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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.

physics
 May 26, 2015

Best Answer 

 #3
avatar+27365 
+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

.

 May 28, 2015
 #1
avatar+27365 
+10

Tank drain 1

 

Tank drain 2

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

 

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

.
 May 27, 2015
 #2
avatar+27365 
+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

.

 May 28, 2015
 #3
avatar+27365 
+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
 #4
avatar+95334 
0

Thank you Alan :)

 May 28, 2015
 #5
avatar
+5

Thats brilliant. Thanks Alan. Very much appreciated.

 May 28, 2015

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