VEGETABLE AND FRUIT DEHYDRATION 105 



3. For a given fan size and a constant static pressure when the specific volume 

 of the air varies: 



(r. p. m, Q, hp.)cc(v^ .A' ^ 



4. For a given fan size, capacity and speed when the specific volume of air 

 varies: 



(hp.,P8)0C^,S, 1) 



5. For a given fan size and a constant amount of air by weight when the specific 

 volume of air varies: 



a. (^,r.p.m.,P S )oc^F,i, T\ 



b. (lip>(7^ 2 ,T 2 ) 



In order to apply these fan laws it is necessary to understand the 

 nature of the static pressure load of the fan as governed by the re- 

 sistance of a given dehydrator system. It is therefore well to add 

 the following laws .: 



6. For a constant specific volume of air, the static pressure of a given 

 dehydrator system varies directly as the square of the air velocity. 



Ps oc U\ when V= constant. 



Ps= static pressure of the dehydrator system and fan, 



inches of water. 

 £7= the air velocity either in the dehydrator or at the 



fan outlet, feet per minute. 

 V= specific volume of air, cubic feet per pound. 



7. For a given dehydrator system and a constant amount of air by weight 

 when the specific volume of air varies, the static pressure of the system changes 

 directly as the specific volume. 



Ps cc V, when W= constant. 



TF=total weight of air per unit of time, pounds per minute. 



A comparison of laws 6 and 7 with the fan laws shows that Nos. 2 and 5 may 

 be directly applied to any given dehydrator system. Their use is best demon- 

 strated by specific examples as follows : 



Example 1. — Assume that a fruit-drying tunnel is to be converted into a vege- 

 table dehydrator. The measured air velocity in the drying section is 400 feet 

 per minute. The observed fan speed and power are 700 r. p. m. and 10 hp., 

 respectively. If the same fan is used and the specific volume of air remains un- 

 changed, what fan speed is needed to furnish an air velocity of 600 feet perminute 

 in the drying section? How much power will the fan absorb at the new operat- 

 ing condition? 



Applying law 6 the static pressure of the dehydrator system and fan both vary 

 with the square of the air velocity. Therefore (from 6 and 2b) the static pres- 

 sure at this new condition will be (600/400) 2 =2.25 times the original static 

 pressure. The new fan speed must be 700X (600/400) =1,050 r. p. m. since (from 

 2a) the capacity is directly proportional to fan speed. Likewise, the fan power 

 for this condition (from 2c) is 10 X ( 500/400) 3 =33.75 hp. 



Example 2. — Suppose 4,000 pounds of air per minute is necessary for the 

 removal of moisture from a given dehydrator ; the specific volume of air at the 

 fan is 16.7 cubic feet per pound and the static pressure of the dehydrator system 

 using standard air conditions is 1.45 inches of water. If standard fan tables are 

 used, what size fan shall be chosen for this dehydrator? What is the correct fan 

 speed and how much power will the fan absorb? Assume there is no air leakage. 



The specific volume of standard air is approximately 13.3 cubic feet per pound 

 of air. Therefore the rated capacity of the fan will be 4,000 X 13.3, or 53,200 

 c. f . m. The corresponding rated static pressure will be 1.45 inches of water. 

 Using the standard fan tables, assume it is found that a fan of suitable dimen- 

 sions operates at a speed of 475 r. p. m. to deliver the rated volume of air against 

 the specified static pressure and the table lists fan power as 19 hp. 



