Energy balances adiabatic operations

The specific enthalpies ia equation 9 can be determined as described earUer, provided the temperatures of the product streams are known. Evaporative cooling crystallizers operate at reduced pressure and may be considered adiabatic (Q = 0). As with of many problems involving equiUbrium relationships and mass and energy balances, trial-and-error computations are often iavolved ia solving equations 7 through 9. 

Estimation of operating data (usually consisting of a mass and energy in which the energy balance decides whether the absorption balance can be considered isothermal or adiabatic)... 

The steady state energy balance for adiabatic operation is determined as follows ... 

The program ENERGY 1 (see Chapter 3) was used to make the balance over on the oxidiser. Adiabatic operation was assumed (negligible heat losses) and the outlet temperature found by making a series of balances with different outlet temperatures to find the value that reduced the computed cooling required to zero (adiabatic operation). The data used in the program are listed below ... 

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. 

For isothermal and adiabatic modes of operation the energy balance equations developed above will simplify so that the design calculations are not nearly as tedious as they are for the other modes of operation. In the case of adiabatic operation the heat transfer rate is zero, so equation 10.2.10 becomes... 

The energy balance equation for adiabatic operation becomes... 

An energy balance on the PFR operating at steady state is given by equation 10.4.6. For adiabatic operation this equation becomes... 

The computational effort required to carry out the design analysis is determined mainly by the magnitude and spatial distribution of the temperature variations that are taken into account. The maximum temperature difference between the inlet and outlet of the reactor occurs when the reactor operates adiabatically. In this case, heat transfer to the reactor wall is neglected so there is no temperature variation in the radial direction. However, the temperature does vary in the axial direction, so the material and energy balance equations are coupled through the dependence of the reaction rate on temperature. If the reactor is well insulated, and/or of large... 

In order to assess the design of both the reactor and the heat exchanger required to control T, it is necessary to use the material balance and the energy balance, together with information on rate of reaction and rate of heat transfer, since there is an interaction between T and /A. In this section, we consider two cases of nonisothermal operation adiabatic (Q = 0) and nonadiabatic (Q = 0). 

From the energy balance (equation 14.3-10) for adiabatic operation [Q = UAC(TC T) =... 

For an endothermic reaction, whether the operation is adiabatic or nonadiabatic, there is no possibility of multiple stationary-states because of the negative slope of the /A versus T relation in the energy balance (see equation 14.3-11). This is illustrated schematically in Figure 14.8. 

To relate fA and T, we require the energy balance (15.2-9). From this equation, for adiabatic operation (the heat transfer term on right side vanishes),... 

Case (2) CSTR T0 for Vmin with adiabatic operation and specified fA, FAo and q The result given by 18.4-4 requires only that the operating temperature within the CSTR be Topt, and implies nothing about the mode of operation to obtain this, that is, nothing about the feed temperature (TJ, or heat transfer either within the reactor or upstream of it. If the CSTR is operated adiabatically without internal heat transfer, T0 must be adjusted accordingly to a value obtained from the energy balance, which, in its simplest integrated form, is, from equation 14.3-10,... 

Substituting the numerical values for To, yAo> C p AHfiQQ in the energy balance in the form applicable to adiabatic operation, i.e. eqn. (74), gives... 

Figure 9.7 Graphical representation of energy balance equation for adiabatic operation. These are adiabatic operating lines.

Modeling of the packed bed catalytic reactor under adiabatic operation simply involves a slight modification of the boundary conditions for the catalyst and gas energy balances. A zero flux condition is needed at the outer reactor wall and can be obtained by setting the outer wall heat transfer coefficients /iws and /iwg (or corresponding Biot numbers) equal to zero. Simulations under adiabatic operation do not significantly alter any of the conclusions presented throughout this work and are often used for verification... 

The last term on the left-hand side of eq. (3.301) corresponds to the heat transfer to the external fixed-bed wall. The overall heat transfer resistance is the sum of the internal, external, and wall resistances. In an adiabatic operation, the overall heat transfer coefficient is zero so the corresponding term in the energy balance expression drops out, while in an isothermal operation this coefficient is infinite, so that 7 f 7 s 7W. 

A gas-liquid contact operation is illustrated in Figure 3.8. Gas is contacted with a liquid from a spray, resulting in both diffusion and heat transfer between the gas and liquid. The gas exits the system at conditions of humidity and temperature quite different from the entrance conditions. Assume the operation to be adiabatic. Perform a material and energy balance for the system. 

Consider an exothermic irreversible reaction with first order kinetics in an adiabatic continuous flow stirred tank reactor. It is possible to determine the stable operating temperatures and conversions by combining both the mass and energy balance equations. For the mass balance equation at constant density and steady state condition,... 

The appropriate energy balance can be written W = AH — Q. Since Q is negative (heat transfer is out of the system), the work of non-adiabatic compression is greater than for adiabatic compression. Note that in order to have the same change in state of the air, i.e., the same AH, the irreversibilities of operation would have to be quite different for the two cases. 

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