Energy balance steady-flow systems

For a steady-flow system with one inlet and one exit, the rate of mass flow into the control volume must be equal to the rate of mass flow out of it. Tliat is, wi n = / oui When the changes in kinetic and potential energies are negligible, which is usually the case, and there is no work interaction, the energy balance for such a steady-flow system reduces to (Fig. 1-17)... 

A surface contains no volume or mass, and thus no energy. Thereore, a surface can be viewed as a fictitious system whose energy content remains constant during a process (just like a steady-slate or steady-flow system). Then the energy balance for a surface can be expressed as... 

In the chemical processing industries, steady-flow systems are common, so the accumulation terms on the left sides of (12.3.1) and (12.3.2) are normally zero, and the rate forms of the balance equations can be used. Then the material and energy balances can be expressed as... 

EXAMPLE 2.7-4. Mechanical-Energy Balance on Pumping System Water with a density of 998 kg/m is flowing at a steady mass flow rate through a uniform-diameter pipe. The entrance pressure of the fluid is 68.9 kN/m abs in the pipe, which connects to a pump which actually supplies 155.4 J/kg of fluid flowing in the pipe. The exit pipe from the pump is the same diameter as the inlet pipe. The exit section of the pipe is 3.05 m higher than the entrance, and the exit pressure is 137.8 kN/m abs. The Reynolds number in the pipe is above 4000 in the system. Calculate the frictional loss the pipe system. 

There are a variety of limiting forms of equation 8.0.3 that are appropriate for use with different types of reactors and different modes of operation. For stirred tanks the reactor contents are uniform in temperature and composition throughout, and it is possible to write the energy balance over the entire reactor. In the case of a batch reactor, only the first two terms need be retained. For continuous flow systems operating at steady state, the accumulation term disappears. For adiabatic operation in the absence of shaft work effects the energy transfer term is omitted. For the case of semibatch operation it may be necessary to retain all four terms. For tubular flow reactors neither the composition nor the temperature need be independent of position, and the energy balance must be written on a differential element of reactor volume. The resultant differential equation must then be solved in conjunction with the differential equation describing the material balance on the differential element. 

Here, h = u + P/p is the enthalpy per unit mass of fluid. Note that the inlet and exit streams include enthalpy (i.e., both internal energy, u, and flow work, P/p), whereas the system energy includes only the internal energy but no P/p flow work (for obvious reasons). If there are only one inlet stream and one exit stream (m, =m0 = m) and the system is at steady state, the energy balance becomes... 

In two-phase flow, most investigations are carried out in one dimension in the steady state with constant flow rates. The system may or may not be isothermal, and heat and mass may be transferred either from liquid to gas, or vice versa. The assumption is commonly made that the pressure is constant at a given cross section of the pipe. Momentum and energy balances can then be written separately for each phase, and with the constraint that the static pressure drop, dP, is identical for both phases over the same increment of flow length dz, these balances can be added to give over-all expressions. However, it will be seen that the resulting over-all balances do not have the simple relationships to each other that exist for single-phase flow. 

Each term in the preceding equations has units of energy/time. Note the signs on each term indicating that heat is removed or added to the reactor. We preserve the minus sign on A Hji because we are more interested in exothermic reactions for which A Hr < 0. The student can recognize each term on the right side from the steady-state enthalpy balance we derived in the previous section from the thermodynamics of a steady-state flow system. 

Turn now to the steady plug-flow energy equation. We neglect kinetic and gravitational potential energy, so we state the system energy balance as... 

The general form of the mechanical energy balance can be derived starting with the open-system balance and a second equation expressing the law of conservation of momentum, a derivation beyond the scope of this book. This section presents a simplified form for a single incompressible liquid flowing into and out of a process system at steady state. 

The energy required to pump a liquid food through a pipe line can be calculated from the mechanical eneigy balance (MEB) equation. The MEB equation can be used to analyze pipe flow systems. For the steady-state flow of an incompressible fluid, the MEB can be written as follows (Brodkey, 1%7) ... 

We will now consider flow systems that are operated at steady state. The steady-state energy balance is obtained by setting dE yJdt) equal to zero in Equation (S-9) in order to yield... 

Flow processes for wliich the accumulationtermof Eq. (2.28), d mU v/dt, is zero are said to occur at steady state. As discussed with respect to tire mass balance, tliis means tliat tire mass of the system within the control volume is constant it also means that no changes occur with time in tire properties of tire fluid witliin the control volume nor at its entrances and exits. No expansion of the control volume is possible under these circumstances. The only work of the process is sliaft work, and the general energy balance, Eq. (2.28), becomes ... 

Summary of Equations of Balance for Open Systems Only the most general equations of mass, energy, and entropy balance appear in the preceding sections. In each case important applications require less general versions. The most common restrictedTcase is for steady flow processes, wherein the mass and thermodynamic properties of the fluid within the control volume are not time-dependent. A further simplification results when there is but one entrance and one exit to the control volume. In this event, m is the same for both streams, and the equations may be divided through by this rate to put them on the basis of a unit amount of fluid flowing through the control volume. Summarized in Table 4-3 are the basic equations of balance and their important restricted forms. 

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... 

For a steady-state flow system, again neglecting changes in the potential and kinetic energies, the energy balance per unit time is given by Equation 1.10. 

The first step in solving any energy flow problem is to choose the thermodynamic system the second step is to write the balance equations for.the system. Here we take the turbine and its contents to be the system. The mass and energy balance equations for this adiabatic, steady-state system are... 

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