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Fundamental of Material Balances

Fundamental of Material Balances. Chapter 4. Batch Continuous Semibatch. Unsteady State Steady State. Process Classification. Balances. General Balance Equation. Input + generation – output – consumption = accumulation. 50,000 people move in. 75,000 people move out. 19,000 die.

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Fundamental of Material Balances

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  1. Fundamental of Material Balances Chapter 4

  2. Batch Continuous Semibatch Unsteady State Steady State Process Classification

  3. Balances • General Balance Equation Input + generation – output – consumption = accumulation

  4. 50,000 people move in 75,000 people move out 19,000 die 22,000 are born Example 4.2-1

  5. Input + generation – output – consumption = accumulation 50,000 + 22,000 – 75,000 – 19,000 = -22,000 Each year the city’s population decreases by 22,000 people

  6. Types of Balances • Integral • describes what happens between two instants of time • Differential • describes what happens at an instant of time • Rates

  7. Simplifications • If the balance is on total mass, there can be no generation or consumption • In – Out = Accumulation • If the balance is on a non-reactive substance there is no generation or comsumption • If the system is at steady state, the accumulation is 0

  8. Balances on Continuous Steady-State Processes • Consider a steady state process where no chemical reactions occur • Generation = 0 • Consumption = 0 • Accumulation = 0 Input = Output

  9. Material Balance on a Continuous Distillation Process 450 kg B/hr m1=(kg T/h) 500 kg B/h 500 kg T/hr m2=(kg T/h) 475 kg T/hr

  10. Three Balances are possible • Benzene Balance • 500 kg B/hr = 450 kg B/hr + m2 • Toluene Balance • 500 kg T/hr = 475 kg T/hr + m1 • Total Balance • 500 kg/hr + 500 kg/hr = 450 + 475 + m1 + m2 Although there are 3 possible balances, there are only two which are independent -- If you add the benzene and toluene balances together, you get the total balance

  11. Integral Balances on Batch Processes • Initial input + generation = final output + consumption

  12. Example balance on a batch mixing process 200 g 0.400 g CH3OH/g 0.600 g H2O/g m g 150 g x g CH3OH/g (1-x) g H2O/g 0.700 g CH3OH/g 0.300 g H2O/g

  13. Again, you could have three balances • Total Balance • 200 + 150 = m • Methanol Balance • 0.4 * 200 + 0.7 * 150 = x * m • Water Balance • 0.6 * 200 + 0.3 * 150 = (1-x) * m • There are only two independent equations • The methanol balance + the water balance = total balance

  14. Flow Charts • Write the values and units of all known streams on the chart • Assign symbols to the unknowns

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