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3. A 100L mixing tank is initially filled with 20 g/1 sugar solution. As more sugar solution is pumped in, & tap is opened, draining some of the tank Sugar wate...

Question

3. A 100L mixing tank is initially filled with 20 g/1 sugar solution. As more sugar solution is pumped in, & tap is opened, draining some of the tank Sugar water enters the tank at the rate of 10 L/ht; but the level of sweetness is constantly increasing; so that the solution has t2 g/L of sugar after t hours The solutions are mixed instantly and the amount drained is 20 L/hr Find the concentration of the sugar in the mixing tank and prove mathematically what the concentration is equal to as

3. A 100L mixing tank is initially filled with 20 g/1 sugar solution. As more sugar solution is pumped in, & tap is opened, draining some of the tank Sugar water enters the tank at the rate of 10 L/ht; but the level of sweetness is constantly increasing; so that the solution has t2 g/L of sugar after t hours The solutions are mixed instantly and the amount drained is 20 L/hr Find the concentration of the sugar in the mixing tank and prove mathematically what the concentration is equal to as the last drops are emptied from the tank The answer to this second part should be physically simple, but we are looking for a proof [10 marks]



Answers

A tank initially contains $20 \mathrm{L}$ of water. A solution containing $1 \mathrm{g} / \mathrm{L}$ of chemical flows into the tank at a rate of $3 \mathrm{L} / \mathrm{min},$ and the mixture flows out at a rate of 2 L/min. (a) Set up and solve the initial-value problem for $A(t),$ the amount of chemical in the tank at time $t$ (b) When does the concentration of chemical in the tank reach 0.5 g/L?

Hello and they will be solving a problem which states That tank initially contains 20 litres of water. The solution flowing and contains 1 g per litre of the chemical, and it's blowing in at a rate of three liters per minute, and the mixture is falling out at a rate of two liters per minute. Part A is asking us to just set up the equation for safety, so the first thing we'll do is we'll draw a picture so we know that just 20 leaders, water and tank and we're assuming that there's no so our solution whatsoever. And but Tink and we know that the solution contains 1 g per litre of the chemical and it's flowing in at a rate of three liters per minute. We know that everything is flowing out at a rate of two liters per minute. So next thing that will do is we get it right out all over information in mathematical terms. So you know that you zero equals 20 zero equals zero are of one is equal to three are too secret to see a one. It's a good one. So first we're gonna find our VT. We know that VT that the slope for BT? It's just poor rating minus four Right out we just dream minus two or one and we know that the initial Mount of the Tank is 20 leaders, so we're gonna add 20 leaders to it. So Viotti, you think what a t plus 20 So not of Ducks parties to find her if t so. We know that a prime sequel to the flow rate and times initial concentration, which is street temps one on regular, subtract everything by a florid out times are final concentration, which is a over a change of volume, which is t plus 20. So after moving the actual outside and simplifying everything, what we're left with a prime plus to over tea plus 20 a all equal to three. So the next thing to do it's a good right down e to the integral of the prohibition of their with just two over t plus 20 be to. So after solving the integral, we get this t eat to the to natural log of T plus 20. After using log rules, we're left with T plus 20 squared. So what we're gonna do is good enough to play the T plus 20 square to our equation and what we'll get after expending everything is plus 20 squared time sick prime plus two Time T plus 20 a, which sequiturs three times two plus 20 squared. So we know that the left side it's just anti derivative of t plus 20 squared time Say pass a quick three times T plus 20 square. So to get a by itself for me to do is take the integral on both sides. So we have Steve plus 20 squared time see the sequel to Integral off three Touch T plus 20 squared to So after solving integral, what we get is t plus 20 squared some see which is equal to a T 20 cubed plus c So we divide both sides by t plus 20 squared So I would love to see a sequel to T plus 20 plus c times t plus 20 All too negative too. So we're not done yet because we still don't know what c s. But we do know Day zero secret zero. So what we're gonna do is plug in zero in for a zero and for tea. That's where we left with zero secret zero plus 20 plus C time zero plus 20 All too negative too. And what we're left with this c is equal to negative. 20 Cubed, sir. Eight to you for party sequel to T plus 20 minus 20 cubed times T plus 20. Too negative too. In a part B is asking us. That's when does the concentration of the chemical the tank reach 0.5 g per liter? So, as remember concentration, it's just the change in amount of chemical, which is safety oh, over the change in volume, which is beauty. So we plug in our knowns, which is that we know that the constraints 0.5 No, the 80. It's just two plus 20 minus 20 Q times T plus 20 too negative too. We know that Vitti t plus 20. So what you do is simplified everything on the left side. So that will be one minus 20. But T plus 20 cubed. You simplify everything on the right side. So no, it starts solving everything. So you subtract one of the upsides. So we're left with no 2.5 equal to negative 20 over T plus 20 cube. So Let's play both sides but negative one lawful 0.5 equal to 20 plus t plus 20 cubed. Now we cubed both sides so left with Cuba of 0.5 just equal 20 over to you plus 20 and what to play both sides by t plus 20. So we're left with the cube root of 0.5 times T plus 20. Would you stick with 20 doing? Divide both sides by the cube root of five So we're left with t plus 20 switch sequitur 20 over to Cuba off 200.5 and Leslie you will subtract 20 both sides So we left with two TZ Quote a 20 of the Cuba off 0.5 minus 20 which is about 5.1 night it for So it's good. Take about five point when I it for minutes Nora for the concentration of the chemical to reach 0.5 g per leader

