I want to attempt it using Newton's Law of cooling but I am not sure how to go about that. Or should I approach it as a first order differential equation mixing problem.
更新1:
Would it be easier if we said the tank was empty at the start.
Stumped on water tank problem. How long will it take for this tank to reach 60°F. Please use Newtons Law of cooling.?
2019-01-26 9:38 pm
A full 300 gallon water tank is initially 70° F. How long will it take the tank to reach 60° F if 10 gallons of water is being replaced every minute and the replacement water is at 55° F and it increases 0.16 degrees per gal, per min.
I want to attempt it using Newton's Law of cooling but I am not sure how to go about that. Or should I approach it as a first order differential equation mixing problem.
I want to attempt it using Newton's Law of cooling but I am not sure how to go about that. Or should I approach it as a first order differential equation mixing problem.
回答 (3)
2019-01-26 11:34 pm
It's not a cooling problem, it's a mixing problem.
Some things in your question are unclear. You say "it increases 0.16 degrees per gal, per min." What is "it"? Is it the temperature of the water being added; i.e., it starts at 55°, then gets warmer. And should that be just 0.16° degrees per minute, not per gallon per minute. If the former, the temp of the water being added is just 55 + 0.16t. If the former, it's more complicated, and requires a measurement of the total number of gallons of the cooler water.
You probably are supposed to assume that the water mixes instantly, so that the water being taken out gets cooler and cooler. So the water in the tank is always a mixture of some original water, and some (but not all) of the cooler water added.
Can you please clarify?
Some things in your question are unclear. You say "it increases 0.16 degrees per gal, per min." What is "it"? Is it the temperature of the water being added; i.e., it starts at 55°, then gets warmer. And should that be just 0.16° degrees per minute, not per gallon per minute. If the former, the temp of the water being added is just 55 + 0.16t. If the former, it's more complicated, and requires a measurement of the total number of gallons of the cooler water.
You probably are supposed to assume that the water mixes instantly, so that the water being taken out gets cooler and cooler. So the water in the tank is always a mixture of some original water, and some (but not all) of the cooler water added.
Can you please clarify?
2019-01-27 2:30 am
If replacement water was supplied at a constant 55° F for t minutes,
and we could assume that the water mixes instantly
300*70 + 10t*55 = (300 + 10t)*60 ..................................(1)
t = 60 minutes.
But that is complicated by “it increases 0.16 degrees per gal, per min”
I tried the assumption that this applies to the input water temperature.
We know that 10 gallons of water is being replaced every minute, so,
perhaps we are to interpret the increase in replacement water temperature after t minutes as 1.6t degrees. This changes equation (1) to
300*70 + 10t*(55 + 1.6t) = (300 + 10t)*60 ........................(2)
Unfortunately, that has no real solution, so assumption wrong.
Could this be the interpretation needed?
300*70 + 10t*55 = (300 + 10t*1.16)*60 .............................(3)
T ~ 20.5 mins
and we could assume that the water mixes instantly
300*70 + 10t*55 = (300 + 10t)*60 ..................................(1)
t = 60 minutes.
But that is complicated by “it increases 0.16 degrees per gal, per min”
I tried the assumption that this applies to the input water temperature.
We know that 10 gallons of water is being replaced every minute, so,
perhaps we are to interpret the increase in replacement water temperature after t minutes as 1.6t degrees. This changes equation (1) to
300*70 + 10t*(55 + 1.6t) = (300 + 10t)*60 ........................(2)
Unfortunately, that has no real solution, so assumption wrong.
Could this be the interpretation needed?
300*70 + 10t*55 = (300 + 10t*1.16)*60 .............................(3)
T ~ 20.5 mins
2019-01-26 10:19 pm
T(t) = Temp at time t = 70
T(0) = initial temp (at time 0)
T(r) = replacement water Temp = 55
T(0) = initial temp (at time 0)
T(r) = replacement water Temp = 55
收錄日期: 2021-04-24 07:32:02
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