Practice Final Exam Solutions

Practice Final Exam Solutions - dt dV βˆ’ βˆ’ = Initial...

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7. Distance to be covered = 435 miles Range of car = miles 5 . 462 fuel gallons 5 . 18 gallon miles 25 = × Without the leak, the student could cover the distance without stopping. Speed = 70 miles/hr Driving time required = 435/70 = 6.214 hr = 373 mins Density of gasoline is constant, so we can do a balance on the volume ( V ) of fuel in the tank: Accumulation = – Output of Fuel to Engine – Output of Fuel through Leakage Rate of fuel output to engine = min / L 177 . 0 gal 17 . 264 L 1000 . s min 60 hr 1 . hr miles 70 . miles 25 gal 1 = Rate of fuel output through leak = 0.001 m = 0.001x(0.8V) = 0.0008V L/min Transient balance: V 0008 . 0 177
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Unformatted text preview: . dt dV βˆ’ βˆ’ = Initial volume = 18.5 gal. = 70 L Integration: ∫ ∫ = = = = βˆ’ = + V 70 V t t t max dt V 0008 . 177 . dV where t max is the time at which the tank will be empty (i.e. V = 0 ). Integrating, we get: s min 344 70 0008 . 177 . 0008 . 177 . ln . 0008 . 1 t max = Γ— + Γ— + βˆ’ = Since this is less than the required driving time, the student will not make it without stopping. He would require 373-344 = 29 mins (0.483 hrs) more, so he would have to stop 0.483x70 = 34 miles short of the destination....
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Practice Final Exam Solutions - dt dV βˆ’ βˆ’ = Initial...

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