Open vs. Closed Systems

Model a Closed Loop System

A closed circulating system can easily be modeled with AFT Fathom by locating a Reservoir junction anywhere in the system and setting the pressure to the known pressure at that point. Inflow and outflow pipes will then need to be connected. The Reservoir junction merely acts as a "reference pressure" against which all other pressures in the system are compared. Because there are no other inlets or outlets to the system, then whatever is delivered by the Reservoir junction to the system will also be received from the other pipes, thus achieving a net zero outflow rate for the system.

When using a Reservoir junction as a reference pressure for a closed system, you should select the Balance Energy in Reservoir option, which will allow the reservoir temperature to vary and be calculated in a manner consistent with the overall energy balance.

Balancing Mass

In order to model a closed system, only one pressure junction is used in the model. Typically this would be either a Reservoir or Assigned Pressure. Pressure type junctions are an infinite source of fluid and do not balance flow. How then can one be used to balance flow in a closed system?

To answer this question, it is worth considering how AFT Fathom views a closed system model. AFT Fathom does not directly model closed systems, and in fact does not even realize a closed system is being modeled.

Consider the system shown in Figure 1. This is an open system. Fluid is taken from J1 and delivered to J4 and J5. Because AFT Fathom's solution engine solves for a mass balance in the system, all flow out of J1 must be delivered to J4 and J5. Because the flow is steady-state, no fluid can be stored in the system; what goes in must go out.

A model of an open system with 3 reservoirs, a pump, a branch junction, and 4 pipes.

Figure 1: Open system - Flow out of J1 equals the sum of J4 and J5

Now consider the systems in Figure 2. The first system appears to be closed, while the second appears to be open. If the same boundary condition (i.e., surface elevation and pressure) is used for J1, J11 and J12 in the second system, it will appear to AFT Fathom as an identical system to the first system. The reason is that AFT Fathom takes the first system and applies the J1 reservoir pressure as a boundary condition to pipes P4, P9 and P10. The second system uses three reservoirs to apply boundary conditions to P4, P9 and P10. But if the reservoirs all have the same elevation and pressure, the boundary conditions are the same as J1 in the first system. Thus the same boundary condition is used for P4, P9 and P10 in both models, and they appear identical to AFT Fathom.

But how is the flow balanced at J1 in the first system? Looking at the first system, one sees that to obtain a system balance, whatever flows into P10 must come back through P4 and P9. Because there is overall system balance by the solver, it will give the appearance of a balanced flow at the pressure junction J1. If there is only one boundary (i.e., junction) where flow can enter or leave the pipe system, then no flow will enter or leave because there isn't anywhere for it to go. Thus the net flow rate will be zero at J1 (i.e., it will be balanced). But recognize that AFT Fathom is not applying a mass balance to J1 directly. It is merely the result of an overall system balance.

An example of a closed system with one pump, one reservoir and two general components. A model that shows an example of an open system.

Figure 2: The first system is closed, and the second open. In both systems the flow into P10 is the same as sum of P4 and P9. If the second system has the same conditions at J1, J11 and J12, the two system will appear identical to AFT Fathom.

Balancing Energy

Here the thermal aspects of open and closed systems are discussed.

Consider the system shown in Figure 3 .The flows are balanced at J1, but how can the energy be balanced? For example, assume the user sets a temperature of 100° F at J1. This temperature will be the inlet pipe temperature for all pipes that flow out of J1. In this case, the 100° F will apply only to pipe P10.

The pipes flowing into the reservoir (P4 and P9) will have their own temperatures that are obtained by balancing energy along their individual flow paths. This will include heat exchanger input and heat transfer to or from pipes.

The only way to obtain an overall system energy balance is for the J1 reservoir temperature to adjust to the mixture temperature of all inflowing pipes. This is the function of the "Balance Energy At Junction" feature in the reservoir and assigned pressure junctions. The junction temperature (input as maybe 100° F) is allowed to "float" and find its own equilibrium. Each iteration the floating reservoir temperature is passed into pipe P10 until convergence. When passed into P10, this will affect the return temperatures of P4 and P9. There is a unique "mixture temperature" (or mixture enthalpy) that will yield an energy balance at J1. This will be the temperature/enthalpy from Equation 3 in the Network Implementation topic.

An image showing energy being balanced in a closed system with a pump, two heat exchangers, and a reservoir.

Figure 3: Example of balancing energy in a closed system