Please copy and paste this embed script to where you want to embed

DRYERS AND DRYING 1. Tobacco in a warehouse, held at 30 C and 40% relative humidity, is placed in a room 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? What is the actual weight of this quantity of tobacco after staying in the working room? Solution:

m1 = 50 kg For tobacco at 40% RH Regain1 = 13.30% at 70% RH Regain2 = 25.00% m1 − Bdw Regain1 0.1330 Bdw = 50 − Bdw Bdw = 44.13 kg

(a) Bdw =

(b) m2 = actual weight m2 = (Regain2 )(Bdw) + Bdw = (0.25)(44.13) + 44.13 = 55.16 kg 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 drier with 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 mass flow rate of material leaving the drier, and (e) the heat requirement of drier per kg of water evaporated. Solution:

DRYERS AND DRYING

at 1, tdb1 = 29 C , t wb1 = 22 C h1 = 64.2 kJ kg W1 = 0.0138 kg kg v1 = 0.874 m3 kg at 2, tdb2 = 80 C , W2 = W1 = 0.0138 kg kg hg 2 = 2643.7 kJ kg h2 = c pt db2 +W2 hg 2 = (1.0062 )(80) + (0.0138)(2643.7 ) = 116.98 kJ kg

at 3, tdb3 = 44 C , φ3 = 80% RH pd 3 = 9.151 kPa ps 3 = φ3 pd 3 = (0.80 )(9.151) = 7.321 kPa W3 =

0.622 ps 3 pt − ps 3

=

0.622(7.321) = 0.0484 kg kg 101.325 − 7.321

(a) Mass of water removed per kg dry air = W3 − W2 = 0.0484 − 0.0138 = 0.0346 kg kg

π 2 (b) Volume flow rate of air entering the reheater = V&1 = (2) (6 ) = 18.85 m 3 s 4 (c) Mass of water evaporated = V& 18.85 (0.0346) = 0.746 kg s m& a (W3 − W2 ) = 1 (W3 − W2 ) = v1 0.875 (d) Mass flow rate of material leaving the dryer = m& 5 m& 5 (1 − 0.08) = m& 4 (1 − 0.24 )

DRYERS AND DRYING m& 4 = 1.21m& 5

but m& 4 − m& 5 = m& a (W3 − W2 ) 1.21m& 5 − m& 5 = 0.746 m& 5 = 3.552 kg s (e) Heat requirement per kg of water evaporated. m& (h − h ) h −h 116.98 − 64.2 = a 2 1 = 2 1 = = 1525 kJ kg water m& a (W3 − W2 ) W3 − W2 0.0346 3. A drier is to be designed to reduce the water content of a certain material from 55% 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 the drier 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, tdb1 = 29 C , W1 = 0.005 kg kg h1 = 42 kJ kg v1 = 0.862 m3 kg

at 2, tdb2 = 50 C , W2 = W1 = 0.005 kg kg h2 = 63.5 kJ kg

at 3, tdb3 = 38 C , φ3 = 70% RH W3 = 0.0298 kg kg

DRYERS AND DRYING m& 5 (1 − 0.10) = m& 4 (1 − 0.55) m& 5 = 1000 kg hr m& 4 =

1000(1 − 0.10 ) = 2000 kg hr 1 − 0.55

m& a =

1 hr m& 4 − m& 5 2000 − 1000 = 11.2 kg s = = (40,323 kg hr ) W3 − W2 0.0298 − 0.005 3600 s

(a) Volume flow rate of air entering the reheater = V&1 = m& av1 = (11.2 )(0.862 ) = 9.65 m3 s (b) Heat supplied in the reheater = = m& a (h2 − h1 ) = 11.2(63.5 − 42) = 240.8 kW

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 is maintained 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 the following: (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, tdb1 = 32 C , t wb1 = 21 C h1 = 60.6 kJ kg W1 = 0.0112 kg kg

at 2, tdb 2 = 45 C , W2 = W1 = 0.0112 kg kg

DRYERS AND DRYING h2 = 74.9 kJ kg v2 = 0.917 m3 kg

at 3, tdb 3 = 45 C , φ2 = 80% RH pd 3 = 9.593 kPa ps 3 = φ3 pd 3 = (0.80)(9.593) = 7.674 kPa W3 =

0.622 ps 3 pt − ps 3

=

0.622(7.674 ) = 0.0510 kg kg 101.325 − 7.674

hg 3 = 2583.2 kJ kg h3 = c ptdb3 + W3hg 3 = (1.0062 )(45) + (0.0510 )(2583.2) = 177 kJ kg

(a) Palay supplied to the dryer in kg/hr m& (1 − 0.10 ) 1000(1 − 0.10) = = 1058.8 kg hr = m& 4 = 5 1 − 0.15 1 − 0.15 (b) Temperature of the humid air from the dryer = tdb 3 = tdryer = 45 C . (c) Air supplied to dryer = V&2 = m& a v2 m& a =

m& 4 − m& 5 1058.8 − 1000 = = 1477.4 kg hr W3 − W2 0.0510 − 0.0112

V&2 = m& a v2 = (1477.4 )(0.917 ) = 1354.8 m3 h

(d) Heat supplied to heater in kW 1477.4 = m& a (h2 − h1 ) = (74.9 − 60.6 ) = 5.87 kW 3600 (e) m& s h fg = 5.87 kW m& s (1890.7 ) = (5.87 )(3600 ) m& s = 11.18 kg hr

