Zero degrees of freedom

Figure 1.1 A fly in a box is essentially confined to a point. It Fias zero degrees of freedom and lives its (depressing) life in a 0-D world. (Drawing by Brian Mansfield.)...

Yes, I understand. Sally s hand darts out at the fly and captures it. You never knew she could move so fast. This fly has three different directions it can travel in the room. Now that it is in my hand, it has zero degrees of freedom. I m holding my hand very still. The fly can t move. It s just stuck at a point in space. It would be the same if I placed it in a tiny box. Now, if I stick the fly in a tube where it can only move back and forth in one direction, then the fly has one degree of freedom (Fig. 1.1 and 1.2). 

Remember, we call the point zero-dimensional because there are zero degrees of freedom. If you lived in such a world, you could not move. Similarly, a 1-D line cuts a 2-D plane into two pieces. ... 

In Figure 14.20c, the pressure is 124 MPa. At this pressure the melting curves for water and for acetonitrile intersect the (liquid + liquid) curve at the same temperature. Thus, both ice and solid acetonitrile are in equilibrium with the two liquid phases. The temperature where this occurs is a quadruple point, with four phases in equilibrium. This quadruple point for a binary system, like the triple point for a pure substance, is invariant with zero degrees of freedom. That is, it occurs only at a specific pressure and temperature, and the compositions of all four phases in equilibrium are fixed. 

Consider the freezing of a ternary eutectic. The pressure is constant. The liquid simultaneously freezes to three solid phases, so there are four phases present during the freezing. One student applies the phase rule and concludes that there are zero degrees of freedom. Another student says that this is wrong because the amounts of the phases are not constant. Who is right Discuss briefly. 

This set of five equations also introduces no new unknown variables. The result is that we have 19 equations and 19 unknowns giving zero degrees of freedom. 

These two examples have illustrated some of the key steps in solving mass balance problems. In particular, the number of mass balance equations that are available for ensuring you have zero degrees of freedom and the use of the system box to limit yourself to the streams that are of interest. These steps are also key points for the analysis of energy balances in processes, the topic of Section 1.3 in this chapter. 

The key to doing process analysis is the identification of the equations that may be used to achieve zero degrees of freedom. These equations will come from a number of sources, including the balance equations themselves (Equations (1) and (19)), process specifications (such as the purity of output streams and the reflux ratio), physical relations (such as the definition of enthalpy for liquid and vapour streams) and other constraints imposed by the problem. Once a full set of equations has been developed, the equations can be solved, usually with little difficulty, and the desired results obtained. 

Voss, D. T. (1988). Generalized modulus-ratio tests for analysis of factorial designs with zero degrees of freedom for error. Communications in Statistics Theory and Methods, 17, 3345-3359. 

As before, prime all the known variables in the equations listed in Table 3.2.1. The table shows that there are twenty-one unknowns and equations, resulting in zero degrees of freedom Thus, we have completely defined the problem. 

The first step in the analysis is to determine if zero degrees of freedom exist in any process unit. In this case, the analysis will be simplified because of the reduction in the number of equations requiring simultaneous solution. After analyzing each process unit, we then combine the equations to determine if the process contains zero degrees of freedom. When analyzing each unit separately, we will repeat some variables and equations. For example, in line 3, the composition and flow rate variables, and the mole fraction summation, are the same for the mixer exit stream and the reactor feed stream. Later, when we combine the various processing units to determine the process degrees of freedom, we will take the duplication of variables and equations into account. 

At this point in the analysis we do not know if the variables are over-specified or under-specified. Table 3.5.3 gives the degrees of freedom for each process unit. As usual prime the specified variables. Except for the splitter, the analysis is straight forward. Since ere is no composition change across the splitter, as stated by Equations 3.5.30 to 3.5.39, only the total mole balance is an independent equation. Also, only the sum of the mole fractions for one of the three streams is an independent equation. Table 3.5.3 shows that no process unit contains zero degrees of freedom. 

