Open systems energy balances

Equation 7.4-12 could be used for all steady-slate open system energy balance problems. As a rule, however, the term Oj + PjVj is combined and written as Hj, the variable previously defined as the specific enthalpy. In terms of this variable. Equation 7.4-12 becomes... 

Consider such a system, letting m be the mass flow rate and V the specific volume of the liquid. If V is replaced by 1/ p, where p is the liquid density, then the open-system energy balance (Equation 7.4-12) may be written... 

Define a system and simplify the open-system energy balance (Equation 7.4-15) for each of the following cases. State when possible whether nonzero heal and shaft work terms are positive or negative. The solution of part (a) is given as an illustration. 

Steam at 260 C and 7.00 bar absolute is expanded through a nozzle to 200 C and 4.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. Tile specific enthalpy of steam is 2974 J/kg at 260°C and 7 bar and 2860 kJ/kg at 200 C and 4 bar. Use the open-system energy balance to calculate the exit steam velocity. 

The open-system energy balance with Wj, A )t, and A p set equal to zero is... 

Remember that we are dealing with a continuous process and hence an open system. [The reason we use n mol) and not (mol/s) is that we took 100 mol CH4 as a basis of calculation.] With A p. and neglected, the open system energy balance yields... 

The open system energy balance neglecting kinetic and potential energy changes and shaft work and selling = 0 for this adiabatic reactor is... 

In this equation, is the extent of reaction (determined from Equation 9.1-3) n, and H,- are respectively the molar flow rate and specific enthalpy of a process species in an inlet or outlet stream and the summations are taken over all species in all of their inlet and outlet stales. Once calculated, AH is substituted in the open-system energy balance, which is solved for Q or whichever other variable is unknown. 

Let us now consider an open system with chemical reaction to illustrate how the previously discussed concepts are incorporated. Without loss of generality, let us consider a steady-state flow process with a reactor having a single inlet and outlet stream. The open-system energy balance then becomes... 

Notice that in the closed-system analysis the surroundings are doing work on the system (the mass element) at the inlet to the compressor, while the system is doing work on its surroundings at the outlet pipe. Each of the.se tenns is a / Pr/V-type work term. For the open. system this work term has been included in the energy balance as a, P V A/V/ term, so that it is the enthalpy, rather than the internal energy, of the flow streams that appears in the equation. The e.xplicit J P dV term that does appear in the open-system energy balance represents only the work done if the system boundaries deform for the choice of the compressor and its contents as the system here this term is zero unless the compressor (the boundary of our system) explodes. B... 

The open-system energy balance (Chap. 4) between any two points R and 1 in such a duct, for steady flow without heat transfer or turbines or compressors, is... 

Explain why it is convenient to use the thermodynamic property enthalpy for (1) streams flowing into and out of open systems and (2) closed systems at constant pressure. Describe the role of flow work and shaft work in open-system energy balances. 

Open-System Energy Balances on Process Equipment 95... 

OPEN-SYSTEM ENERGY BALANCES ON PROCESS EQUIPMENT... 

SOLUTION First, the process is sketched in Figure E5.4A. To find the heat in, we will apply the first law (i.e., do an energy balance). Assuming steady-state, the open-system energy balance with one stream in and one stream out can be written ... 

First of all, we will touch a widely believed misunderstanding about impossibility of using the second law of thermodynamics in the analysis of open systems. Surely, the conclusion on inevitable degradation of isolated systems that follows from the second law of thermodynamics cannot be applied to open systems. And particularly unreasonable is the supposition about thermal death of the Universe that is based on the opinion of its isolation. The entropy production caused by irreversible energy dissipation is, however, positive in any system. Here we have a complete analogy with the first law of thermodynamics. Energy is fully conserved only in the isolated systems. For the open systems the balance equalities include exchange components which can lead to the entropy reduction of these systems at its increase due to internal processes as well. 

The interface itself has negligible mass compared to the masses of the phases, and during processes, states of the interface may be undefined or undefinable. We will treat the interface as an open system and interpret each phase as a "port" for the other phase that is, the open-system energy and entropy balances from 2.4 will apply. In what follows, we first derive the combined first and second laws ( 7.2.1). Then we find limits on the directions ( 7.2.2) and magnitudes ( 7.2.3) of mass and energy transfers between phases a and p. 

For an open system with energy exchange across its boundaries, as shown in Fig. 1.20, the energy balance can be written as... 

The energy balance and individual components are illustrated in Figure 3.1. The energy balance shown in the figure is for an open flow system. For a nonflow (or closed) system, the energy balance would appear as in Figure 3.2. 

Example 1 Application of the total mechanical-energy balance to noncom-pressible-flow systems. Water at 61°F is pumped from a large reservoir into the top of an overhead tank using standard 2-in.-diameter steel pipe (ID = 2.067 in.). The reservoir and the overhead tank are open to the atmosphere, and the difference in vertical elevation between the water surface in the reservoir and the discharge point at the top of the overhead tank is 70 ft. The length of the pipeline... 

We will take a closer look at one of the simplest systems conceivable, a constant-volume and -density, cooled CSTR with a first-order, irreversible reaction A - B. While this model is quite simple it still contains most of the relevant issues surrounding an open-loop, nonlinear reactor. Referring to Fig. 4.5, this system can be described by one component balance and one energy balance ... 

Thermodynamics is the area of science that relates to the interplay of heat and other forms of energy. At the molecular level, this science describes the balance between two generally opposing thermodynamic forces the natural tendency for mechanical systems to move toward lower energies and the equally natural tendency for thermal Brownian motion to perturb this mechanical order. For open systems at constant pressure, this balance is expressed by the classic Gibbs free energy change (AG) ... 

Define the terras closed process system, open process system, isothermal process, and adiabatic process. Write the first law of thermodynamics (the energy balance equation) for a closed process system and state the conditions under which each of the five terms in the balance can be neglected. Given a description of a closed process system, simplify the energy balance and solve it for whichever term is not specified in the process description. 

Given a description of any nonreactive process for which tabulated specific internal energies or specific enthalpies are available at all input and output states for all process species, (a) draw and completely label a flowchart, including Q and W (or Q and for an open system) if their values are either specified or called for in a problem statement (b) perform a degree-of-freedom analysis and (c) write the necessary equations (including the appropriately simplified energy balance) to determine all requested variables. 

Starting with the open system balance equation, derive the steady-state mechanical energy balance equation (Equation 7.7-2) for an incompressible fluid and simplify the equation further to derive the Bernoulli equation. List all the assumptions made in the derivation of the latter equation. 

Given fluid conditions (pressure, flow rate, velocity, elevation) at the inlet and outlet of an open system and values of friction loss and shaft work within the system, substitute known quantities into the mechanical energy balance (or the Bernoulli equation if friction loss and shaft work can be neglected) and solve the equation for whichever variable is unknown. 

A system is termed open or closed according to whether or not mass crosses the system boundary during the period of time covered by the energy balance. A batch process system is, by definition, closed, and semibatch and continuous systems are open. 

ENERGY BALANCES ON OPEN SYSTEMS AT STEADY STATE... 

An open process system by definition has mass crossing its boundaries as the process occurs. Work must be done on such a system to push mass in, and work is done on the surroundings by mass that emerges. Both work terms must be included in the energy balance. 

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