1848.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



19 



■ The mean 1-008 



By this paper it is intended to be shown that there are certain 

 velocities witli wliieh tlie tips of the vanes of a fan should move 

 according to the recpiired density of air, and that there are certain 

 laws whidi p;overn these velocities. 



First. — W'ater is 827 times lieavier than air ; mercury is 13'5 

 heavier than water: consequently, mercury is 1116+ heavier than 

 air. A column of mercury, one inch in height, would tlierefore 

 balance a column of air 1116+ inches, or 930-3 feet in heiglit. Let 

 A be a column of mercury equal in height to any given density, 

 and let B represent 930-3, and C 64.'' ; then V (A X BX C) = V or 

 the velocity that a body would acquire in falling the heiglit of a 

 column of air equivalent to the density. 



Scrotifl. — The centrifugal force of air coincides with the results 

 obtained hv the laws of falling bodies, that is when the velocity is 

 the same as the velocity which a body will acquire in falling the 

 height of a homogeneous column of air equivalent to any gi\en 

 density. To obtain the centrifugal force or density of air apply 

 the following general rule. 



Having given the ^ elocity of the air, and the diameter of the 

 fan, to ascertain the centrifugal force, — 



RrLE.^ — Divide the velocity by 401, and again divide the square 

 of the quotient by the diameter of the fan. This last quotient 

 multiplied by the weight of a cubic foot of air, at 60^ Fahrenheit, 

 is equal to the force in ounces per square foot, which, divided by 

 144, is ecjual to the density of air per square inch. Or, substituting 

 the following formula, we ha\e 



D = N V X -000034 

 where D is the density of the air in ounces per square inch, and 

 N the number of revolutions of fan per minute, and V the velocity 

 of the tips of the fan in feet per second. 



Let us now compare the results of the foregoing table. To do 

 this, we will first take the velocity of the tips of vanes per second, 

 and the power necessary to drive the fan. We will first take Nos. 

 1, 2, 3, 4, 5, and 6, and' we shall find by inspecting tlie table that 

 the corresponding velocities to these numbers are 236-8, 220-8, 202-1, 

 185-2, 171-5. and 144-1, and the corresponding densities of air per 

 square inch are 9-4, 7-9, 6-9, 5-G, 4-5, and 3-5 ounces. Tlie fan, it 

 must be understood, is discharging no air ; the velocity of the fan 

 is merely keeping the air at a certain density or pressure per square 

 inch. Under these circumstances, it requires a certain \elocity of 

 the tips of the fan to maintain a certain density of air, but not in a 

 direct ratio. 



The law which governs the velocity of the tips of the fan ap- 

 pears from tliese experiments to be -^ of the velocity a body would 

 acquire in falling the height of a homogeneous column of air equi- 

 valent to the density. This we have called the theoretical velo- 

 city, and by comparing Nos. 1, 2, 3, 4, 5, and 6 experiments as 

 above, that is, by comparing the velocity of the tips of the fan per 

 second with ^ of the theoretical velocity, we shall find them to 

 agree toleralily near. Thus, if the velocity of the tips of the fan 

 per second be represented by 1, then ^ of the theoretical \elocity 

 will be represented by 



1004 in No. 1 experiment, 



-986 2 



1-008 3 



•990 4 



•960 5 



1^0007 6 



But we shall not only find that the ^a. of theoretical veloei^j' go- 

 verns the fan when it is not discharging air, but that the theoreti- 

 cal velocity governs it also when the outlet ]iipe is open ; that is, 

 that the maximum eifect of the fan is when tlie vanes nuive from 

 the theoretical velocity to ^ of that velocity due to the density of 

 the air, that tlie greatest quantity of air is discharged by the fan 

 under tliese conditions with the least expenditure of power. To 

 illustrate this more fully, let us refer to the table of experiments, 

 and for our example we 'will take Nos. 9, 10, and 11 ; here the den- 

 sity in each is six ounces. In No. 10 the velocity of the tips of 

 the vanes is 213-33 feet per second, w hile the theoretical velocity 

 is 211 feet per second, being nearly the same. The quantity of air 

 discharged is 77-9 cubic feet per second, and the power employed 

 in this case amounts to 12-5 horses. 



