Spray Dryer
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Transcript of Spray Dryer
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Energy Balance
PROCESS CONDITIONS
Solid Components at 303.15 K
Total Mass Flow Rate 10987.18592 kg/day
Mass of BaSO4 6776.159157 kg/day
Mass of ZnO 50.07507506 kg/day
Mass of ZnS 3188.78078 kg/day
Inlet Temperature (T1) 50 0C 323.15 K
Outlet Temperature (T2) 125 0C 398.15 K
Air at 523.15 K
Mass Flow Rate 1269266300 kg/day
Constant Pressure Heat Capacity 1.0407388 kJ/kg-K
Inlet Temperature (t1) 515 0C 788.15 K
Outlet Temperature (t2) 335 0C 608.15 K
For the calculation of wet bulb temperature, for spray dryers, Nt is in general between 0.25 to 1. (Land,
2012). Assume the value of Nt =1
= ln (
)
1 = ln (788.15 608.15
)
= 503.3941928
Calculating the logarithmic mean temperature difference:
=( ) ( )
ln (
)
=(788.15 503.3942) (608.15 503.3942)
ln ((788.15 503.3942)608.15 503.3942 )
= 180
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CALCULATION OF HEAT DUTY
Calculation of Heat Duty:
i. Sensible heat required to bring the solids to the final temperature:
Assume: Solids before entering the dryer are at 500C:
Solving for the calorific temperature of the solid components:
=1 + 2
2=
323.15 + 398.15
2
= 360.65
Using Perrys Handbook Cp solid components @ 360.65 K: (Perry & Green, 2008)
Cp of BaSO4 6.144707242 kJ/kg-K
Cp of ZnO 2.445455874 kJ/kg-K
Cp of ZnS 2.709484455 kJ/kg-K
1 = ( )
1 = (6776.159157
) (6.144707242
kJ
kg K) (398.15 323.15)
+ (50.07507506
) (2.445455874
kJ
kg K) (398.15 323.15)
+ (3188.78078
) (2.709484455
kJ
kg K) (398.15 323.15)
3717841.093
= 154910.0455
ii. Sensible heat to raise the temperature of water from initial (30 C) to vaporization temperature (50C
323.15 K)
Using Perrys Handbook, Cp of liquid water
T Cp
323.15 K 4.185257778 kJ/kg K
2 = ( )
-
2 = (972.1709048
) (4.185257778
kJ
kg K) (373.15 323.15)
2 = 203439.292
= 8476.637168
iii. Latent heat to vaporize the moisture:
Using Perrys Handbook, latent heat of vaporization at 373.15 K (1000C):
T Latent heat
373.15 K 2258.234 kJ/kg
3 = ().
3 = (971.6848193
) (2258.234
kJ
kg)
3 = 2194291
= 91428.80268
iv. Sensible heat to raise temperature of water from vaporization temperature to final product temperature
(1250C):
Solving for the calorific temperature of solid component:
=1 + 2
2=
373.15 + 398.15
2= 385.65 = 112.50
Using Perrys Handbook, Cp of liquid water
T Cp
385.25 K 4.235077 kJ/kg K
4 = ( ).
4 = (972.1709048 971.6848193)(4.235077)(398.15 373.15)K
4 = 51.46523735
= 2.144384889
Total heat requirement of the system:
= 1 + 2 + 3 + 4
= (3717841 + 203439 + 2194292 + 51.4652) = 6115623.547
-
Using Ambient Temperature of 26 C and RH = 89% (Yesterday's and last week's weather in manila, 2015)
Using Psychometric Chart from Perrys Chemical Engineers Handbook:
Humidity = 0.01901 kg of water per kg of dry air
Taking the humid heat of Air:
= 1.005 + 1.88
= 1.005 + 1.88(0.01901)
= 1.0407388
Calculating for the amount of air needed.
=
=
6115623.547
1.0407388
(788.15 608.15)
= 1057721917
For the humidity of outlet air:
=971.6848193
1057721917+ 0.01901
kg water
kg air
= 00.019010919 kg water
kg air
Allotting an allowance of 20%, for the possible heat losses and due to startup, shutdown and cleaning:
(Land, 2012)
= 1.20 (1057721917
)
= 1269266300