Question 3.31: A steady-state distillation column is designed to separate b...

A) Energy balance on Reboiler (toluene)

\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{N \left(\underline{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{n}}_{\text {in }}\left(\underline{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\text {in }}\right)-\dot{\mathrm{n}}_{\text {out }}\left(\underline{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right)+\dot{W}_{S}+\dot{W}_{\text {EC }}+\dot{\mathrm{Q}}\end{aligned}

Cancelling terms

\begin{gathered}0=\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\mathrm{out}}\left(\underline{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\\dot{\mathrm{Q}}=-\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)+\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{L}}\left(\underline{\mathrm{H}}_{\mathrm{out}, \mathrm{L}}\right)+\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{V}}\left(\underline{\mathrm{H}}_{\mathrm{out}, \mathrm{V}}\right)\end{gathered}

The following values of the molar enthalpy were found from the NIST webbook (accessed May 22, 2013):

\underline{\mathrm{H}}_{\mathrm{in}}=\underline{\mathrm{H}}_{\mathrm{out}, \mathrm{L}} \rightarrow Saturated liquid toluene at 1.1~ \mathrm{bar} \rightarrow 0.541 \frac{\mathrm{kJ}}{\mathrm{mol}}

\underline{\mathrm{H}}_{\mathrm{out}, \mathrm{V}} \rightarrow Saturated vapor toluene at 1.1 ~\mathrm{bar} \rightarrow 33.60 \frac{\mathrm{kJ}}{\mathrm{mol}}

Find mole flow rates

0=-\dot{\mathrm{n}}_{\mathrm{in}}+\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{V}}+\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{L}}

We know that \dot{\mathrm{n}}_{\mathrm{in}} \rightarrow 10 \frac{\mathrm{mol}}{\mathrm{sec}}

We also know that 80 % of \dot{\mathrm{n}}_{\mathrm{in}} becomes vapor. Therefore, \dot{\mathrm{n}}_{\mathrm{out}, \mathrm{V}} \rightarrow 8 \frac{\mathrm{mol}}{\mathrm{sec}}

\begin{array}{ll}\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{L}}=-\dot{\mathrm{n}}_{\mathrm{out}, \mathrm{V}}+\dot{\mathrm{n}}_{\mathrm{in}}=2 \frac{\mathrm{mol}}{\mathrm{sec}}\\\dot{\mathrm{Q}}=-\left(10 \frac{\mathrm{mol}}{\mathrm{sec}}\right)\left(0.541 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)+\left(8 \frac{\mathrm{mol}}{\mathrm{sec}}\right)\left(33.60 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)+\left(2 \frac{\mathrm{mol}}{\mathrm{sec}}\right)\left(0.541 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)\\\dot{\mathrm{Q}}=\bf 264.46 \frac{kJ}{sec}\end{array}

B) Set an energy balance around the water passing through the reboiler

\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}

Cancelling terms

\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}\end{gathered}

The heat transferred into the toluene from part A ) is the opposite of the heat exiting the steam.

For steam at 1 bar:

Find the enthalpies from Appendix A

\widehat{\mathrm{H}}_{\mathrm{I}} \rightarrow Saturated Steam at 1 bar \rightarrow 2674.9 \frac{\mathrm{kJ}}{\mathrm{kg}}

\widehat{\mathrm{H}}_{\mathrm{J}} \rightarrow Saturated Liquid at 1 bar \rightarrow 417.5 \frac{\mathrm{kJ}}{\mathrm{kg}}

\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{J}}-\widehat{\mathrm{H}}_{\mathrm{I}}\right)}=\frac{-264.46 \frac{\mathrm{kJ}}{\mathrm{sec}}}{\left(417.5 \frac{\mathrm{kJ}}{\mathrm{kg}}-2674.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)}=\mathbf{0 . 1 1 7} \frac{\mathbf{k g}}{\mathbf{s e c}}

For steam at 3 bar:

Find the enthalpies from Appendix A

\widehat{\mathrm{H}}_{\mathrm{I}} \rightarrow Saturated Steam at 3 bar \rightarrow 2724.9 \frac{\mathrm{kJ}}{\mathrm{kg}}

\widehat{\mathrm{H}}_{\mathrm{J}} \rightarrow Saturated Liquid at 1 bar \rightarrow 561.4 \frac{\mathrm{kJ}}{\mathrm{kg}}

\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{J}}-\widehat{\mathrm{H}}_{\mathrm{I}}\right)}=\frac{-264.46 \frac{\mathrm{kJ}}{\mathrm{sec}}}{\left(561.4 \frac{\mathrm{kJ}}{\mathrm{kg}}-2724.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)}=\mathbf{0 . 1 2 2} \frac{\mathbf{k g}}{\mathbf{s e c}}

