Time to Empty a Tank with Water Flowing at 2h kg/s

Manshah
Messages
1
Reaction score
0
a tank is filled with water up to its brim a hole was made at the bottom of tank find time taken to empty tank if water flows at rate of 2h kg/s where h is height of liquid column and is equal to 20m radius is equal to h/2

[Moderator's note: moved from a technical forum.]
 

Attachments

  • 16478353816156896100703618581842.jpg
    16478353816156896100703618581842.jpg
    51.7 KB · Views: 121
Physics news on Phys.org
Differential equation for change of mass in the tank for infinitesimal time change dt is
\rho dV=\rho Adh =-2h c dt
where ##\rho## is density of water ##1000 kg/m^3##, A is horizontal cross section area ##2*20*\pi ##m^2, c is constant to adjust physical dimension c = 1 kg /(sec meter). h=40 m when t=0.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top