What is the most accurate way to calculate heating in a waterjet catcher tank?

[Roboto]

Here is a fun question for you heat transfer experts.
I have a high pressure waterjet firing into a open tank of water. The hydraulic power of the jet entering the tank is known, along with the flow rate of the jet entering into the tank. All of the energy of the jet is dissipated into the large volume of water in the tank. The volumetric flow rate of water leaving the tank is equal to the volumetric flow rate of the water entering the tank. Tank volume of water remains constant.
Typically the heating input is the mass flow rate of the water times its specific heat and temperature is the input. But with a high pressure waterjets, there is the very high kinetic energy aspect to the incoming water.

On the one hand one can just assume the all of the hydraulic power is converted into heat and that is the sole input. But we are also adding water to the tank that is mixing with the water in the tank and a then leaving the tank of water. So then, does one add to the input to the tank the mass flow rate of the water and its specific heat and incoming temperature with the heat of the hydraulic power? Or does one use the water heating component of the jet, and then using the difference between the water fraction and the total hydraulic power entering the tank as additional energy to be added to the tank with the water component.

The puzzling point is that the input is a single water input that has a single mass flow rate, but also has an extremely high kinetic energy. How one handles this yields very different heating rates of the tank of water. I am trying to avoid double counting something or miss accounting for something.

Intuition says to use the hydraulic power plus the heating from the water's mass flow rate and temperature. but I keep wrestling with myself thinking that I am double counting energy sources.

 

[Chestermiller]

Here is a fun question for you heat transfer experts.
I have a high pressure waterjet firing into a open tank of water. The hydraulic power of the jet entering the tank is known, along with the flow rate of the jet entering into the tank. All of the energy of the jet is dissipated into the large volume of water in the tank. The volumetric flow rate of water leaving the tank is equal to the volumetric flow rate of the water entering the tank. Tank volume of water remains constant.
Typically the heating input is the mass flow rate of the water times its specific heat and temperature is the input. But with a high pressure waterjets, there is the very high kinetic energy aspect to the incoming water.

On the one hand one can just assume the all of the hydraulic power is converted into heat and that is the sole input. But we are also adding water to the tank that is mixing with the water in the tank and a then leaving the tank of water. So then, does one add to the input to the tank the mass flow rate of the water and its specific heat and incoming temperature with the heat of the hydraulic power? Or does one use the water heating component of the jet, and then using the difference between the water fraction and the total hydraulic power entering the tank as additional energy to be added to the tank with the water component.

The puzzling point is that the input is a single water input that has a single mass flow rate, but also has an extremely high kinetic energy. How one handles this yields very different heating rates of the tank of water. I am trying to avoid double counting something or miss accounting for something.

Intuition says to use the hydraulic power plus the heating from the water's mass flow rate and temperature. but I keep wrestling with myself thinking that I am double counting energy sources.

Let’s see your balance equations for the two options.

 

[russ_watters]

I'm having trouble understanding your description of both the setup and your approach. A diagram with the energies listed would help.

To me though it sounds like you have two inputs and one output and it's a mixing problem. The two inputs are:
1. The water jet, with thermal and kinetic energy.
2. Make-up water(?) with thermal energy.

Just add them together and the output is all thermal.

 

[Chestermiller]

@Roboto Are you familiar with the open system (control volume) version of the 1st law of thermodynamics for a system operating at steady state? If so, it is perfect for solving this problem.

 

[Roboto]

Ultimately I am working on the change in temperature within the tank as the high velocity waterjet is dissipated in the tank.

m*Cp*dT/dt = Qin - mdot*Cp*Tout - Q_sides - Q_surface
where
m = mass of the water in the tank
Cp = specific heat of the water
dT/dt = rate of Temperature change in the tank
mdot = mass flow rate of the water leaving the tank. It is the same as the mass flow rate of water entering the tank. Mass of the water in the tank remains constant.
Q_sides = is the heat loss through free convection from the sides of the tank. Air speed is zero since the tank is in a building. Radiative heat loss from the side walls are negligible. This changes based on the temperature of the sides of the tank. It is not insulated, 1/8 inch thick steel. Temperature of the room is assumed to remain relatively constant. The coefficient of convection is iterated from tables.
Q_surface = is the heat loss from the open surface of the tank of water. It consists of convection, evaporation, and radiation. Again these values are iterated from tables.

