Could you help me??

[Ingnataly]

Hi all,
I'm looking for a calculation procedure to find the minimum flow through the coil and the minimum heat loss required to avoid freezing of water in the coil for a closed circuit cooling tower. as a maintance procedure during winter season

inlet water coil temperature of 50ºF, outside temperature of -10 ºF ambient with 45 MPH wind

Many thanks in advance.

 

[IRstuff]

Do a search on this site. There have been many threads on this subject, like: thread378-295195

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[2brains]

This seems fairly simple. The amount of heat lost would need to be dQ= 80cal/gm + 50cal/gm because if I remember water has to lose 80cal at zero C to freeze and lose 1 cal for each gm to lower the temp to zero. Now heat x time or Ht = dQ = kA(T1-T2)/thickness of the pipe and you just figure out how long you want the water to stay in the coil from one end to the other. Its common knowledge that allow water to run it won't freeze. I'm neglecting the wind which could have some effect.

 

[zekeman]

You "worst case" it to find the outside film coefficient (" google surface film coefficient with wind").

Next use the film coefficient as the overall conductance to give a heat loss of

hA*delta T

Next you write the differential equation that describes the process,

rho*a*c*v*dT/dx=h*p*(T0-T)

rho density of water
a cross sectional area
p perimeter of pipe
c specific heat=1
T0=ambient temperature=

d(T-T0)/dx+ hp/rho*a*c*v[T-T0]=0

for convenience

lamda=hp/rho*a*c*v

d(T-T0)/dx+lamda*[T-T0]=0



with a solution

T-T0=(Ti-T0)exp(-x/lamda)


If we take the case of the water just reaching 32 at the outlet the solution is

32--10=(50--10)exp(-L/lambda)
42=60*exp(-L/lambda)
l= length of coil

we now take the ln of both sides

-L/lambda=Ln(42/60)=-.356
lamda=l/.356

now from the definition of lamda which contains v, you can evaluate v.

I think I gave you enough to get a good answer, but I would at least triple the velocity I get for safety reasons.

Or if you don't like differential equations, use

rho*a*c*v*deltaT= Lph(T0-50)

rho*a*c*v*(32-50)=T0-50=-60

get v from this.

conservative but not as accurate as the ODE.

 

[ione]

Go with Zekeman's suggestion using 100 W/(m^2*K) as order of magnitude for the heat transfer coefficient (h). Be aware that considering a constant heat transfer coefficient implies a simplifying assumption as the heat transfer coefficient will vary as flowing water temperature drops