2. (30 points) I make my personal Windex solution which is 50 wt% ammonia in wat
ID: 707473 • Letter: 2
Question
2. (30 points) I make my personal Windex solution which is 50 wt% ammonia in water. My increasingly naughty daughter thought it was funny to leave a 10 lbm bucket of the solution open in our empty tool shed and lock the door! The termperature inside the shed is maintained constant at 80F and the humidity was 0% when she closed the door. The size of the shed is 10ft X 10ft X 5ft, and there are no leaks or vents. As you imagine, when I opened the shed the air was smelling of ammonia. Assume everything is at equilibrium in the shed well before I opened the shed door. Using your skills in mass and energy balance, help me find the following (25 points) How would you analyze this problem? What is the system? Is it open or closed? Draw the 2 engineering flow chart or figure to analyze the situation. - What is my new ammonia concentration in the solution. - How much of my liquid windex evaporated - What is the concentration of ammonia in the vapor phase within the shed in lbm/ft3. - Is the above concentration more or less than the occupational limit of 25ppm. -It costs me 1 dollar for every 1 Btu of heat added or removed to maintain the shed temperature rL 1 constant. How much dollars did I lose from this windex evaporation prank.Explanation / Answer
-System is closed because no mass is allowed to enter or leave of the system
-80 F = 26.7 deg C
at 26.7 C, Solubility of Ammonia in water = 33%
=> New concentration = 33%
-
Total Mass initially = 10 lb
Initial mass contain 50% NH3
=>Total Initial NH3 in solution = 10*50/100=5 lb
=>Total water in solution = 10-5=5 lb
At 80 F after reaching equilibrium, solubility = 33%
Let x lb NH3 evaporated
Then, Ammonia concentration = Ammonia mass in solution/(Ammonia mass in solution+Water mass in solution)=(5-x)/(5+5-x)=(5-x)/(10-x)=0.33
=> Total NH3 evaporated, x = 2.5 lb
- Total Ammonia in shed = 5-2.5=2.5 lb
Total shed volume = 10*10*5=500 ft3
=>Ammonia concentration in shed = Total Ammonia in shed/Total shed volume=2.5/500=5*10^-3 lb/ft3
-25 ppm = 25 mg/litre (1 ppm=1mg/litre)
1 mg = 10^-6/0.453 lb
1litre=10^-3*(3.3)^3
=>25 ppm = 10^-6/0.453*10^-3*(3.3)^3=7.9*10^-8 lb/ft3
=>current concentration(5*!0^-3lb/ft3) is more than occupational limit (7.9*10^-8 lb/ft3)
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