Question 8.1.1: Energy Balance on a Condenser Acetone (denoted as Ac) is par......

We will follow the procedure given preceding this example.
1. Perform required material balance calculations. None are required in this example.
2. Write and simplify the energy balance.
For this open steady-state system, \dot{Q} + \dot{W}_{s}=\Delta \dot{H} + \Delta \dot{E}_{k} + \Delta \dot{E}_{p}. There are no moving parts in the system and no energy is transferred by electricity or radiation, so \dot{W}_{s} = 0. No significant vertical distance separates the inlet and outlet ports, so \Delta\dot{E}_{p} ≈ 0. Phase changes and nonnegligible temperature changes occur, so \Delta\dot{E}_{k} ≈ 0 (relative to \Delta \dot{H} ). The energy balance reduces to

\dot{Q}=\Delta \dot{H} = \sum\limits_{out}{\dot{n}_{i}\hat{H}_{i}} – \sum\limits_{in}{\dot{n}_{i}\hat{H}_{i}}

3. Choose reference states for acetone and nitrogen.
The reference states may be chosen for computational convenience, since the choice has no effect on the calculated value of \Delta \dot{H} . We will arbitrarily choose the inlet stream condition for nitrogen (65°C, 1 atm) as the reference state for this species and one of the two outlet stream conditions for acetone (l, 20°C, 5 atm) as the reference state for acetone, which will enable us to set the corresponding values for \hat{H} in the enthalpy table equal to zero instead of having to calculate them.
4. Construct an inlet—outlet enthalpy table.
We first write the chosen reference states, then construct the table shown below:

Note the following points about the table:
• Nitrogen has only one inlet state (gas, 65°C, 1 atm) and one outlet state (gas, 20°C, 5 atm), so we need only one row in the table for N_{2}. Acetone has one inlet state (vapor, 65°C, 1 atm) but two outlet states (vapor and liquid, each at 20°C and 5 atm), so we need two rows for this species.
• We mark out (using dashes) the two cells corresponding to \dot{n}_{in} in and \hat{H}_{in} for liquid acetone, since no liquid acetone enters the system.
• The \dot{n} values are obtained from the flowchart. The flow rate of acetone vapor at the inlet, for example, is determined as (100 mol/s)[0.669 mol Ac(v)/mol] = 66.9 mol Ac(v)/s.
• Since the nitrogen entering the system and the liquid acetone leaving the system area are at their reference states, we set their specific enthalpies equal to zero.
• Three unknown specific enthalpies have been labeled and must be determined in Step 5.

5. Calculate all unknown specific enthalpies.
To calculate the three unknown specific enthalpies in the table, we construct hypothetical process paths from the reference states to the states of the species in the process and evaluate Δ\hat{H} for each path. This is the part of the calculation you have not yet learned to do. We will show you the calculation of \hat{H}_{1} to illustrate the method, give the results of the other calculations, and go into detail about the required procedures in Sections 8.2–8.5.

When choosing a process path for the determination of Δ\hat{H}, it helps to know that formulas and data are given in this chapter for enthalpy changes corresponding to certain types of processes:
• Section 8.2 gives the formula Δ\hat{H} = \hat{V} ΔP for a change in pressure (ΔP) undergone by a liquid or solid with constant specific volume \hat{V}. The value of \hat{V} for liquid acetone may be determined as 0.0734 L/mol from the specific gravity (0.791) given in Table B.1.

• Section 8.3 shows that \Delta \hat{H} = \int_{T_{1}}^{T_{2}}{C_{p}(T)} dT for a change from T_{1} to T_{2} at constant P. Formulas for C_{p} (T), the heat capacity at constant pressure, are given in Table B.2. The formulas for acetone liquid and vapor are as follows:

where T is in °C.
A time-saving alternative for evaluating the integral by hand is to use the APEx function Enthalpy(“species”,T1,T2,[temp. units],[state]), beginning the cell entry with an equal sign. The function integrates the heat capacity formula in Table B.2 for the specified species from temperature T1 to temperature T2 and returns Δ\hat{H} in kJ/mol. If a temperature unit is not specified as “K,” “R,” or “F,” a unit of “C” is assumed. If the state of the species is not specified as “l,” “g,” or “c” (for solid), the first entry in Table B.2 is assumed.
For example, the integrals of the heat capacity of acetone liquid and acetone vapor needed in this example would be calculated using the following formulas:

= Enthalpy(“acetone”,20,56)
= Enthalpy(“acetone”,56,65,,“g”)

Notice that internal arguments that go to their default values, like the temperature unit in the first formula, may be left blank but the commas around them must still be supplied. Default values of arguments at the end, such as the temperature unit and state in the second formula, may be omitted.
• Section 8.4 defines the heat of vaporization Δ\hat{H}_{v} ( T_{b}) as Δ\hat{H} for a change from liquid to vapor at the normal boiling point, T_{b}. Table B.1 lists T_{b} for acetone as 56.0°C and Δ\hat{H}_{v} ( T_{b}) as 30.2 kJ/mol.
Normal boiling points in °C and heats of vaporization at the normal boiling point in kJ/mol can also be retrieved using the APEx functions Tb(“species”) and Hv(“species”).
The following process path from the reference state [Ac(l), 20°C, 5 atm] to the process state [Ac(v, 65°C, 1 atm)] enables us to use all this information in the determination of \hat{H}_{1}:³

Ac(l,20°C,5  atm) \overset{\Delta \hat{H}_{1a} }{\longrightarrow }Ac(l,20°C,1  atm) \overset{\Delta \hat{H}_{1b} }{\longrightarrow }Ac(l,56°C,1  atm) \\ \overset{\Delta \hat{H}_{1c} }{\longrightarrow }Ac(v,56°C,1  atm) \overset{\Delta \hat{H}_{1d} }{\longrightarrow }Ac(c,65°C,1  atm) \\ \left. \Large{\Downarrow} \right.\\ \hat{H}_{1}=\Delta\hat{H}_{path}\\ = \Delta \hat{H}_{1a} + \Delta \hat{H}_{1b}+ \Delta \hat{H}_{1c}+\Delta \hat{H}_{1d} \\ = \hat{V}_{Ac(l)}(1  atm  –  5  atm) + \int_{20°C}^{56°C}{(C_{p})_{AC(l)}dT} + (\Delta \hat{H}_{v} )_{Ac}+\int_{56°C}^{65°C}{(C_{p})_{AC(v)}dT}

When we substitute the values of \hat{V}_{Ac(l)} and Δ\hat{H}_{v} and the formulas for C_{p}(T) into the expression for \hat{H}_{1} and carry out the necessary unit conversions and integrations, we obtain \hat{H}_{1} = (0.0297 + 4.68 + 30.2 + 0.753) kJ/mol = 35.7 kJ/mol.
Proceeding in a similar manner, we obtain the values for \hat{H}_{2} and \hat{H}_{3} shown in the following revised enthalpy table:

6. Calculate Δ\dot{H}.

The factors in the last equation come directly from the inlet–outlet enthalpy table.
7. Calculate nonzero work, kinetic energy, and potential energy terms.
Since there is no shaft work and we are neglecting kinetic and potential energy changes, there is nothing to do in this step.
8. Solve the energy balance for \dot{Q} .

\dot{Q} = \Delta \dot{H} = –  2320  kJ/s = \boxed{-  2320  kW}

Heat must be transferred from the condenser at a rate of 2320 kW to achieve the required cooling and condensation.

³ To be completely accurate, we would include a step in which the acetone and nitrogen are mixed since the references are the pure species; however, enthalpy changes when gases are mixed are generally negligible (Section 8.5).