Mass and Energy Balances on a Nonreacting System A continuous-flow steam-heated mixing kettle will be used to produce a 20 wt % sulfuric acid solution at 65.6°C from a solution of 90 wt % sulfuric acid at 0°C and pure water at 21.1◦C. Estimate a. The kilograms of pure water needed per kilogram of

Question

Mass and Energy Balances on a Nonreacting System

A continuous-flow steam-heated mixing kettle will be used to produce a 20 wt % sulfuric acid solution at 65.6°C from a solution of 90 wt % sulfuric acid at 0°C and pure water at 21.1◦C. Estimate

a. The kilograms of pure water needed per kilogram of initial sulfuric acid solution to produce a mixture of the desired concentration

b. The amount of heat needed per kilogram of initial sulfuric acid solution to heat the mixture to the desired temperature

c. The temperature of the kettle effluent if the mixing process is carried out adiabatically

Accepted Answer

We choose the contents of the mixing kettle as the system. The difference form of the equations of change will be used for a time interval in which 1 kg of concentrated sulfuric acid enters the kettle.

a. Since there is no chemical reaction, and the mixing tank operates continuously, the total and sulfuric acid mass balances reduce to

[latex]0=\sum_{k=1}^{3} \dot{M}_{k} \quad ext { and } \quad \sum_{k=1}^{3}\left(M_{ H _{2} SO _{4}} ight)_{k}=0[/latex]

Denoting the 90 wt % acid stream by the subscript 1 and its mass flow by [latex]\dot{M}_{1}[/latex], the water stream by the subscript 2, and the dilute acid stream by subscript 3, we have, from the total mass balance,

[latex]0=\dot{M}_{1}+\dot{M}_{2}+\dot{M}_{3}=\dot{M}_{1}+Z \dot{M}_{1}+\dot{M}_{3}[/latex]

or

[latex]\dot{M}_{3}=-(1+Z) \dot{M}_{1}[/latex]

where Z is equal to the number of kilograms of water used per kilogram of the 90 wt % acid. Also, from the mass balance on sulfuric acid, we have

[latex]0=0.90 \dot{M}_{1}+0 \dot{M}_{2}+0.20 \dot{M}_{3}=0.90 \dot{M}_{1}-0.20(1+Z) \dot{M}_{1}[/latex]

Therefore,

[latex]1+Z=\frac{0.90}{0.20}=4.5 \quad ext { and } \quad Z=3.5[/latex]

so that 3.5 kg of water must be added to each 1 kg of 90 wt % acid solution to produce a 20 wt % solution.

b. The steady-state energy balance is

[latex]0=\sum_{k}(\dot{M} \hat{H})_{k}+\dot{Q}[/latex]

since [latex]W_{s}=0 ext { and } \int P d V=0[/latex]. From the mass balance of part (a),

[latex]\begin{aligned}\dot{M}_{2} &=3.5 \dot{M}_{1} \\dot{M}_{3} &=-4.5 \dot{M}_{1}\end{aligned}[/latex]

From the enthalpy-concentration chart, Fig. 8.1-1, we have

[latex]\begin{aligned}\hat{H}_{1} &=\hat{H}\left(90 ext { wt } \% H _{2} SO _{4}, T=0^{\circ} C ight)=-183 kJ / kg \\hat{H}_{2} &=\hat{H}\left( ext { pure } H _{2} O , T=21.1^{\circ} C ight)=91 kJ / kg \\hat{H}_{3} &=\hat{H}\left(20 ext { wt } \% H _{2} SO _{4}, T=65.56^{\circ} C ight)=87 kJ / kg\end{aligned}[/latex]

so that

[latex]\begin{aligned}Q &=(4.5 imes 87-3.5 imes 91-1 imes(-183)) \&=(391.5-318.5+183)=256 kJ / kg ext { of initial acid solution }\end{aligned}[/latex]

c. For adiabatic operation, the energy balance is

[latex]0=\sum_{k}(\dot{M} \hat{H})_{k}[/latex]

or

[latex]\begin{array}{c}0=4.5 \hat{H}_{3}-3.5(91)-1(-183) \4.5 \hat{H}_{3}=135.5 \quad ext { and } \quad \hat{H}_{3}=30.1 kJ / kg\end{array}[/latex]

Referring to the enthalpy-concentration diagram, we find that [latex]T \sim 50^{\circ} C[/latex].

Question 8.4.2: Mass and Energy Balances on a Nonreacting System A continuou...

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