How would you calculate the time it would take to drain down a cyclinder that is 250mm diameter and 10 meters long with an orifice .025mm. Just using atmospheric pressure. Then I would like to know how much of an increase in flow would i get if I introduce some extra pressure. i.e compressed air of 2 bar. Thank you.

Guest May 26, 2015

#3**+15 **

The corresponding expression for a 2 bar applied pressure is:

$$\tau=\frac{\sqrt2A}{ag}(\sqrt{\frac{\Delta p}{\rho}+gh_0}-\sqrt{\frac{\Delta p}{\rho}})$$

Δp = 2 - 1 = 1 bar = 10^5 N/m^2 assuming 1 atmosphere = 1 bar

Assuming the liquid is water with a density of 1000 kg/m^3 then:

$${\mathtt{time}} = {\frac{\left({\frac{{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}}{{\mathtt{9.8}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{10}}}}{\mathtt{\,-\,}}{\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}}}\right)}{\left({\mathtt{3\,600}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}\right)}} \Rightarrow {\mathtt{time}} = {\mathtt{1.862\: \!986\: \!880\: \!688\: \!817\: \!1}}$$

time ≈ 1.9 years

.

Alan May 28, 2015

#1**+10 **

My expression for τ above is incorrect. It should be:

$$\tau=\frac{A}{a}\sqrt{\frac{2h_0}{g}}$$

.Alan May 27, 2015

#2**+15 **

A = pi*(0.25/2)^2 m^2

a = pi*(0.000025/2)^2 m^2

A/a = 10^8

g = 9.8 m/s^2

h0 = 10 m

so

$${\mathtt{time}} = {\frac{{{\mathtt{10}}}^{{\mathtt{8}}}{\mathtt{\,\times\,}}{\sqrt{{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{10}}}{{\mathtt{9.8}}}}}}}{\left({\mathtt{3\,600}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}\right)}} \Rightarrow {\mathtt{time}} = {\mathtt{4.529\: \!970\: \!283\: \!394\: \!941}}$$

time ≈ 4.5 years

.

Alan May 28, 2015

#3**+15 **

Best Answer

The corresponding expression for a 2 bar applied pressure is:

$$\tau=\frac{\sqrt2A}{ag}(\sqrt{\frac{\Delta p}{\rho}+gh_0}-\sqrt{\frac{\Delta p}{\rho}})$$

Δp = 2 - 1 = 1 bar = 10^5 N/m^2 assuming 1 atmosphere = 1 bar

Assuming the liquid is water with a density of 1000 kg/m^3 then:

$${\mathtt{time}} = {\frac{\left({\frac{{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}}{{\mathtt{9.8}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{10}}}}{\mathtt{\,-\,}}{\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}}}\right)}{\left({\mathtt{3\,600}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}\right)}} \Rightarrow {\mathtt{time}} = {\mathtt{1.862\: \!986\: \!880\: \!688\: \!817\: \!1}}$$

time ≈ 1.9 years

.

Alan May 28, 2015