water to water heat exchange 2

[1RMD]

Hello All,
I'm trying to determine the distance that water, with an inlet temp of +- 50*F, will have to travel to reach the desired temp of 120*F. This potable water, traveling in 3/4 copper tubing, will be submerged in transiet water of 140*F inside a PVC tube. I will assume that the temps of the inlet potable water and the transiet water will remain constant. I have tried to calculate this useing Fouriers Law and Thermal Conductivity of Copper being 400 but I'm afriad I'm lost here. I suppose I should be able to calculate this with a formula from a post by "zekeman" on a similar problem from "cobra 007": "Time for water to cool in a pipe". Any help would be greatly appreciated.
Merry Christmas,
Mark(1RMD)

 

[1RMD]

I forgot to post a flow rate. Lets assume 10gpm. Thanks again.

 

[vpl]

1RMD

I believe the temperature of the "transient water" would need to decrease as it went along, otherwise no heat transfer would occur.

You might take a look at a heat transfer book for a tube-in-tube heat exchanger.

I suspect that the length of tubing will be rather long.

Patricia Lougheed

******

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[1RMD]

Patricia,

Thanks for your post. To clarify, the temp of the transient water would decrease as the heat is trasferred to the potable water but the volume and flow rate of the transient water should make this temp drop negligible, at least for the calculation I'm trying to make. I guess I should have stated 140*F as the average temp of the transient water. My thoughts are for a coil of 3/4 copper or several sections joined together to form the required length for the heat transfer to take place inside perhaps a 10' section of 4" pvc pipe. Any ideas? Thanks

 

[IRstuff]

You could, but that would normally be absurdly inefficient. For every deg*gpm the potable goes up, that's got to be made up by the transient flow. So, 70deg*10 gpm = 700 deg*gpm, and if you want say, 1 deg drop on the transient, then you need 700 gpm flow. So, you'll spend gobs of horsepower just cranking water around, and you still have to make up the same amount of heat, i.e., 700deg*gpm.

TTFN

FAQ731-376

 

[25362]

Why not take an overall heat transfer coefficient U of, say, 300 W.m^-2K^-1 with an LTD of, say, 26^oC. The surface area of the pig tail cooler tube could be calculated, for the transferred heat Q.

Q is estimated from:
Q = water mass flow rate x [Δ]t x Cp
[Δ]t = 70^oF ~ 39^oC
Cp for water taken as = 1.0
Q = U.A.LTD

Just take care of using consistent units.

 

[zekeman]

I'm not sure about how you can insure constant external water temperature but if you can, the formula can be developed as follows:
Take a slug of water dx wide moving in the copper pipe and as a 1st assumption let the pipe just act as a transfer medium;then you can write the instantaneous energy equation,
Rho*Ac*C*dT/dt=h*As*(T0-T)
(dx)rho*Pi D^2/4*c*dT/dt=h*Pi*D*dx*(T0-T)
dx drops out
so
62*1*D/4*dT/dt=h*(T0-T)
The solution to this differential equation with a time constant of
62*.75/12/h= tau
is:
T=T0-(T0-Ti)*e^-t/tau and
the time t in transit required for T to reach 120 is next obtained from
120=140-(140-50)*e^-t/tau
t=[ln(90/20)]*tau=1.5 tau
I evaluated tau using 150BTU/Hr/ft^2/deg for h, the external film, which swamps the internal one
so tau=62*.75/12/150=.025hrs=1.5 minutes and
t=1.5*1.5=2.25 minutes
The water at 10 gpm is traveling at about
360'/minute the distance would be
2.25*360=810 feet
Barring mathe errors ,an 810 foot long coil doesn'tlook too good.

 

[1RMD]

Thanks All for your posts.

I can understand how "25362" is leading me to solve the problem, however I cant seem to arrage the equation with consistant units so that only A="x"m^2 is left after canceling etc.
Zekeman, how does your equation differ? 810ft. does not look good. But If I'm reading your solution correctly, Your saying that the gpm of the potable water heated to the desired temp of 120*F is directly proportional to the distance it must travel? So theoretically I could have 1gpm with 81ft. of pipe? I suppose I will have use a storage tank to offset difference for what I need. Thanks Again.

 

[25362]

The heat transfer accounts for 39,000 W. After you get the surface in m² convert to ft² and divide it by 0.275 ft²/ft for the 3/4" tube. It should result in about 70 m length.

 

[zekeman]

Oops, left out a factor of 4 in getting Tau, So the good news is that translates into tau now being 1.5/4 minutes and the required length of coil is now only 200 feet and agrees with 25362 coincidentally, but the methodology appears flawed since you don't know the average temperatures differences of the water and what the average h employed.

IRMD,
That is true since each slug of water requires 2.25 minutes to achieve the final temperature so slowing it down by a factor of ten reduces the distance by the same factor. The problem is you get 1/10 of the output you need. A clearer way of doing this is to convert the time equation into a distance equation,as follows:
Since dT/dt is equal to dT/dx*dx/dt and dx/dt=v, the velocity of the water, the solution is a function of the length and now looks like
T-T0=(Ti-T0)*e^-x/vTau
taking the ln of both sides and rearranging
x/vtau=ln[(T0-Ti)/T0-T)]=ln[(140-55)/(140-120)]=1.5
x=v*tau*1.5=v*1.5/4*1.5=0.562v and since v=360'/min
x=0.562*360= 202 feet


x/vtau=