For steam at 5 bar:

Find the enthalpies from Appendix A

\widehat{\mathrm{H}}_{\mathrm{I}} \rightarrow Saturated Steam at 3 bar \rightarrow 2748.1 \frac{\mathrm{kJ}}{\mathrm{kg}}

\widehat{\mathrm{H}}_{\mathrm{J}} \rightarrow Saturated Liquid at 1 bar \rightarrow 640.1 \frac{\mathrm{kJ}}{\mathrm{kg}}

\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{J}}-\widehat{\mathrm{H}}_{\mathrm{I}}\right)}=\frac{-264.46 \frac{\mathrm{kJ}}{\mathrm{sec}}}{\left(640.1 \frac{\mathrm{kJ}}{\mathrm{kg}}-2748.1 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)}=\bf 0.125 \frac{\mathbf{k g}}{\mathbf{s e c}}

C) Since the flow rates are so similar, there is no obvious advantage to using the more expensive, high pressure steam. However, these calculations don’t account for the temperature driving force required in a heat exchanger. According to the NIST Webbook, the boiling temperature for toluene at P=1.1 bar is 386.66 K, or 113.41°C. Thus, it is warmer than saturated water at P=1 bar; heat will not transfer from the steam to the toluene. Notice that even though we know realistically that heat cannot transfer from a cold fluid to a hotter fluid, such a process doesn’t violate the second first law of thermodynamics; we were able to do energy balances and obtain answers for this impossible process. Chapter 4 introduces the property entropy which is helpful in assessing realistic limitations on energy transfer.

The higher pressure steam has a higher temperature (~133°C at 3 bar and ~151°C at 5 bar), which means either of these could be used in the reboiler. While the higher-pressure steam is more expensive, the calculations in part B don’t account for the cost of equipment. A higher the temperature difference between the steam and the toluene means faster heat exchange, which means a smaller heat exchanger is needed. The trade-off between operating costs and equipment costs are typically addressed in design courses in the chemical engineering curriculum.

D) Set an energy balance around the Benzene entering and leaving the condenser. Note that it does not matter for this calculation how much of the benzene is removed as product and how much is returned to the column as reflux- it is all condensed, so there is only a single vapor stream entering and a single liquid stream leaving.

\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{N \left(\underline{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\text {in }}\right)-\dot{\mathrm{n}}_{\text {out }}\left(\underline{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right)+\dot{W}_{\mathrm{S}}+\dot{W}_{E C}+\dot{\mathrm{Q}}\end{aligned}

Cancelling terms

\begin{gathered}0=\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\mathrm{out}}\left(\underline{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\\dot{\mathrm{Q}}=-\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)+\dot{\mathrm{n}}_{\text {out }}\left(\underline{\mathrm{H}}_{\text {out }}\right)\end{gathered}

The values of the molar enthalpy were determined from the NIST webbook (accessed May 22, 2013):

\underline{\mathrm{H}}_{\mathrm{out}} \rightarrow Saturated liquid benzene at 0.9 bar \rightarrow-0.559 \frac{\mathrm{kJ}}{\mathrm{mol}}

\underline{\mathrm{H}}_{\mathrm{in}} \rightarrow Saturated vapor benzene at 0.9 bar \rightarrow 30.483 \frac{\mathrm{kJ}}{\mathrm{mol}}

\begin{gathered}\dot{\mathrm{Q}}=-\left(8 \frac{\mathrm{mol}}{\mathrm{sec}}\right)\left(30.483 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)+\left(8 \frac{\mathrm{mol}}{\mathrm{sec}}\right)\left(-0.559 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)\\\dot{\mathrm{Q}}=-248.3 \frac{\mathrm{kJ}}{\mathrm{sec}}\end{gathered}

The heat transferred from the benzene is the opposite as the heat entering the coolant,

\dot{\mathrm{Q}}=+248.3 \frac{\mathrm{kJ}}{\mathrm{sec}}

Set an energy balance around the water entering and leaving the condenser.

\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\mathrm{in}}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}

Cancelling terms

\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}\end{gathered}

We know that the water leaving the condenser is 10°C  cooler than the benzene in the condenser. The NIST webbook gives a boiling temperature of 349.44 K for benzene at 0.9 bar. (The vapor pressure could also be estimated using the Antoine equation and data in Appendix E).