By saying that teach you divided by DT in is equal to queue in times d v d t. In. So solving we have deep u T T in is equal to in is equal to Q. One times are one. So basically we're calculating the rate of input and output in pounds per gallons, and that is doing the same thing. But for the output. So we have Dutt out is equal to q R. Two divided by the Okay, So the next thing I'm going to do is we're gonna solve for V. So we have V is equal to p ay plus DVD tee times t Um and then we can solve for this to get DVD T is equal to our want minus R two or V is equal to the zero plus our one minus R two times teeth. Applying equation of the equation we get teach U D T is equal to be cute. Dp in minus d Q d t out. So this is just equal to of doing what we did before. Q one are one minus. Q r. Two divided by p And since we have a new equation for V We can just substitute that in to get minus. Q R two divided by the zero plus on one minus r two times T, which can be rearranged in order to get deep You d t plus que or two divided by V zero plus our one minus R two times T is equal to Q one are one. So now we will begin solving the required. So if we have V f is equal to V, I, I'd plus are one minus R two titles. T playing in the numbers we get. T is equal to 50 minutes. All right, so now we will use the differential equation acquired in the integrating factor techniques. To get the value of Q A T is equal to 50 minutes, substituting it into the equation. You get that D. Q. D T plus three Q. Divided by 100 plus two. T is equal to 2.5 um, re of anything you're pfft. P E of T is equal to three divided by 100 plus two t So s of t is equal to 2.5 extra unified i to be e tends the integral of p f. T d t. So we take the integral, so we have the integral off p of T. D t is equal to three over two times ln off 50 plus tea or I is equal to 50 plus t to the power of three over two. Next. Uh, if we saw for a q of t, we start with the integral of I S d t is equal to ay you flooding in from what we know to solve. For Q, we end up Q of t is equal to 50 plus three to the power of three over to minus 50 to the power off. Five over two, divided by 15 plus t to the power off, three over two. So for part B Q. Of 50 easy to 82.32 pounds, and you are 50 years also equal to 163.64 pounds.

Let you be the amount off concentrating the solution. Um, then Q divided by the total volume off the tank. Uh, is how many, Um, how many pounds off concentrate per gallon are in the town? Great. Now, if we multiply this by the flow rate off the water that's living off the solution that's leaving the tank, it's going to give us the rate at which the amount off the concentrate up decreases. So, in other words, Q Prime AT T is equal to negative Q over V multiplied by Q A T. This is a very simple linear differential equation, and the solution is, Is this where you not is also known as kyu at zero uh, the amount of concentrate in pounds in the solution at time. Zero. These completes part A. Let's solve part B. Find the time at which the amount of concentrate in the tank at each is £15. Let Q one. Sorry, let let Q one be £15 and Kyu Wan is reached at some time. T one. Let's isolate t one. Divide both sides by Q by Q Not then take natural log on both sides. Nature log and exploited constant each other. Then let's multiply both sides by V and divide by negative view. Sorry, this gives us t one. Let's substitute the numbers. Um V is equal to 200 gallons. Que is equal to 10 gallons per minute. Q one equals £15 and you're not equal to £25. The unions council Substituting the numbers gives us approximately 10.2 minutes. Let's solve party. Find the amount of concentrate in pounds in the solution. S t approaches infinity. Let's take a look at how cute depends on time. S t goes toe positive. Infinity e toe negative power to negative. Large power is almost zero. It's a it approaches zero. So, um, the limit off this factor is zero. So the the limit of the entire function is also zero. So the amount of concentrate in pounds approaches zero. As time approaches, infinity


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