# Heat Transfer calculation for Tank and piping system

## 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

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Chestermiller
Mentor
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?

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).

Chestermiller
Mentor
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

Ahhhh, thank you so much. Finally jogged my memory and got it figured out!

CWatters
Homework Helper
Gold Member
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.