Heat Transfer Energy Balance 4

[Josh2008]

If I have a container with hot water coming in and mixed water exiting at the same flow rate, what is the energy balance for the system.

Tank
200 gal (0.76 m3)
k = 235 W/mk
surface area conducting heat = 4.98 m2
ambient air temp at 60F

because the water temperature varies coming into the tank, i calculated the heat transferred in instead. Assume constant at 148.4 kW.

8GPM (0.505 kg/s) into the tank
8GPM out of the tank

Tank water initially at 70F.

Please write the energy balance, or help me calculate a formula with the Tavg water in tank varied with time.

Thanks!

 

[Josh2008]

Assume water at 1000kg/m3 and 4200 J/kgK. I understand these are variables with respect to time but assume they stay constant for arguement sake.

 

[gruntguru]

You say the water temp coming into the tank varies, but you don't describe the variation.

Energy flow into the tank is:
dE/dt = [mass flow x Cp x (T1-Tavg)]-[235 x 4.98 x (Tavg-Tamb)]

Tank temp will increase as a result as follows:
dE/dt = 760kg x Cp x dTavg/dt

Assumes perfect instant mixing ie Tavg = Tout

Engineering is the art of creating things you need, from things you can get.

 

[vpl]

I think gruntguru's formula might be a bit simplistic. It doesn't account for the heat added by the mixing. Nor does it address heat loss through the tank walls.

Unfortunately, these type of problems seem easy on the surface and aren't. There have been several threads in this forum about these types of problems that might give you more ideas.

If you really need to know, with any exactitude from a design perspective, you're looking at computer modeling addressing the differing temperatures in, the heat losses out, and the variations in flow rates.

If you're looking for "quick and dirty" go throw a "color chart" thermometer on the outlet pipe -- one of those where the material changes color as the temperature changes.

Patricia Lougheed

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[786392]

Dear Patricia Hello/Good Morning,

I am 100% seconding your thoughts. as sometimes a "shot-cut" practically and in lifetime consequentially proves to be "Longest Cut " eating up lots of valuable time and resources.


Best Regards
Qalander(Chem)

 

[gruntguru]

It doesn't account for the heat added by the mixing. Nor does it address heat loss through the tank walls.

- "Mixing"? are you talking about work done by an agitator?

- The "-[235 x 4.98 x (Tavg-Tamb)]" term accounts for heat loss through the walls.

I may have not been sufficiently clear. The two equations in my post need to be solved simultaneously.

Engineering is the art of creating things you need, from things you can get.

 

[zekeman]

"dE/dt = [mass flow x Cp x (T1-Tavg)]-[235 x 4.98 x (Tavg-Tamb)]

The heat transfer term out of the tank , 235 x 4.98 x (Tavg-Tamb)] is incorrect.
235 is a thermal conductivity factor and should be the convection and radiation to the outside, a much smaller number.

Insofar as the instant mixing assumption, especially since the incoming temperature varies, I , like Patricia, am not persuaded.

This problem can be done assuming instant lateral mixing by
taking a sliver of incoming liquid and following it transfer heat to the outside. Then you can write the ODE rather than the more complicated partial DE. I get
rho*A*cp*dT/dt+hl*(T-Tamb)=0 and the solution is
T=T(0)Exp-(hl/rhoAcp)t
where
t= time to position x
x=vt
T(0)temperature at entrance
l=2*Pi* radius of tank
rho*A=Mass flow/v
v=fluid velocity at entrance * entrance diameter/tank diameter
h= convective plus linearized radiative transfer coefficients
So at any position x and time t the
T(0)=T(t-x/v)
T(x,t)=T(t-x/v)*Exp-(hl/rhoAcpv)x