# Energy BalanceChange in T or P

### Upon studying this section, you should be familiar with the following:

• DH = Q, it's derivation and why it is so prevenlent from here on.
• Finding DH for changes in pressure and temperature
• By using Cp or by tabulated enthalpy tables
• Relationship between DH and DU, and between Cv and Cp

## Let's Get a Perspective

 Lot's of E-balance topics, where are we? Thus far, we have covered four energy topics. First, we covered some basic energy definitions (Ek, Ep, DH, and DU), then we learned to write energy balances on closed systems (those without streams) and open systems (those with streams), and finally for mechanical systems (those where kinetic and potential energy dominate over internal energy, a system such as a waterfall). Just about from now on, DH = Q. Uh, what? In 201 and in thermo (316), this expression is first derived from an energy balance and then it is used extensively in material to follow, where it's connection with energy balances slipped. That is, we have learned a few energy balance expressions thus far, and for the most part, we will only apply this reduce expression for the energy balances to come. The topics to come deal with how to find DH (and thus Q) with energy balances on systems of change, that is, that involve chemical reactions, as well as changes in state, temperature, pressure. So, before we begin with finding enthalpy for these changes, I would like to solidify the connection between chapter 7 and this equation, resulting in a strong connection between the energy topics of this course. If you have just said to yourself, this is obvious, the link is clear. Chemical engineering processes that we will solve usually involve flows, resulting in the use of the open energy balance. And these processes often don't usually involve valves (or other components that significantly change velocity and thus kinetic energy), and nor do pipes involve great height changes (no potential energy), and there is often no stirrer. Hence, the connection between the expressions learned in chapter 7 and the DH = Q expression that will be usually used from here on out, is obvious. If this was your response then bravo, and you probably can skip on to the interrelations of this section, if your response was "hmm", or "huh?", then take a moment, and read on to solidify the connection. Make the connection Keeping in mind the basic equations we learned in chapter 7, and the energy balance equations for closed, open, and mechanical systems, let's consider a most elementary chemical process. What equations would you apply to the following system? How? As your experience thus far in the course would tell you, the following process is as simple as they come, a single unit, two compontent process with one inlet stream and two outlet streams. So, consider your energy balance tool box? Which tool do you use, and how? Well, we have a closed, an open, and a mechanical system energy balance equation. There are streams, thus the system is not closed. There are no valves or huge height changes, thus the system is not a mechanical system. Thus, it is clearly an open system in which only the energy balance equation for open systems applies. And thus, the equation is: DH + DEk + DEp = Q - Ws There clearly is no change in flow velocity (no valve), nor any significate change in height, nor is there a stirrer, therefore, respectively, the kinetic energy, potential, and the shaft work can be neglected. DH + DEk + DEp = Q - Ws And now we are left with the popular expression relating enthalpy to heat.DH = Q

## DH for changes in pressure and/or temperature

 Now we understand the expression DH = Q, where it comes from and why it recurs often in chemical engineering. Now, we learn how to find DH (and thus Q) for various scenarios. In this page, for changes in temperature and pressure changes, and later for changes in state and for chemical reactions. DH for a Temperature change: Here, the following expressions are used to calculate the change in enthalpy for solids, liquids, and gases. or, DH = Hfinal - Hinitial where, Hfinal and Hinitial for several different compounds are found directly from tables, such as table B8, B9, or for water, from steam tables, such as tables B6 and B7. For the specific system of air and water vapor, the course text offers a psychometric plot (a cross-plot that relates several properties), from which we can get Hinitial and Hfinal. For a tutorial on how to read a water vapor - air psychrometric chart, click here. So, we would opt to find the enthalpy change for a change in temperature by consulting one of these tables or charts. If the table does not have the information for our compound or for our temperature change, then we would use the enthalpy change relation with heat capacity in it. The heat capacity could be given in the problem statement as a constant, or may need to be looked up in a table (Appendix B.2) where it is given as a simple polynominal in terms of Temperature. Either way, whether constant or a polynominal, the intergal can be easily determined and we get DH for a temperature change. DH for a Pressure change: Here, the interrelations to calculate enthalpy change are very straight forward: for liquids and solids, DHhat = Vhat*DP for ideal gases, DHhat = 0 As you can see, the new interrelations introduce in this section are very simple. DH for a Pressure and Temperature change: All we do for a system in which there is a pressure change and a temperature change is simply calculate the DH for the temperature change, and then DH for the pressure change, and then add them up to get the total DH. DHchange in T & P = DHfrom T1 to T2, at P1 + DHfrom P1 to P2, at T2 Where the relations for DHfrom T1 to T2 and DHfrom P1 to P2 for solids, liquids, and gases have been given above. or, for steam, we could use steam tables, B6 and B7, DHchange in T & P = Hat T2 & P2 - Hat T1 & P1 Where the relations for DHfrom T1 to T2 and DHfrom P1 to P2 for solids, liquids, and gases have been given above. The interrelations given here are very simple, as will be the interrelations given in the last few sections. The most important aspect of using them is clearly not math, it is reading the problem statement and recognizing that you have a temperature or pressure change between the inlet and outlet streams, and that you need to find DH. Following, the definitions of DU and Cv are given, as are the relationships between DH and DU, and between Cv and Cp.

