Heating water flowing through Copper tube

[seanist5890]

Hey guys,

Im designing a system where water flows through a copper tube or coil where it is heated and sprayed out the other end. My question is, how does the length of the tube/coil play a role in this heat transfer problem?

What I would like to do is pass the water through the inlet at a given temperature (likely room temp) and heat it using the copper tubing inside of a handheld device, and spray it out the other end at a temperature around 37 °C. What would be a good equation or set of equations to use? Thanks.

 

[rktpro]

Hey guys,

Im designing a system where water flows through a copper tube or coil where it is heated and sprayed out the other end. My question is, how does the length of the tube/coil play a role in this heat transfer problem?

What I would like to do is pass the water through the inlet at a given temperature (likely room temp) and heat it using the copper tubing inside of a handheld device, and spray it out the other end at a temperature around 37 °C. What would be a good equation or set of equations to use? Thanks.

Because H=i^2rt and r=rho L/A ; more the length, more the resistance and more is the heat generated.
But remember that H is also proportional to i^2
Therefore, you must also aim for amount of current to be more.

 

[Mech_Engineer]

What you're asking for is more complex than you realize, but you basically need to utilize heat transfer to calculate the convective coefficient of the fluid flowing through the tube, and based on this you can find the temperature rise per unit length of the tube.

To a first approximation, if you assume that heat transfer between the fluid and the tube is high (high reynold's number e.g. turbulent, etc.) and the length of the tube is long compared to the thermal entry length, your basic calculation will depend on:

• The mass flow rate of the fluid (kg/s).
• Initial Temperature (*C or K)
• Final Temperature (*C or K)
• Fluid specific heat capacity (J/(kg*K))

Multiplying the mass flow rate by water's heat capacity and the temperature difference will net the first-order required power in watts.

$Power = c_{p}*m_{dot}*\Delta T$

I've attached a PDF with a sample calculation in it.

 

[seanist5890]

Thanks both of you for your contributions.

 

[alphy]

I would first like to commend you on your design idea. Using copper tubing to heat water and spray it out at a specific temperature is a great concept.

When it comes to heat transfer, the length of the tube/coil does play a role in the process. The longer the tube/coil, the more surface area there is for heat transfer to occur. This means that longer tubes/coils will have a higher heat transfer rate compared to shorter ones. However, there are other factors that can also affect heat transfer, such as the flow rate of the water and the temperature difference between the inlet and outlet.

To determine the specific equations to use for your design, it would be helpful to know more about the specifications of your handheld device and the flow rate of the water. Generally, the heat transfer rate can be calculated using the equation Q = U x A x ΔT, where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the surface area, and ΔT is the temperature difference between the water and the surroundings. However, this equation may need to be modified based on the specific conditions of your design.

I would also recommend conducting some experiments or simulations to determine the best length for your copper tube/coil in order to achieve the desired outlet temperature of 37 °C. This will help ensure that your design is efficient and effective.

Overall, your idea shows great potential and with the right equations and experimentation, you can successfully design a handheld device that heats and sprays water at a specific temperature. Good luck with your project!