View Full Version : math seems right but principal is confusing
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi there,\n\n\nCould someone explain (in layman terms, please...\nWhy is it that an air fan producing 2 times greater cubic feet per\nminute (c.f.m.), using 2.5 times more electicity and having 50% greater\ndiameter blades will produce LOWER wind speed (mile per hour) than a\nsmaller fan. I know the math works but i don\'t understand the\nprincipal as to why.\nbigger fan, more electicity, more air movement BUT lower wind speed\ncoming out of the fan, huh?!?\n\n\nhere are my numbers, are they ok?\nfan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20\nfan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17\n\n\nAny explanation or clarification (be gentle, i\'m new at this!) would be\nmost helpful and very\nappreciated. thanks Peter.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi there,
Could someone explain (in layman terms, please...
Why is it that an air fan producing 2 times greater cubic feet per
minute (c.f.m.), using 2.5 times more electicity and having 50% greater
diameter blades will produce LOWER wind speed (mile per hour) than a
smaller fan. I know the math works but i don't understand the
principal as to why.
bigger fan, more electicity, more air movement BUT lower wind speed
coming out of the fan, huh?!?
here are my numbers, are they ok?
fan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20
fan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17
Any explanation or clarification (be gentle, i'm new at this!) would be
most helpful and very
appreciated. thanks Peter.
Igor Khavkine
Apr4-05, 06:23 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <1112466895.588152.170310@f14g2000cwb.googlegroups .com>,\nmavava wrote:\n> Hi there,\n>\n> Could someone explain (in layman terms, please...\n> Why is it that an air fan producing 2 times greater cubic feet per\n> minute (c.f.m.), using 2.5 times more electicity and having 50% greater\n> diameter blades will produce LOWER wind speed (mile per hour) than a\n> smaller fan. I know the math works but i don\'t understand the\n> principal as to why.\n> bigger fan, more electicity, more air movement BUT lower wind speed\n> coming out of the fan, huh?!?\n>\n> here are my numbers, are they ok?\n> fan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20\n> fan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17\n>\n> Any explanation or clarification (be gentle, i\'m new at this!) would be\n> most helpful and very\n> appreciated. thanks Peter.\n\nIf you already know the flow rate (cubic feet per minute) you don\'t need\nto know how much electricity is used. You already have all the\ninformation that you need.\n\nImagine a cylinder extending from the face of the fan in the direction\nthe air is blown. It represents the column of air pushed out by the fan.\nThe cross section of the cylinder is fixed it\'s given by the area of the\nfan. A = pi*d^2/4 where d is the diameter. The flow rate is the volume\nof air that enters this cylinder per unit time T. If you know the\nvolume, you know what length of the cylinder will be occupied by it,\nV = A*L. But L is then also the length traveled by the air in the same\namount of time, hence its velocity is v = L/T.\n\nPutting it all together, you find that v = (V/T)/A = 4*(V/T)/pi/d^2,\nwhere d is the diameter of the blades and V/T is the flow rate. You\nshould be able to see that to keep the same air velocity, whenever you\ndouble the diameter, you need to quadruple the flow rate (V/T).\n\nIn your case, you increased the diameter by a factor of 1.5 while the\nflow rate increased only by a factor of 2. Hence the velocity changed by\na factor of 2/(1.5)^2 = 2/2.25 (< 1 hence a decrease). I have a certain\ndistaste for the imperial system of units, so I hope you\'ll forgive me\nfor not working with your actual numbers.\n\nHope this helps.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <1112466895.588152.170310@f14g2000cwb.googlegroups. com>,
mavava wrote:
> Hi there,
>
> Could someone explain (in layman terms, please...
> Why is it that an air fan producing 2 times greater cubic feet per
> minute (c.f.m.), using 2.5 times more electicity and having 50% greater
> diameter blades will produce LOWER wind speed (mile per hour) than a
> smaller fan. I know the math works but i don't understand the
> principal as to why.
> bigger fan, more electicity, more air movement BUT lower wind speed
> coming out of the fan, huh?!?
>
> here are my numbers, are they ok?
> fan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20
> fan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17
>
> Any explanation or clarification (be gentle, i'm new at this!) would be
> most helpful and very
> appreciated. thanks Peter.
If you already know the flow rate (cubic feet per minute) you don't need
to know how much electricity is used. You already have all the
information that you need.
Imagine a cylinder extending from the face of the fan in the direction
the air is blown. It represents the column of air pushed out by the fan.
The cross section of the cylinder is fixed it's given by the area of the
fan. A = \pi*d^2/4 where d is the diameter. The flow rate is the volume
of air that enters this cylinder per unit time T. If you know the
volume, you know what length of the cylinder will be occupied by it,
V = A*L. But L is then also the length traveled by the air in the same
amount of time, hence its velocity is v = L/T.