## DU for changes in pressure and/or temperature

 Understand, DU can be found for an open system, and that DU, DH, Cp, and Cv are all related. This should serve as a reminder that enthalpy and internal energy are related, as are the heat capacities. Thus, we use the open energy balance because it is convenient (tailor made for flow problems, see the derivation), but the reminder is that in problems in which DH or DU is given or can be found, the other can be readily obtain using the following interrelations. Relating DH to DU DH = DU + P*DV This applies solids, liquids, and ideal gases. Simply combine this relation with the simple relations for DH given above to determine DH from DU and vise versa. Relating the Cp and CvCp = Cv (for solids and liquids)Cp = Cv + R (for ideal gases) Get DU from Cv This interrelations applies for a temperature change (at constant volume), and is analogous to the expression relating DH and Cp. The relations given here are done so primarily because the exist, and thus should be understood, but again, the open energy balance is used for a reason. Most of our systems are open, an emphasis was first placed on expressions of DH and open systems. Secondly, the relationships in this section are given and their relationship should be kept in mind.

## Examples

Example 1:

Calculate the internal-energy and enthalpy changes that occur when air, at a constant molar volume of 35.00 ft3/lbmol, is changed from an initial state of 50 °F and 20 atm, to a final state of 140 °F and 1 atm. Assume for air that CV = 5 and Cp = 7 (Btu)/(lbmol*°F).
Hint: Break calculations into two steps, the first for the temperature change (keep the pressure at 10 atm), and the second step for the pressure change (keep the temperature at 140 °F), as given above. Thus, we had a change of states for which the volume was constant, and both the pressure and temperature changed. Because both the change in temperature and the change in pressure effected the change in molecular motion, both must be taken into account in determination of DU and DH.

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Example 2:

For 100 kmol/min of a mixture of 85 % air and 15% carbon dioxide is heated and compressed, from 100 °C and 1 atm to 500 ° C and 5 atm. Consider the gaseous mixture an ideal gas. Determine the required heat for this process with the heat capacities defined as a function of temperature, found in literature (appendix B2). Now, consider a tabulated enthalpy table, and redetermine the required heat rate (appendix B7, B8).

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Example 3:

Determine the heat required to drive the following process, where the gaseous mixture is considered ideal, and the flowrate is constant.

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Example 4:

Given an air water system, and three different initial and final states, i, ii, iii, determine a)for the initial state, determin all of the following properties, Tdb, Twb, Tdp, hr, ha, and V-hat.
b) determine the heat required for the process.
c) one of the three processes is adiabatic, which is it?, which is a pure heating of a vapor?

given 100 moles of air,

State 1State 2
iT = 41 °C
hr = 10%
T = 25 °C
hr = 10%
iiV-hat = .85 m3/kg-DA
ha = .0028 m3/kg-DA
T = 19.5 °C
ha = .0064 m3/kg-DA
iiiTwb = 28.5 °C
hr = 40%
T = 27 °C
hr = 60%

For i, Goto | Check Answer | See Solution

For ii, Goto | Check Answer | See Solution

For iii, Goto | Check Answer | See Solution

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