Hello,
For first estimation of flash dryer design,I see this in Perry's (6th ed.) 20-54 (with refer to 20-53):
Q = ua.V.(delta tm) (eq. 20-35,but applicable for flash dryer,as mentioned)
Ua = 0.5 G**0.67/D (eq. 20-36,but applicable for flash dryer,as mentioned)
(first of page 20-54)For estimating purposes...(in summary):
1.Assume Vair and Qair then gain Dpipe. ( I:Vair=20-30 m/s but what is the rough number for Qair?)
2. By use of eq. 20-35 ,gain Ua.(ok)
3. By use of eq. 20-36 , gain V ,volume of dryer.(Do I gain Q with required evaporation rate?)
Finally, in page 20-53,I think figures are incompatible:
G=4.8 Kg/s. sqm
D=0.3 m
Then: Ua= 2235 J/cu m .s.K
But in SI eq. 20-36 became: Ua= (26)G**0.67/D
Can anybody clarify me?(with a good example)
Thanks
For first estimation of flash dryer design,I see this in Perry's (6th ed.) 20-54 (with refer to 20-53):
Q = ua.V.(delta tm) (eq. 20-35,but applicable for flash dryer,as mentioned)
Ua = 0.5 G**0.67/D (eq. 20-36,but applicable for flash dryer,as mentioned)
(first of page 20-54)For estimating purposes...(in summary):
1.Assume Vair and Qair then gain Dpipe. ( I:Vair=20-30 m/s but what is the rough number for Qair?)
2. By use of eq. 20-35 ,gain Ua.(ok)
3. By use of eq. 20-36 , gain V ,volume of dryer.(Do I gain Q with required evaporation rate?)
Finally, in page 20-53,I think figures are incompatible:
G=4.8 Kg/s. sqm
D=0.3 m
Then: Ua= 2235 J/cu m .s.K
But in SI eq. 20-36 became: Ua= (26)G**0.67/D
Can anybody clarify me?(with a good example)
Thanks