Putting it all together, you find that v = (V/T)/A = 4*(V/T)/\pi/d^2,
where d is the diameter of the blades and V/T is the flow rate. You
should be able to see that to keep the same air velocity, whenever you
double the diameter, you need to quadruple the flow rate (V/T).
In your case, you increased the diameter by a factor of 1.5 while the
flow rate increased only by a factor of 2. Hence the velocity changed by
a factor of 2/(1.5)^2 = 2/2.25 (< 1 hence a decrease). I have a certain
distaste for the imperial system of units, so I hope you'll forgive me
for not working with your actual numbers.
Hope this helps.
Igor
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>mavava wrote:\n> Hi there,\n>\n>\n> Could someone explain (in layman terms, please...\n> Why is it that an air fan producing 2 times greater cubic feet per\n> minute (c.f.m.), using 2.5 times more electicity and having 50%\ngreater\n> diameter blades will produce LOWER wind speed (mile per hour) than a\n> smaller fan. I know the math works but i don\'t understand the\n> principal as to why.\n> bigger fan, more electicity, more air movement BUT lower wind speed\n> coming out of the fan, huh?!?\n>\n>\n> here are my numbers, are they ok?\n> fan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20\n> fan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17\n>\n>\n> Any explanation or clarification (be gentle, i\'m new at this!) would be\n> most helpful and very\n> appreciated. thanks Peter.\n\nWe\'ll check numbers in a second. The law that pertains is sometimes\ncalled the "law of continuity", more familiarly known as the\n"conservation of stuff" or "gozinta equals gozouta".\n\nThe flux of air (c.f.m.) is the area of the fan times the velocity.\n(The area is, say, in square ft, and the velocity can be written [or\nconverted to] ft per minute, and so the product will be cubic feet per\nminute.)\n\nBut let\'s just look at ratios.\nFan 2 has a diamter that is 1.5 times the diameter of fan 1. This means\nthat it has an area 2.25 times the area of fan 2, because 1.5 squared\nis 2.25.\nFan 2 has a velocity that is 0.85 times the velocity of fan 1.\nThis means that fan 2 has a flux that is (2.25)x(0.85) = 1.9 times that\nof fan 1. This is close to the advertised flux of 2 times that of fan\n1.\n\nPut another way, sure the velocity of fan 2 is a bit lower, but the\narea of the fan is quite a bit bigger (more than twice as much), and\nthat compensates. Said yet another way, a bigger fan doesn\'t *need* to\npush the air faster to get more volume pushed through; in fact, it can\npush it somewhat slower and still get more air pushed through.\n\nThat wasn\'t so hard, was it?\n\nPD\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>mavava wrote:
> Hi there,
>
>
> Could someone explain (in layman terms, please...
> Why is it that an air fan producing 2 times greater cubic feet per
> minute (c.f.m.), using 2.5 times more electicity and having 50%
greater
> diameter blades will produce LOWER wind speed (mile per hour) than a
> smaller fan. I know the math works but i don't understand the
> principal as to why.
> bigger fan, more electicity, more air movement BUT lower wind speed
> coming out of the fan, huh?!?
>
>
> here are my numbers, are they ok?
> fan 1: c.f.m. = 15,000; blade = 40" diameter; mph = 20
> fan 2: c.f.m. = 30,000; blade = 60" diameter; mph = 17
>
>
> Any explanation or clarification (be gentle, i'm new at this!) would be
> most helpful and very
> appreciated. thanks Peter.
We'll check numbers in a second. The law that pertains is sometimes
called the "law of continuity", more familiarly known as the
"conservation of stuff" or "gozinta equals gozouta".
The flux of air (c.f.m.) is the area of the fan times the velocity.
(The area is, say, in square ft, and the velocity can be written [or
converted to] ft per minute, and so the product will be cubic feet per
minute.)
But let's just look at ratios.
Fan 2 has a diamter that is 1.5 times the diameter of fan 1. This means
that it has an area 2.25 times the area of fan 2, because 1.5 squared
is 2.25.
Fan 2 has a velocity that is .85 times the velocity of fan 1.
This means that fan 2 has a flux that is (2.25)x(0.85) = 1.9 times that
of fan 1. This is close to the advertised flux of 2 times that of fan
1.
Put another way, sure the velocity of fan 2 is a bit lower, but the
area of the fan is quite a bit bigger (more than twice as much), and
that compensates. Said yet another way, a bigger fan doesn't *need* to
push the air faster to get more volume pushed through; in fact, it can
push it somewhat slower and still get more air pushed through.
That wasn't so hard, was it?
PD
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