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Dimensional Consistency in Equations: Understanding Units and Concentrations

Learn the importance of dimensional consistency in equations, including the correct units for each term. Explore concepts of moles, mass, flow rates, concentrations, and dilutions. Calculate concentrations, flow rates, and analyze gas mixtures. Convert between mass and mole fractions and understand the relationship between molality and mass percentages.

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Dimensional Consistency in Equations: Understanding Units and Concentrations

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  1. Equations must be dimensionally consistent . • What does this mean? • Let us take the eq p = gh + p0 Is it correct ? How do we make it correct? • Same for units, each term must have the same unit(s) as all other terms

  2. The mole unit • What is 1 mole of Na ? Units? Are we using grams for mass ? or Kg? or lb? Then we call the mole by the mass we use • What is the mass of lbmole of H2SO4 ? • 98 lb H2SO4 = 44.45 Kg • How many gms in 6 gmoles of H3PO4? How many moles of H are available? Of P ? of O ?

  3. = 438 g H3PO4 = 18 gmol H = 24 gmol O • 100 gmoles of O2 = gms • = lbmoles O2

  4. Density, Specific gravity, sp. Volume, mole fraction, mass fraction, volume fraction • = mass/unit vol sp volume= 1/ = in units ? •  it is volume/unit mass ( vol of one unit of mass of material. • sp. gravity = densityA/ densityref • Does change with Temp? YES

  5.  We need to specify the temp used for A and for the ref. • Example • sp. gr = 0.73 20o What does this mean ? • 4o

  6. In the Petroleum Industry • sp gr is measured in oAPI • oAPI = 141.5 . • sp gr 60o • 60o - 131.5 • EXAMPLE (mass fr. , mole fr and Average MWt ) • What is the avg MWt of a gas mixture that contains: • Mass, Kg • H2 3 • CH4 8 • C2H4 56 • Total 67

  7. = 16.75 kg/kgmol DO YOU HAVE QUESTIONS ?

  8. Concentrations of solutions, • mass/unit vol, PPM, molar, molal , etc • Concentration is : Qty of solute per specified amount of solvent or solution • How do we specify concentration? 1.Mass of solute per unit volume solution For example 0.10 kgNaCl/L soln .

  9. 2.Moles of solute per unit volume solution.For example : 0.10 kgmolNaCl/L soln 3.Molarity (mol solute/L soln) 4.Molality (mol solute/Kg solvent) 5.For dilute soln : we use PPM (parts per million) .

  10. Q1: 700 mg salt /m3 water is PPM 0.7 PPM . Note: total mass of solution = 1000 kg+ 700 mg  1000 kg

  11. FLOW RATES • How do we measure the flow rate of liquid in the pipe ? • L/s (Volumetric flow rate) • Kg/s (Mass flow rate) • How do we change from one to the other? We use the density • How do we calc the velocity ? • Example: • 40% NaCl soln is flowing at 10L/min in a pipe. If the sp gr of the soln is 1.4, • calculate the concentration of NaCl in Kg/L • calculate the flow rate of soln in kgmol soln/min

  12. How do we calc the velocity ? • Example: • 40% NaCl soln is flowing at 10L/min in a pipe. If the sp gr of the soln is 1.4, then • calculate the concentration of NaCl in Kg/L • calculate the flow rate of soln in kgmol soln/min Q = A . v

  13. 0.56

  14. Example A gas mixture contains : 40% N2, 45% O2 and 15%CH4.This mixture is hazardous, so to make it safe, it was diluted by adding equal amount of N2 to it. What is the analysis of the new safe mixture ?

  15. 40 mol% N2 45 mol% O2 15 mol% CH4 ? mol% N2 ? mol% O2 ? mol% CH4 • +Equal amount • of N2 ? • How do we solve? • What should we start with?

  16. BASIS: 1000 gmoles of starting gas mixture BASIS: 100 gmoles of starting gas mixture BASIS: 1 lbmole of starting gas mixture BASIS: 50 kgmoles of starting gas mixture

  17. BASIS: 100 gmoles of starting gas mixture 40 gmol N2 45 gmol O2 15 gmol CH4 80 gmol N2 45 gmol O2 15 gmol CH4 140gmoles +40 gmol N2  % 80 gmol N2 57 45 gmol O2 32 15 gmol CH411 140gmoles 100%

  18. SOLUTION How do we solve? What should we start with?

  19. Basis: ? SOLUTION Basis 100 kg of the solution , thus kg NaCl 5 KCl 10 HCl 5 H2SO4 soln80 100 H2SO4 solution (water + acid ) How much H2SO4 and how much water in the solution ?

  20. Let us change the molality to mass % H2SO4 Molality = mol H2SO4 /Kg water  If we have 1 Kg water, there is 1.5 gmol of H2SO4 Therefore, in 1 Kg water there is 1.5 gmol x 98g gmol = 147 g H2SO4 1.0 kg water 1.5 gmol acid Basis: ? fraction 1000 g water 0.128 147 g acid 0.872 1147 1.000

  21. kg NaCl 5 KCl 10 HCl 5 H2SO4 soln80 100 H2SO4 80x 0.128 =10.24 kg H2O 80x 0.872 =69.76 kg 80.00 kg How much H2SO4 and how much water in the solution ?

  22. THEREFORE,

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