Thermodynamic processes in open systems

In open systems various processes may happen. Fig. 8 shows a black-box open system. The open system has inflows and outflows. Furthermore, energy in the form of heat or work may be transported across the system’s borders. For assessing system performance, we need to create a relation between the flow of mass and the energy streams. To do that, we can set up an energy balance equation connecting the change of the fluid streams’ states with the energy transported across the system border. Therefore, unlike in closed systems where we can assess the state of the system in different points in time, we want to assess the state of the fluid while entering and while leaving the open system at the same time. This assumes the system is operated in steady-state, which means that the operation of the system does not change with time, e.g. due to effects of thermal inertia.

Fig. 8 Black box open system with inflowing and outflowing mass and transfer of energy.

Energy balance equation

For the energy balance equation we separate the transport of energy with mass from the energy transferred via work or heat. Furthermore, since we want to balance systems in steady state, the derivative to time is used, i.e.:

• mass flow \(\dot m\) in kg/s

• heat transfer \(\dot Q\) in W or J/s

• work rate \(\dot W\) in W or J/s

• dissipation energy rate \(\dot W_\text{diss}\) in W or J/s

Heat and work may be transferred to the system and that changes the states of the inflowing and outflowing fluids. We will assume, that neither kinetic nor potential energy of the fluid will change, when the fluid passes the system, or the change is negligibly small in comparison to the change of enthalpy. In Fig. 8 \(o\) denotes outlet streams and \(i\) denotes inlet streams. The general energy balance of the thermodynamic open system in steady-state operation can then be written as follows:

\[\dot Q + \dot W + \dot W_\text{diss} = \sum_o \left( \dot m_o \cdot h_o \right) - \sum_i \left( \dot m_i \cdot h_i \right) \;\forall o \in \text{outlets};\forall i \in \text{inlets}\]

Note

By substracting the energy of the inflowing mass from the energy of the outflowing mass we are implying a sign convention for heat and work:

• A positive sign for heat or work means that energy is transferred into the system.

• A negative sign indicates that energy is transferred out of the system.

• Dissipation energy is always positive \(\dot W_\text{diss}\geq 0\).

Dissipation of energy …

Many thermodynamic components will only have a single fluid stream, therefore simplifying the equation:

\[\dot Q + \dot W + \dot W_\text{diss}= \dot m \cdot \left( h_\text{out} - h_\text{in}\right)\]

Simple processes

One of the main challenges now lies in the identification of heat and work transfer given a change of state. The following section will outline, how this can be achieved:

If we observe a change of state in a mass flow between the inlet and the outlet of a system, we cannot make any conclusions back to the process taking place in the system. For example, we might see enthalpy not changing between inlet and outlet. Without further knowledge on the system’s attributes, we cannot know what processes happened. Work could be transferred into the system and the same amount of heat could be withdrawn. However, if we knew that we have an adiabatic system, we could conclude that no work is transferred as well.

Therefore, we have a closer look at simple processes typical for open systems, which may help us create a direct connection between a change of state and process energy. Work rate can be calculated:

\[\frac{\dot W}{\dot m} = w = \int_1^2 v \text{d} p\]

Heat rate is

\[\frac{\dot Q}{\dot m} = q = \int_1^2 T \text{d}s\]

Dissipation of energy is

\[\frac{\dot W_\text{diss}}{\dot m} = w_\text{diss} = \int_1^2 T \text{d}s\]

Caution

Without process knowledge we cannot distinguish between heat transfer and dissipation of energy, when work rate is 0.

Isobaric

A process is called isobaric, if pressure remains unchanged, therefore:

\[v \text{d}p = 0\]

Since pressure does not change, that means the work rate of such a process must be zero. Typical examples for such processes are heat exchangers or junctions.

Isentropic: adiabatic and reversible

If a fluid is undergoing a process, in which does not transfer heat to or from the fluid and which is reversible (thermodynamically ideal), the entropy of the fluid cannot change, since entropy only changes with heat transfer or can be generated through irreversibility.

\[s_2 - s_1 = 0\]

Such a process cannot happen in reality, but it may serve as a benchmark for real processes, i.e. in compression or expansion machine, since this is the best possible process possible.

Polytropic

• heat transfer with pressure loss

• adiabatic and irreversible