Consequently the coolant enters the condenser at 25°C and leaves the condenser at 339.44 K or 66.29°C. This is true regardless of which coolant is used. Using the heat capacity of liquid water from Appendix D:

\begin{gathered}\widehat{H}_{\text {out }}-\widehat{\mathrm{H}}_{\mathrm{in}}=\int_{\mathrm{T}_{\mathrm{in}}}^{\mathrm{T}_{\text {out }}} \mathrm{C}_{\mathrm{P}} \mathrm{dT} \\\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=(\mathrm{R})(\mathrm{MW}) \int_{\mathrm{T}=298.15 \mathrm{~K}}^{\mathrm{T}=339.44 \mathrm{~K}} \mathrm{~A}+\mathrm{BT}+\mathrm{CT}^{2} \mathrm{dT} \\\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=\left.\left(8.314 \frac{\mathrm{J}}{\mathrm{molK}}\right)\left(\frac{1 \mathrm{~mol}}{18.02 \mathrm{~g}}\right)\left(\mathrm{AT}+\frac{\mathrm{BT}^{2}}{2}+\frac{\mathrm{CT}^{3}}{3}\right)\right|_{\mathrm{T}=298.15 \mathrm{~K}} ^{\mathrm{T}=339.44 \mathrm{~K}}\end{gathered}

From Appendix \mathrm{D}: \mathrm{A}=8.712, \mathrm{~B}=1.25 \times 10^{-3}, \mathrm{C}=-0.18 \times 10^{-6}

\begin{gathered}\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=173.2 \frac{\mathrm{J}}{\mathrm{g}}=173.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\\\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}=\frac{+248.3 \frac{\mathrm{kJ}}{\mathrm{sec}}}{173.2 \frac{\mathrm{kJ}}{\mathrm{kg}}}=\mathbf{1 . 4 3} \frac{\mathbf{k g}}{\mathbf{s e c}}\end{gathered}

E) The energy balance and value of Q are the same as in part d:

\dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}=\frac{+248.3 \frac{\mathrm{kJ}}{\mathrm{sec}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}

Use the ideal gas heat capacity to determine \widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}. The molecular mass of air is 28.84 if we are modeling air as 79 % nitrogen and 21 % oxygen.

\begin{array}{ll}\widehat{H}_{\text {out }}-\widehat{H}_{\mathrm{in}}=\int_{\mathrm{T}_{\mathrm{in}}}^{\mathrm{T}_{\text {out }}} \mathrm{C}_{\mathrm{P}}^{*} \mathrm{dT} \\\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=(\mathrm{R})(\mathrm{MW}) \int_{\mathrm{T}=298.15 \mathrm{~K}}^{\mathrm{T}=339.44 \mathrm{~K}} \mathrm{~A}+\mathrm{BT}+\mathrm{CT}^{2}+\mathrm{DT}^{3}+\mathrm{ET}^{4} \mathrm{dT} \\\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=\left.\left(8.314 \frac{\mathrm{J}}{\mathrm{molK}}\right)\left(\frac{1 \mathrm{~mol}}{28.84 \mathrm{~g}}\right)\left(\mathrm{AT}+\frac{\mathrm{BT}^{2}}{2}+\frac{\mathrm{CT}^{3}}{3}+\frac{\mathrm{DT}^{4}}{4}+\frac{\mathrm{ET}^{5}}{5}\right)\right|_{\mathrm{T}=298.15 \mathrm{~K}} ^{\mathrm{T}=339.44 \mathrm{~K}}\end{array}

From Appendix D:

\begin{aligned}& \widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=41.8 \frac{\mathrm{J}}{\mathrm{g}}=41.8 \frac{\mathrm{kJ}}{\mathrm{kg}} \\& \dot{\mathrm{m}}=\frac{\dot{\mathrm{Q}}}{\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)}=\frac{+248.3 \frac{\mathrm{kJ}}{\mathrm{sec}}}{41.8 \frac{\mathrm{kJ}}{\mathrm{kg}}}=\bf 5.94 \frac{\mathbf{kg}}{\mathbf{s e c}}\end{aligned}

F) The mass flow rate of the air is ~4x larger than the mass flow rate of the water, for the same amount of cooling. Again, these calculations don’t take into account cost of equipment. The volume of 1.43 kg of liquid water is orders of magnitude smaller than the volume of 5.94 kg of air at ambient pressure, which has implications for the size of the equipment. Heat transfer to liquids is much more efficient than heat transfer to lowpressure gases, which also influences the size, and therefore the cost, of the heat transfer equipment. Thus, there are incentives to use liquid water as the coolant, even if the air is free and the water isn’t.