View more...
m1 = 50 kg For tobacco at 40% RH Regain1 = 13.30% at 70% RH Regain2 = 25.00% m1 − Bdw Regain1 0.1330 Bdw = 50 − Bdw Bdw = 44.13 kg

(a) Bdw =

(b) m2 = actual weight m2 = (Regain2 )(Bdw) + Bdw = (0.25)(44.13) + 44.13 = 55.16 kg 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 drier with 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 mass flow rate of material leaving the drier, and (e) the heat requirement of drier per kg of water evaporated. Solution:

DRYERS AND DRYING

at 1, tdb1 = 29 C , t wb1 = 22 C h1 = 64.2 kJ kg W1 = 0.0138 kg kg v1 = 0.874 m3 kg at 2, tdb2 = 80 C , W2 = W1 = 0.0138 kg kg hg 2 = 2643.7 kJ kg h2 = c pt db2 +W2 hg 2 = (1.0062 )(80) + (0.0138)(2643.7 ) = 116.98 kJ kg

at 3, tdb3 = 44 C , φ3 = 80% RH pd 3 = 9.151 kPa ps 3 = φ3 pd 3 = (0.80 )(9.151) = 7.321 kPa W3 =

0.622 ps 3 pt − ps 3

=

0.622(7.321) = 0.0484 kg kg 101.325 − 7.321

(a) Mass of water removed per kg dry air = W3 − W2 = 0.0484 − 0.0138 = 0.0346 kg kg

π 2 (b) Volume flow rate of air entering the reheater = V&1 = (2) (6 ) = 18.85 m 3 s 4 (c) Mass of water evaporated = V& 18.85 (0.0346) = 0.746 kg s m& a (W3 − W2 ) = 1 (W3 − W2 ) = v1 0.875 (d) Mass flow rate of material leaving the dryer = m& 5 m& 5 (1 − 0.08) = m& 4 (1 − 0.24 )

DRYERS AND DRYING m& 4 = 1.21m& 5

but m& 4 − m& 5 = m& a (W3 − W2 ) 1.21m& 5 − m& 5 = 0.746 m& 5 = 3.552 kg s (e) Heat requirement per kg of water evaporated. m& (h − h ) h −h 116.98 − 64.2 = a 2 1 = 2 1 = = 1525 kJ kg water m& a (W3 − W2 ) W3 − W2 0.0346 3. A drier is to be designed to reduce the water content of a certain material from 55% 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 the drier 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, tdb1 = 29 C , W1 = 0.005 kg kg h1 = 42 kJ kg v1 = 0.862 m3 kg

at 2, tdb2 = 50 C , W2 = W1 = 0.005 kg kg h2 = 63.5 kJ kg

at 3, tdb3 = 38 C , φ3 = 70% RH W3 = 0.0298 kg kg

DRYERS AND DRYING m& 5 (1 − 0.10) = m& 4 (1 − 0.55) m& 5 = 1000 kg hr m& 4 =

1000(1 − 0.10 ) = 2000 kg hr 1 − 0.55

m& a =

1 hr m& 4 − m& 5 2000 − 1000 = 11.2 kg s = = (40,323 kg hr ) W3 − W2 0.0298 − 0.005 3600 s

(a) Volume flow rate of air entering the reheater = V&1 = m& av1 = (11.2 )(0.862 ) = 9.65 m3 s (b) Heat supplied in the reheater = = m& a (h2 − h1 ) = 11.2(63.5 − 42) = 240.8 kW

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 is maintained 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 the following: (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, tdb1 = 32 C , t wb1 = 21 C h1 = 60.6 kJ kg W1 = 0.0112 kg kg

at 2, tdb 2 = 45 C , W2 = W1 = 0.0112 kg kg

DRYERS AND DRYING h2 = 74.9 kJ kg v2 = 0.917 m3 kg

at 3, tdb 3 = 45 C , φ2 = 80% RH pd 3 = 9.593 kPa ps 3 = φ3 pd 3 = (0.80)(9.593) = 7.674 kPa W3 =

0.622 ps 3 pt − ps 3

=

0.622(7.674 ) = 0.0510 kg kg 101.325 − 7.674

hg 3 = 2583.2 kJ kg h3 = c ptdb3 + W3hg 3 = (1.0062 )(45) + (0.0510 )(2583.2) = 177 kJ kg

(a) Palay supplied to the dryer in kg/hr m& (1 − 0.10 ) 1000(1 − 0.10) = = 1058.8 kg hr = m& 4 = 5 1 − 0.15 1 − 0.15 (b) Temperature of the humid air from the dryer = tdb 3 = tdryer = 45 C . (c) Air supplied to dryer = V&2 = m& a v2 m& a =

m& 4 − m& 5 1058.8 − 1000 = = 1477.4 kg hr W3 − W2 0.0510 − 0.0112

V&2 = m& a v2 = (1477.4 )(0.917 ) = 1354.8 m3 h

(d) Heat supplied to heater in kW 1477.4 = m& a (h2 − h1 ) = (74.9 − 60.6 ) = 5.87 kW 3600 (e) m& s h fg = 5.87 kW m& s (1890.7 ) = (5.87 )(3600 ) m& s = 11.18 kg hr

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.