Question

The figure shows a cylindrical tank of base radius r=7/10 m. The height h (in m) of water in the tank is maintained by controlling the inlet volume flow rate Vi (in m^3/min) and outlet volume flow rate Vo (in m^3/min) of water. Vi and Vo are recorded during time t∈(0,6) (in min) as Vi=1/50(t^2+14t) and V0=1/50(−t^2-6t) respectively. Given that for a small time interval
V′=rate of change of volume of water in the tank
h′=rate of change of height of water in the tank
V′=π×r^2h′=inlet rate of water−outlet rate of water
If h′=k(a2t^2+a1t+a0), then ht would be h=k(a2t^3/3+a1t^2/2+a0t), where k is a constant, a0,a1,a2 are real numbers.
Find h (till two decimal places) at t=2. (Take π=22/7)

Answer 1

Answer:

sorry but need point I don't know answer

Step-by-step explanation:

sorry but need point I don't know answer

sorry but need point I don't know answer

Answer 2

given inlet and outlet volume flow rate for a cylindrical tank with radius 0.7m

Explanation:

-> given inlet volume flow and outlet volume flow denoted by Vi and Vo as,

$Vi=\frac{t^{2} +14t}{50} \\Vo=\frac{-t^{2}-6t }{50}$                                                  

$r=0.7m$            

-> given rate of change of volume of water in the tank denotes by V' is,

$V'=Vi-Vo\\from\ above\ point\ we\ know,\\V'= \frac{t^{2}+14t+t^{2}+6t }{50} \\ V'= \frac{t^{2}+10t }{25}\ \ \ \ \ \ \ \ \ \ \----(i)$

-> given rate of change of height of water in tank denoted by h' is,

$V'=\pi r^{2}h'\\h'=\frac{V'}{\pi r^{2} } \\from\ (i)\ we\ get\ ,\\h'=\frac{t^{2}+10t }{25(\frac{22}{7} )(\frac{7}{10} )}\ \ \ \ \ \ \ \ \ \ \ (taking\ \pi =\frac{22}{7},\ r=\frac{7}{10})\\h'=\frac{t^{2}+10t }{55}\ \ \ \ \ \------(ii)$

-> given is h' as,

$h'= k(a_{2}t^{2}+a_{1}t+a_{0})\ \ \ \ \ \ \ \ \ \ \ \ -----(iii)\\comparing\ (i)\ and\ (iii) we\ get,\\k=\frac{1}{55} \\a_{0}=0,\ \ \ \ a_{1}=10,\ \ \ \ a_{2}=1$

-> given height of water in the tank at a given time 't' denoted by h is,

$h=k(a_{2}\frac{t^{3} }{3}+ a_{1}\frac{t^{2} }{2} +a_{0}t)\\from\ above\ point\ we\ get,\\h= \frac{1}{55} (\frac{t^{3} }{3}+10\frac{t^{2} }{2}+0 ) \\value\ of\ h\ at\ t=2\ is,\\h=0.41m$