Question

Let m > 0 be a constant. Suppose a tank starts with 10kg of salt water with 1% salt by mass. Suppose further that every hour, m kg of salt water with a concentration of 10% salt is added, and 2kg of the water in the tank is drained. The solution is kept well mixed at all times. (a) (2 points) Write a differential equation that describes the amount of salt in the water.

Answer 1

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sont from the given data mo- Constant m mars lokg of salt water ly. salt by mas - Concentication of salt 10% cidded. akg of water drained a) let us consider t=0, mass of salt in tants = 10x1 z 1 kg 100 l Solution may = lokg Now, at time t, mass of salt is given as e(t) Mass of water = m-a)t +10 24.04 concentration is lessed from mas Concentration of salt at t time in tank is given by = 2Ct). 10+(m-2)t By considering time t d (4). is the change ing mass Of Salt in the soution for at time period dait Amount of Calt being added in at time - mass of salt being drained m dt period of tror m alt) amel Cact) = 1 0 10(-2)

a .. The differential equation for the given data is given by dult = m - X(t) at 10 10+(m-2)= dxCt) + 2(E) Em 10 10+ m2) E let y't f(x)y= g(x) Soin is lify = g(x) 2.f.de Pnlegrad function - Sfan) dx Pl . dt Totmzt dt . In I lo+m-2) E m-2 (b) for Colving m-3 kutogial function = (10 + (3-a) ) 13-2 = (10 716) = lott (10+t)y = r. 10+t)dt + (1074) 94€t)

+C (0+t) )=S. C10H8) dt +c (to++) 260) = Rot +)+c xlt) = 3+ (20+t) + E 20 (10+4) Ciort) Y at t=0, alo) = 1 gives c=1. is the X(t) = 3+(20tt) + 20(1077) 1 Tort 0 final solution for solution alt) at time t = Concentration of selt) lott - 2 . Given Given 10+t = ult) 20 alt) = 3 (2071+ 20(1070) lott 10+t - 3+ (20+t) + 20(107t) 20 L lott lott t 2p . 3t (@ott) + 20 20(6+t)

(10++)?= 6ot + 3t+20 100+206 +ť = 3t' + 6ot +20 34_t? + Get-aot +20-400 = 0 at +40t –80 =0 ť +20t -40 =0 Gires ta -218 & 1.832 t= 1.832 hr.