Dryers and Drying

8/3/2019 Dryers and Drying

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DRYERS AND DRYING

1. Tobacco in a warehouse, held at 30 C and 40% relative humidity, is placed in aroom at 32 C and 70% relative humidity preparatory to being worked on. For each

50 kg of tobacco moved from the warehouse, what is the bone-dry weight? Whatis the actual weight of this quantity of tobacco after staying in the working room?

Solution:

kgm 501 =

For tobacco at 40% RH

%30.131 =gain Re

at 70% RH

%00.252 =gain Re

(a)1

1

gain Re

Bdwm

Bdw

−

=

Bdw Bdw −= 501330.0

kg Bdw 13.44=

(b) 2m = actual weight

( )( ) ( )( ) kg Bdw Bdwgain Rem 16.5513.4413.4425.022 =+=+=

2. Air enters an adiabatic drier at 6 m/s through a 2-m diameter duct at 29 C dry bulb

and 22 C wet bulb temperatures. It is heated to 80 C before reaching the material

to be dried and leaves the drier at 44 C and 80% RH. The material enters the drierwith a moisture content of 24%, and leaves with a moisture content of 8%.Determine (a) the mass of water removed per kg of dry air, (b) the volume flow of

air entering the reheater, (c) the kg of water evaporated per second, (d) the massflow rate of material leaving the drier, and (e) the heat requirement of drier per kg

of water evaporated.

Solution:



DRYERS AND DRYING

at 1, C t db 291= , C t wb 22

1=

kgkJ h 2.641 =

kgkgW 0138.01 =

kgmv 3

1 874.0=

at 2, C t db 802= , kgkgW W 0138.0

12 ==

kgkJ hg 7.26432 =

( )( ) ( )( ) kgkJ hW t ch gdb p 98.1167.26430138.0800062.1222 2=+=+=

at 3, C t db 443= , %80

3 =φ RH

kPa pd 151.93=

( )( ) kPa p p d s 321.7151.980.033 3 ===φ

( )kgkg

p p

pW

st

s0484.0

321.7325.101

321.7622.0622.0

3

3

3 =

−

=

−

=

(a) Mass of water removed per kg dry air = kgkgW W 0346.00138.00484.023 =−=−

(b) Volume flow rate of air entering the reheater = ( ) ( ) smV 32

1 85.18624

==π

&

(c) Mass of water evaporated =

( ) ( ) ( ) skgW W v

V W W ma 746.00346.0

875.0

85.1823

1

123 ==−=−

&

&

(d) Mass flow rate of material leaving the dryer = 5m&

( ) ( )24.0108.0145 −=− mm &&



DRYERS AND DRYING

5421.1 mm && =

but ( )2354 W W mmm a −=− &&&

746.021.155 =−mm &&

skgm 552.35=&

(e) Heat requirement per kg of water evaporated.

( )( )

water kgkJ W W

hh

W W m

hhm

a

a 15250346.0

2.6498.116

23

12

23

12=

−=

−

−=

−

−=

&

&

3. A drier is to be designed to reduce the water content of a certain material from55% to 10%. Air at 29 C dry bulb temperature and with a humidity ratio of 0.005

kg/kg is heated to 50 C in a reheater before entering the drier. The air leaves thedrier at 38 C with 70% relative humidity. On the basis of 1000 kg of product per

hour, calculate (a) the volume flow rate of air entering the reheater, and (b) the

heat supplied in the reheater.

Solution:

At 1, C t db 291= , kgkgW 005.01 =

kgkJ h 421 =

kgmv 3

1862.0=

at 2, C t db 502= , kgkgW W 005.0

12 ==

kgkJ h 5.632 =

at 3, C t db 383= , %70

3 =φ RH

kgkgW 0298.03 =



DRYERS AND DRYING

( ) ( )55.0110.0145 −=− mm &&

hr kgm 10005 =

&

( )hr kgm 2000

55.01

10.0110004 =

−

−=&

( ) skgs

hr hr kg

W W

mmma 2.11

3600

1323,40

005.00298.0

10002000

23

54=

=

−

−=

−

−=

&&&

(a) Volume flow rate of air entering the reheater = ( )( ) smvmV a

3

11 65.9862.02.11 === &&

(b) Heat supplied in the reheater = ( ) ( ) kW hhma8.240425.632.11

12 =−=−= &

4. A dryer is to deliver 1000 kg/hr of palay with final moisture content in the feed is

15% at atmospheric condition with 32 C dry bulb and 21 C wet bulb. The dryer ismaintained at 45 C while the relative humidity of the hot humid air from the dryer

is 80%. If the steam pressure supplied to the heater is 2 MPa, determine thefollowing:

(a) Palay supplied to the dryer in kg/h.(b) Temperature of the hot humid air from the dryer in C.

(c) Air supplied to dryer in cu m/h.(d) Heat supplied by the heater in kW.

(e) Steam supplied to heater in kg/h.

Solution:

at 1, C t db 321= , C t wb 21

1=

kgkJ h 6.601 =

kgkgW 0112.01 =

at 2, C t db 452= , kgkgW W 0112.0

12 ==



DRYERS AND DRYING

kgkJ h 9.742 =

kgmv 3

2917.0=

at 3, C t db 453= , %80

2 =φ RH

kPa pd 593.93=

( )( ) kPa p p d s 674.7593.980.0

33 3 ===φ

( )kgkg

p p

pW

st

s0510.0

674.7325.101

674.7622.0622.0

3

3

3 =

−

=

−

=

kgkJ hg 2.25833 =

( )( ) ( )( ) kgkJ hW t ch gdb p 1772.25830510.0450062.1333 3=+=+=

(a) Palay supplied to the dryer in kg/hr

= ( ) ( ) hr kgmm 8.105815.01

10.01100015.01

10.0154 =

−

−=

−

−= &&

(b) Temperature of the humid air from the dryer = C t t dryer db 453

== .

(c) Air supplied to dryer = 22 vmV a&& =

hr kgW W

mmma

4.14770112.00510.0

10008.1058

23

54=

−

−=

−

−=

&&&

( )( ) hmvmV a

3

228.1354917.04.1477 === &&

(d) Heat supplied to heater in kW

( ) ( ) kW hhma 87.56.609.743600

4.147712 =−

=−= &

(e) kW hm fgs 87.5=&

( ) ( )( )360087.57.1890 =sm&

hr kgms18.11=&

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