X-ray study of the ion-exchanged solid shows that partial solid solution occurs, the composition range increasing with decreasing crystallinity. The pH curve resolved for amorphous ZrP, however, exhibits a positive slope to show that the zero degrees of freedom restriction no longer applies to this system. 

The degree-of freedom analysis tells us that there are five unknowns and that we have five equations to solve for them [three mole balances, the density relationship between V2 (= 225 Uh) and hj, and the fractional condensation], hence zero degrees of freedom. Hie problem is therefore solvable in principle. We may now lay out the solution—still before proceeding to any algebraic or numerical calculations—by writing out the equations in an efficient solution order (equations involving only one unknown first, then simultaneous pairs of equations, etc.) and circling the variables for which we would solve each equation or set of simultaneous equations. In this... 

If the number of unknowns equals the number of equations relating them (i.e., if the system has zero degrees of freedom), write the equations in an efficient order (minh mizing simultaneous equations) and circle the variables for which you will solve (as in Example 4.3-4). Start with equations that only involve one unknown variable, then pairs of simultaneous equations containing two unknown variables, and so on. Do no algebra or arithmetic in this step. 

The procedure for material balance calculations on multiple-unit processes is basically the same as that outlined in Section 4.3. The difference is that with multiple-unit processes you may have to isolate and write balances on several subsystems of the process to obtain enough equations to determine all unknown stream variables. When analyzing multiple-unit processes, carry out degree-of-freedom analyses on the overall process and on each subsystem, taking into account only the streams that intersect the boundary of the system under consideration. Do not begin to write and solve equations for a subsystem until you have verified that it has zero degrees of freedom. 

We will not go through the detailed solution but will simply summarize. The degree-of-freedom analysis leads to the results that the overall system has one degree of freedom, the evaporator has zero, and the crystsillizer-filter has one. (Verify these statements,) The strategy is therefore to begin with the evaporator and solve the balance equations for mi and m2. Once m2 is known, the crystallizer has zero degrees of freedom and its three equations may be solved for m3, rhi, and ms. The rate of production of crystals is... 

Overall system. 7 unknowns nQ,XQc,n >n4,np,X5c> 5H) + 1 reaction - 5 independent balances (C02,H2,I,CH30H, H2O) 3 degrees of freedom. Since we do not have enough equations to solve for the number of unknowns in the overall system, we check subsystems to see if one can be found with zero degrees of freedom. 

Draw and completely label a process flowchart. Include in the labeling the molar flow rates and SO2 mole fractions of the gas streams and the mass flow rates and SO2 mass fractions of the liquid streams. Show that the scrubber has zero degrees of freedom. 

Whether you do process analysis manually or with a spreadsheet or simulation program, you can only determine all unknown process variables associated with a process if the process has zero degrees of freedom. It is therefore always a good idea to perform a degree-of-freedom analysis before attempting to solve the system equations. 

After counting all the equations and variables in Tables 3.4.1 and 3.4.2, we find that we now have zero degrees of freedom. Thus, we have defined the problem, and we can now outline the solution procedure. The twenty-two equations are decoupled, i.e., it is not necessary to solve all them simultaneously. By inspection we find that we can solve the mole balance equations independently of the energy balance. This frequently occurs, usually when the temperatures in some of the lines are known. Furthermore, in this case, we do require an iterative calculation procedure. We again obtained a solution procedure by inspection, which is given in Table 3.4.3. 

The parameters required to define the operation of an absorber are discussed in relationship to a single stage. A single stage with a feed of fixed rate, composition, and thermal conditions has two degrees of freedom (Chapter 2). It is completely defined (zero degrees of freedom) if its pressure and heat duty are fixed. 

A column section is a vertical vapor-liquid countercurrent adiabatic column section with no heaters or coolers, and whose only feeds and products are a liquid feed and a vapor product at the top, and a vapor feed and a liquid product at the bottom. The feeds to the column section are considered fixed or determined by their points of origin. Assuming the number of stages and the pressure in the column section are also fixed, this unit has no variables that can be controlled independently and therefore has zero degrees of freedom. 

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