The question comes into the appropriate handling of Qin.
The mass flow rate of the water entering into the tank of water is known. The actual temperature of the jet is unknown. There is no way to measure the actual temperature of the jet. We know the hydraulic power of the jet entering into the tank. The velocity of the jet is almost 3 times the speed of sound. The purpose of the tank of water is to adsorb the high kinetic energy of the jet. All that goes into heating of the water into the tank.

Now do we just take the hydraulic power of the jet and convert that into heat that is the sole input.

But then there is a mass flow rate that is entering the tank. The temperature of the jet is traditionally accepted to be a few degrees warmer than the supply water to the high pressure pump. One can take that temperature and multiply it by the specific heat and mass flow rate. The heat value is around 40 to 45 % of the hydraulic power. So from a conservation of mass principle I need to account for the mass flow rate of the water entering into the tank.

But that mass flow rate has an extremely high kinetic energy component that needs to be accounted for.

One approach would be to ignore the mass flow rate entering into the system, and just assume the volume remains constant, and that the heat input is just the Hydraulic power entering the system (converted to the appropriate heat units)
Or, assume a water inlet temperature, multiply it by the incoming mass flow rate and specific heat, and then add the heat from the hydraulic input power
or, assume a water inlet temperature, multiply it by the incoming mass flow rate and specific heat, and take a fraction of the hydraulic input power as additional heating from the input water, and the fraction is such that the total heat entering the tank is equal the heating from the input water flow rate and fraction of the heating from the hydraulic power is equal to the total heating.

My heat transfer books don't get into this type of question. Everything is broken out into convenient numbers. In my case, I just have a single input, but that input has two components, a temperature, a mass flow rate, and an extremely high kinetic energy.

These tanks get hot, very hot after dumping 100 hp into them all day long. I am trying to get a reasonable model for this as opposed to the brute force method that just throws a massive heat exchanger and see what happens.

 

[jrmichler]

You can simplify the problem considerably by assuming that 100% of the power going into the motor shows up as heat in the water. Then just decide what you want the steady state water temperature to be, and size your heat exchanger and/or water flow accordingly.

The exact mechanism of heating the water is not important. Some of the heat is from pump efficiency losses, some from line friction, most from dissipating the kinetic energy of the water jet. Since the individual sources will add up to the total power output from the motor to the pump, and since 100 hp induction motors are 93 to 95% efficient, you can just use the total power into the motor and know that your final answer will be about 10% conservative.

And 10% conservative is far better than throwing in a massive heat exchanger to find what happens.

 

[Chestermiller]

You kinda have the right idea, but what you've done so far needs a little improvement. If the mass in the tank is truly constant, then the overall mass balance should read: $$\dot{m}_{in}=\dot{m}_{out}+\dot{m}_{evap}$$where $\dot{m}_{out}$ is the liquid mass flow rate out of the tank and $\dot{m}_{evap}$ is the rate of mass exiting by evaporation.

If we assume that the fluid in the tank is well-mixed, then we can apply the open system (control volume) version of the first law of thermodynamics to obtain the following energy balance equation:
$$mC_p\frac{dT}{dt}=\dot{m}_{in}\left[C_p(T_{in}-T)+\frac{v^2}{2}\right]-\dot{m}_{evap}\lambda(T)-Q_{sides}-Q_{surface}$$where $\lambda(T)$ is the heat of vaporization at temperature T, and $Q_{surface}$ includes only the convection and radiation (but not evaporation). The term $\frac{v^2}{2}$ represents the kinetic energy per unit mass of the entering stream, which, when multiplied by $\dot{m}_{in}$ gives the rate of kinetic energy entering the tank; this kinetic energy is dissipated by viscous friction, and becomes part of the enthalpy of the exit stream plus the increase in enthalpy of the tank contents. Please be aware that, in deriving this energy balance equation from the 1st law, I have made use of the mass balance equation to bring about some simplification.

 

[Roboto]

Thanks, I like the way you got the velocity component worked into this. Awesome, thanks.