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Conservation of Mass

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Presentation on theme: "Conservation of Mass "β€” Presentation transcript:

1 Conservation of Mass π‘š 𝑖𝑛 π‘š π‘œπ‘’π‘‘
π‘š 𝑖𝑛 βˆ’ π‘š π‘œπ‘’π‘‘ + π‘š π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ βˆ’ π‘š π‘π‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ = βˆ† π‘š π‘ π‘¦π‘ π‘‘π‘’π‘š

2 DISCLAIMER & USAGE The content of this presentation is for informational purposes only and is intended for students attending Louisiana Tech University only. The authors of this information do not make any claims as to the validity or accuracy of the information or methods presented. Any procedures demonstrated here are potentially dangerous and could result in damage and injury. Louisiana Tech University, its officers, employees, agents and volunteers, are not liable or responsible for any injuries, illness, damage or losses which may result from your using the materials or ideas, or from your performing the experiments or procedures depicted in this presentation. The Living with the Lab logos should remain attached to each slide, and the work should be attributed to Louisiana Tech University. If you do not agree, then please do not view this content. boosting application-focused learning through student ownership of learning platforms

3 Material Balance Engineers design products, and we make them out of different raw materials. change the shape of a body by adding or removing material change the organization of atoms in a material combine components to create new substances through chemical reactions removing material Blacksmith hot forging a steel tool using a hammer and an anvil. rearranging atomic structure OH- H2O H e- 2 𝐻 2 𝑂(𝑙)+2 𝑒 βˆ’ β†’ 𝐻 2 𝑔 +2 𝑂𝐻 βˆ’ (π‘Žπ‘ž) chemical reactions We use material balance to analyze all sorts of physical processes (industrial, biological, environmental) mass can neither be created nor destroyed just rearranged short of nuclear reactions (𝐸=π‘š 𝑐 2 )

4 Material Balance - Keeping Track of the Mass
2 𝐻 2 𝑂(𝑙)+2 𝑒 βˆ’ β†’ 𝐻 2 𝑔 +2 𝑂𝐻 βˆ’ (π‘Žπ‘ž) If chemical reactions occur, new system components may be generated π‘š 𝑖𝑛 βˆ’ π‘š π‘œπ‘’π‘‘ + π‘š π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ βˆ’ π‘š π‘π‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ = βˆ† π‘š π‘ π‘¦π‘ π‘‘π‘’π‘š accumulation of mass in the system massfinal – massinitial While others are consumed π‘š 𝑖𝑛 Example Applications ( π‘š 𝑖𝑛 βˆ’ π‘š π‘œπ‘’π‘‘ =βˆ† π‘š π‘ π‘¦π‘ π‘‘π‘’π‘š ) water in a lake water-flowingin + rain – water-flowingout - evaporation β‰ˆβˆ† water mass laundry dryer wet-laundryin – partially-dry-laundryout - evaporation = 0 fishtank project NaCl initially in system + NaCl added - NaCl leaving through overflow = βˆ† NaCl π‘š π‘œπ‘’π‘‘

5 Batch Problems Start with nothing and end with nothing in system
making a batch of homemade ice cream mixing a batch of concrete in a mixer Example Problem: A 10-gallon aquarium contains 2% salt by weight. How much salt would you need to add to bring the salt concentration to 3.5% salt by weight? 1. Draw a diagram to represent the system 3. Convert volumes to mass 2. Label all inputs and outputs, assigning variables to unknowns 𝑋 lbs dry salt π‘Š 𝐻 2 𝑂 =10π‘”π‘Žπ‘™ βˆ™ 𝑓 𝑑 3 π‘”π‘Žπ‘™ βˆ™62.3 𝑙𝑏 𝑓 𝑑 3 =83.3 𝑙𝑏 10 gal salt water 2% NaCl 98% H2O π‘Œ lbs salt water 3.5% NaCl 96.5% H2O

