Heat Transfer calculation for Tank and piping system

  • #1

Main Question or Discussion Point

Having difficulty remembering how to model a simple heat transfer equation for a liquid tank/piping system and wondering if anyone can provide some quick help.

I have a 1000L tank of water that is heated by a 1kW heater (target temp of 43°C). The piping system (outside of the tank), holds 100L of water at any given time and circulates the fluid from and back into the tank at 30 liters/minute. The temperature of the liquid on return is 36°C (so a delta T of about 7°C attributed to heat loss of the piping).

So basically, I have 900 kg reservoir of water at 43°C that is being heated by a 1kW heater, while also being subject to cooling by a return line of the fluid at a mass flow rate of 30 kg/min at a temperature of 36°C. Assuming the Qout of the tank is zero as the tank is insulated and sealed.

Looking for help to get a heat transfer model on this system. Looking to calculate the time it will take to heat and run the tank reservoir at a temperature of 53° C (so, 10°C above the current temp)


Thanks for all and any help!

Ethan
 

Answers and Replies

  • #2
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OK. You start out by assuming that the water in the tank is well-mixed so that its temperature is uniform. If T is the temperature in the tank, what is the temperature of the exit stream?
 
  • #3
The exit stream temperature is the same temperature as inside the tank. Also assuming that the exit stream vs return stream delta T is constant (so, 7 degrees cooler upon return).
 
  • #4
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OK. You're going to do a transient heat balance on the tank. The internal energy in the tank is U=MC(T-T0), where M is the mass of water in the tank, C is the heat capacity, and T0 is some arbitrary reference temperature. How is M related to the volume and the density of the water? Let Q represent the heating rate of the heater. Let w represent the rate that water exits the tank and enters the tank. Let's see if you can write down a transient heat balance on the tank.

Chet
 
  • #5
Ahhhh, thank you so much. Finally jogged my memory and got it figured out!
 
  • #6
CWatters
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Just for info.. I made the loss in the pipe work far more than 1kW...

P (in Watts) = mCΔT

where
m is the mass flow rate in kg/S (30L/min = 0.5Kg/S)
C = specific heat capacity (4181J/Kg/C)
ΔT = 7 degrees

P = 0.5 * 4181 * 7 = 14KW

So tank won't be heating up.
 

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