We take now No. 11 experiment. Here the velocity of the tips 

 of the fan is 192 feet per second, and -^ of the theoretical velocity 

 190 feet per second. Now these two experiments are in proportion 

 to each other nearly, viz., in No. 11 the quantity of air discharged 

 amounts to 35-7 cubic feet per second, and takes 6-4 horse power, 

 while No. 10 discharges 77-9 cubic feet per second, and takes 12-5 



* The space which a gravitating body will pass through in one second is IByj feet; 

 but by the principle of accelerating forces, the velocity of a falling body in any given 

 time is equal to tivice the space through which it has passed in that lime, or the velocity 

 is equal to the square root ol the number obtained by multiplying 64 by the height in feet. 



horse-power. Thus the discharge of air is nearly 2 to 1, and the 

 horse-power emjiloyed in the same proportion. 



In the following examples we shall call the theoretical velocity 

 per second uinty, beginning with No. 15. In this example we 

 shall also call the quantity of air discharged in cubic feet per second 

 unity, and also the horse-power. 



No. 



To give a further illustration of this part of our subject, we will 

 take Nos. 7, 9, 12, and 16 experiments. Here tlie velocity of the 

 tips of the fan is the same, which we shall denote unity. The cor- 

 responding densities are 7, 6, 5, and 4 ounces ; we shall call the 

 highest unity, also the cubic feet discharged per second, and the 

 horse-power. 



Nearly all the preceding examples justify our conclusion, that 

 the greatest results are obtained when the theoretical velocity and 

 the tips of the vanes are nearly equal. It carries its own convic- 

 tion that if we increase the velocity of the tips of the vanes, and 

 only double the cubic quantity of air delivered, that it must take 

 more than double the expenditure of power, the density of air re- 

 maining the same. 



We shall now give examples of the data dictated by our table of 

 experiments. And first, having given the density of air per square 

 inch to determine the velocity of the tips of the vanes per second ; 

 also the horse power requisite to drive the fan under these circum- 

 stances, the fan not discharging air, but its velocity merely keeping 

 the air at a certain density. 



Let D denote the density of the air in ounces per square inch, 

 and A a column of mercury equivalent in height to that densitv. 

 Then by the laws of falling bodies V (A X 930-3 X 64) = V the ve- 

 locity acquired by a body falling through a column of air of the 

 corresponding density. 



qo V T) 



Then =r P the number of pounds acting on the vanes, 



9 V V 60 P 



and ^- — ^ '- ^ H. P. or horse-power required. 



33000 '^ 



The constant number 38 is obtained by the following formula. 



H P X 33000 



P. Then 



PX 16 



; 38 



^ V X 60 ■ D 



Example. — Let D := 9-4 oz. per square in., and A =i 1-175 in. of 

 mercury, to determine the velocity of the tips of the vanes per 

 second, and also the horse-power. 



Then V (930-3 X 64 X 1-175) = 264-4, the theoretical velocity, 

 ^ of which is =: 237-96 = V, or velocity of tips of vanes per sec. 



qo V Q.J, 



Now ^^ — = 22-32 =: P, or pounds acting on the vanes of fan. 



Then- 



16 

 237-96 X 60 



X 22-32 



:= 9-6 the horse-power reciuired. 



33000 



Having given the velocity of the air in feet per second (or as it 

 has been termed the theoretical velocity) to determine the density 

 of the air in accordance with the laws of centi-ifugal force. 



Let the velocity be 264-4 feet per sec, and the diameter of the 

 fan 3-9 feet. Then by former rules we have 



264-4 ^„ ,66-2^ „„. 11169 X 1-209 _ 



= 66-2 and--— = 11169 = and —, = 



4-01 3-9 144 



9' ounces density, the answer required. 



Or by the second rule, take the velocity of the fan in feet per 

 second, multiplied by the number of revolutions of the fan per 

 minute, the product multiplied by -000034 — the density required. 

 Here we must remark, that according to our table of experi- 

 ments, that when the tips of the vanes are to move at .f-^ of the 

 theoretical velocity, that not more than 220 lb.[of air are discharged 

 per minute ; but this is without any. attenuation in the density. 



4* 