6 Example Problem continued
𝑋 lbs dry salt π‘š=83.3 𝑙𝑏 10 gal salt water 2% NaCl 98% H2O π‘Œ lbs salt water 3.5% NaCl 96.5% H2O 4. Apply conservation of mass to each component (salt & water) and for mixture: π‘š 𝑖𝑛 βˆ’ π‘š π‘œπ‘’π‘‘ + π‘š π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ βˆ’ π‘š π‘π‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ = π‘š π‘“π‘–π‘›π‘Žπ‘™ βˆ’ π‘š π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ overall: π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ salt: π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ water: 83.3 𝑙𝑏 + 𝑋 =π‘Œ 0.02(83.3 𝑙𝑏) + 𝑋 =0.035βˆ™π‘Œ 𝑙𝑏 =0.965βˆ™π‘Œ (1) (2) (3) 4. Solve for the unknowns (you can use any of the three equations above) From (3): π‘Œ= 0.98 (83.3 𝑙𝑏) =84.6 𝑙𝑏 Is equation (2) useful??? why?? Plug this into (1): 𝑋=π‘Œβˆ’83.3 𝑙𝑏 =84.6 𝑙𝑏 βˆ’83.3 𝑙𝑏=1.29 𝑙𝑏 use it to check your work dry salt to add

7 Problem Solving Tips Draw a picture of the system. Sometimes it’s not easy to determine the boundaries of your system. (a large river flowing into the ocean for example where does river end and ocean begin?) Label all inputs and outputs, listing all known quantities & concentrations and assigning variables to the unknowns. This key step is where errors usually occur. Think about the problem a little bit. What is happening with the overall system? Are components generated or consumed? Revise (1) and (2) if needed. Write conservation of mass (or weight) for each component and for the entire system. Modify the diagram as new information is uncovered. Solve for the unknowns. Reflect on your solution. Do the concentrations or quantities make sense? Avoid trying to just solve these problems in your head. Use the systematic approach above.

8 Class Problem: A 245 quart aquarium contains 3% salt by weight
Class Problem: A 245 quart aquarium contains 3% salt by weight. How much 10% salt by weight water would you need to add to bring the salt concentration to 4.5% salt by weight? Useful Conversions: 1L = 0.001m3 = quarts = 0.264gal = 61.02in3 𝑋 kg salt water 10% NaCl 90% H2O Convert quarts into mass: 245π‘žπ‘’π‘Žπ‘Ÿπ‘‘π‘ βˆ™ 1𝐿 π‘žπ‘’π‘Žπ‘Ÿπ‘‘ βˆ™ 1π‘˜π‘” 1𝐿 =231.85π‘˜π‘” kg salt water 245 quarts salt water π‘Œ kg salt water 3% NaCl 4.5% NaCl 97% H2O 95.5% H2O Apply Conservation of Mass Equations: π‘š 𝑖𝑛 βˆ’ π‘š π‘œπ‘’π‘‘ + π‘š π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ βˆ’ π‘š π‘π‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ = π‘š π‘“π‘–π‘›π‘Žπ‘™ βˆ’ π‘š π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ overall: π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ salt: π‘š 𝑖𝑛 = π‘š π‘œπ‘’π‘‘ water: 231.85kg + 𝑋 =π‘Œ 0.03(231.85π‘˜π‘”) + 0.1𝑋 =0.045βˆ™π‘Œ π‘˜π‘” +0.9X =0.955βˆ™π‘Œ (1) (2) (3) From (3): π‘Œ= π‘˜π‘” +0.9X Plug this into (1): 𝑋= π‘˜π‘” +0.9X βˆ’231.85π‘˜π‘” 𝑋=235.49π‘˜π‘” π‘‹βˆ’231.85π‘˜π‘” 0.0576𝑋=3.641π‘˜π‘” 𝑋=63.23 π‘˜π‘” Water with 10% salt by weight


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