A Dryer is to deliver 1000 kg/hr of palay with a final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric condition with 32°C dry bulb and 21 degrees centigrade wet bulb. The dryer is maintained 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 heat supplied by heater in kw. C. 4.23 A. 3.23 B. 5.46 D. 6.23

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Draw the schematic diagram completely

A Dryer is to deliver 1000 kg/hr of palay with a final moisture content
of 10%. The initial moisture content in the feed is 15% at atmospheric
condition with 32°C dry bulb and 21 degrees centigrade wet bulb. The
dryer is maintained 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 heat supplied by heater in kw.
A. 3.23
C. 4.23
D. 6.23
B. 5.46
SOLUTION:
Let m amount of palay in wet feed
Solid in wet feed = solid in product
0.85(m) 0.90(1000)
=
m = 1,058.832 kg/hr
From psychrometric chart:
h₁ 60.5 KJ/kg
h₂ = 74 KJ/kg
h3
196 KJ/kg
W₁ = W₂ = 0.0111 kg/kg
W3 = 0.0515 kg/kg
V₂ = 0.915 m³/kg
=
Amount of moisture removed 1058.823 1000
Amount of moisture removed
=
58.823 kg/hr
=
ma(w3w₂) 58.823
m₂(0.0515-0.0111) = 58.823
-
ma 1456.015 kg/hr
Heat supplied by heater = m, (h₂ - h₁)
Heat supplied by heater = (1456.015/3600)(74 - 60.5)
Heat supplied by heater 5.46 kw
Transcribed Image Text:A Dryer is to deliver 1000 kg/hr of palay with a final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric condition with 32°C dry bulb and 21 degrees centigrade wet bulb. The dryer is maintained 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 heat supplied by heater in kw. A. 3.23 C. 4.23 D. 6.23 B. 5.46 SOLUTION: Let m amount of palay in wet feed Solid in wet feed = solid in product 0.85(m) 0.90(1000) = m = 1,058.832 kg/hr From psychrometric chart: h₁ 60.5 KJ/kg h₂ = 74 KJ/kg h3 196 KJ/kg W₁ = W₂ = 0.0111 kg/kg W3 = 0.0515 kg/kg V₂ = 0.915 m³/kg = Amount of moisture removed 1058.823 1000 Amount of moisture removed = 58.823 kg/hr = ma(w3w₂) 58.823 m₂(0.0515-0.0111) = 58.823 - ma 1456.015 kg/hr Heat supplied by heater = m, (h₂ - h₁) Heat supplied by heater = (1456.015/3600)(74 - 60.5) Heat supplied by heater 5.46 kw
A Dryer is to deliver 1000 kg/hr of palay with a final moisture content
of 10%. The initial moisture content in the feed is 15% at atmospheric
condition with 32°C dry bulb and 21 degrees centigrade wet bulb. The
dryer is maintained 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 heat supplied by heater in kw.
A. 3.23
C. 4.23
D. 6.23
B. 5.46
SOLUTION:
Let m amount of palay in wet feed
Solid in wet feed = solid in product
0.85(m) 0.90(1000)
=
m = 1,058.832 kg/hr
From psychrometric chart:
h₁ 60.5 KJ/kg
h₂ = 74 KJ/kg
h3
196 KJ/kg
W₁ = W₂ = 0.0111 kg/kg
W3 = 0.0515 kg/kg
V₂ = 0.915 m³/kg
=
Amount of moisture removed 1058.823 1000
Amount of moisture removed
=
58.823 kg/hr
=
ma(w3w₂) 58.823
m₂(0.0515-0.0111) = 58.823
-
ma 1456.015 kg/hr
Heat supplied by heater = m, (h₂ - h₁)
Heat supplied by heater = (1456.015/3600)(74 - 60.5)
Heat supplied by heater 5.46 kw
Transcribed Image Text:A Dryer is to deliver 1000 kg/hr of palay with a final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric condition with 32°C dry bulb and 21 degrees centigrade wet bulb. The dryer is maintained 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 heat supplied by heater in kw. A. 3.23 C. 4.23 D. 6.23 B. 5.46 SOLUTION: Let m amount of palay in wet feed Solid in wet feed = solid in product 0.85(m) 0.90(1000) = m = 1,058.832 kg/hr From psychrometric chart: h₁ 60.5 KJ/kg h₂ = 74 KJ/kg h3 196 KJ/kg W₁ = W₂ = 0.0111 kg/kg W3 = 0.0515 kg/kg V₂ = 0.915 m³/kg = Amount of moisture removed 1058.823 1000 Amount of moisture removed = 58.823 kg/hr = ma(w3w₂) 58.823 m₂(0.0515-0.0111) = 58.823 - ma 1456.015 kg/hr Heat supplied by heater = m, (h₂ - h₁) Heat supplied by heater = (1456.015/3600)(74 - 60.5) Heat supplied by heater 5.46 kw
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