(Uc*Ac+ Us*Ac)  ..............(3)  <= not depend upon time.  (singularity point)

For other cases,    (K1-K2T)/(K1-K2To) =  exp(-K2 t/MC), or

T  = {(K1-K2To)/K2 }*{(K1/(K1-K2To)-exp(-K2 t/MC)} = K1/K2 - (K1-K2To)/K2*exp(-K2 t/MC)....(4)   <== T vs t equation

MC = Mp*Cpp+Mst*Cst,  K1 = Uc*Ac*Th + Us*Ac*Ta,  K2 =Uc*Ac+ Us*Ac

if t = 0,  T = To,   if t = ∞, T= K1/K2

Time required to get T3 and also the maximum temperature can be calculated from equation (4).

~Stefano