BME200_23FA_EnergyBalance_with_Reaction_and_Work_vCanvas

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Mechanical Engineering

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Dec 6, 2023

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BME 200 Conservation Principles Fall 2023 1 / 3 Using Heat effects to determine conversion Lactose is broken down into monosaccharides in the stomach through the following reaction: lactose + H 2 O glucose + galactose ∆𝑯 r º = 41.8 kJ The above reaction is usually catalyzed by the enzyme lactase. However, lactose-intolerant people are not able to produce lactase and become ill when they consume too much lactose. Suppose that you have developed a new treatment for lactose intolerance. To test its effectiveness, you decide to run a simulation using a bioreactor. Lactose dissolved in water is added to the bioreactor at a rate of 150 g lactose/min. The bioreactor is operated at steady-state and its contents are mixed using an impeller which consumes power at the rate of 50 J/s. Careful measurements show that 280 J/s heat is removed to keep the bioreactor temperature constant. The reactants and products are both at standard conditions (25°C, 1 atm). The MW of lactose is 342. a) What percentage of the lactose is broken down into glucose and galactose? Hint: The extent of reaction will be useful here. b) Do you need to calculate the input and output enthalpies to answer (a)? Explain briefly why or why not. Impeller Heat Bioreactor lactose, water 25°C, 1 atm 4.5 kg/min glucose, galactose lactose, water 25°C, 1 atm 4.5 kg/min
BME 200 Conservation Principles Fall 2023 2 / 3 SOLUTION Energy Balance: This is an open system with chemical reaction , so the Energy Balance is applied with the Bioreactor as the Control Volume: 𝑑𝐸 cv 𝑑𝑡 = 𝑄 ̇ − 𝑊 ̇ cv − 𝛏 ̇ ∆𝑯 ? ? + ∑ 𝑛̇ 𝑖 (? ? + V 𝑖 2 2 + 𝑔𝑧 𝑖 ) 𝑖? − ∑ 𝑛̇ 𝑒 (? ? + V 𝑒 2 2 + 𝑔𝑧 𝑒 ) ??? Since the systems is at steady-state and there is no information on kinetic and potential energy changes, the energy balance above reduces to ? = 𝑸 ̇ − 𝑾 ̇ cv − 𝛏 ̇ ∆𝑯 ? ? − ∑ ?̇ ? ??? 𝑯 ̂ ? + ∑ ?̇ ? ?? 𝑯 ̂ ? We take the following as reference for enthalpy: lactose (25°C, 1 atm), galactose (25°C, 1 atm), glucose (25°C, 1 atm), H 2 O ( l , 25°C, 1 atm) Since these are the conditions at both input and output, all the specific enthalpies are zero. We therefore do not need to calculate the input and output enthalpies. The energy balance equation becomes: 𝛏 ̇ ∆𝑯 ? ? = 𝑸 ̇ − 𝑾 ̇ cv From the given, 𝑸 ̇ = −??? 𝑱 ? 𝑾 ̇ cv = −?? 𝑱 ? ∆𝑯 ? ? =– ??. ? ?𝐉 = −??, ??? 𝐉 Plugging into the Energy Equation: 𝛏 ̇ = (𝑸 ̇ − 𝑾 ̇ cv ) ∆𝑯 ? ? = (−??? 𝑱 ? − (−?? 𝑱 ? )) −??, ??? 𝐉 = ?. ???? ? −? From given : in = ( ??? ? ?𝒂????? ??? ) ( ? ???? ??? ?/???? ) = ?. ??? ???? ?𝒂????? ???
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