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Board Exam
on Dryers
DRYER Board exam problems
 Drier 5 booklet BP Oct 181 A tower-type, moist air dryer is to deliver
1000 kg/hr of cassava flour with 2% residual moisture of 20% in the feed.
The air to be heated in the heating chamber is a mixture of fresh air at 92
deg.F dry bulb and 65% relative humidity is heated by steam coil to 200
deg.F. The dryer is properly insulated so that moisture absorption can be
adiabatic. Compute: (1) Required flow of heated air mixture to dryer in
pounds per minute. ; (2) Capacity forced draft fan on dryer in cu.mhr ; (3)
Heat in kcal/hr for heating the the air mixture in heating chamber. ; (4)
Percentage of weight of fresh air in mixture.
 D-2 BP APRIL , 1982
 Eight hundred pounds of dry air per hour is saturated at 115 deg.F, then
heated to 224 deg.F, passed through an adiabatic drier and discharged at
142 deg.F. Determine each of the following: (1) Total pounds of water in air
before heating, per hour. ; (2) Final pounds of water in air when
discharged, per hour. ; Water evaporated from material in the drier per
hour ; (4) Relative humidity of air leaving the drier ; (5) Dewpoint of air
leaving the drier ; (6) Percentage of saturation of air leaving the drier ; (7)
Number of cubic feet of air per min. before drying
 D-3 BP APRIL , 1983
 The temperature of air in the dryer is maintained constant by the use of
steam coils within the dryer. The product enters the dryer at the rate of
one metric ton per hour. The initial moisture content is 3 kilograms
moisture per kilogram of dry solid. Air enters the dryer with a humidity
ratio of 0.016 kilogram moisture per kilogram dry air and leaves with a
relative humidity 100% while the temperature remains constant at 60
deg.C. If the total pressure of the air is 101.3 Kpa, determine: (a) Total
required amount of air in kh/hr under entrance condition. ; (b) Capacity of
the forced draft fan to handle this air in cu.m/min.
 D-4 BP NOVEMBER , 1983
 A rotary dryer fired with bunker oil of 10,000 kcal/kg HHV is to produce 20
metric tons per hour of dried sand with 0.5% moisture from wet feed
containing 7% moisture, specific heat of sand is 0.21 BTU/lb per deg.F,
temperature of wet feed is 30 deg.C and temperature of dried product is 115
deg.C.
 QUESTION:
 1. Calculate the weight in kg. of wet feed per hour
 2. Calculate the weight of water to be removed per hour in kilograms.
 3. Calculate heat in BTU/hour required
 4. Calculate liters of Bunker oil per hour if specific gravity of bunker oil is 0.90
and dryer efficiency is 60%
 D-6 BP APRIL , 1985
 A rotary dryer produces 20 metric tons per hour of dried sand containing 0.5% moisture
from a wet feed containing 10% moisture. Temperature of wet sand is 30 deg.C and that
of dried sand is 114 deg.C. Fuel used per hour of bunker oil is 275 liters, with a HHV of
39,000 BTU/liter. Specific heat of sand is 0.21 BTU/lb-deg.Centrigrade Neglecting
radiation loss, calculate the efficiency of the sand dryer. Specific gravity of oil is 0.9.
 Required:
 Feed rate in kg/hr
 Volume flow rate in the dryer
 Heat Supplied to the dryer
 Efficiency of the sand dryer
 D-7 BP OCTOBER , 1985
 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 deg.C dry bulb and 21 deg.C wet bulb. The dryer is maintained at 45 deg.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.abs., determine the following:
 1] Palay supplied to the dryer in kg/hr
 2] Temperature of the hot humid air from the dryer in deg.C
 3] Air supplied to dryer in cu.m/hr
 4] Heat supplied by the heater in KW
 5] Steam supplied to heater in kg/hr
 D-9 BP OCTOBER , 1990
 Wet material containing 215% moisture (dry basis) is to be dried at the rate
of 1.5 kg/sec in a continuous dryer to give a product containing 5%
moisture (wet basis_. The drying medium consist of air heated to 373 K and
containing water vapor equivalent to a partial pressure of 1.40 kPa. The air
leaved the dryer at 310 K and 70% saturated. Calculate how much air will
be required to remove the moisture.
 D-10 BP APRIL , 1988
 An assembly hall was to have an air conditionaing unit installed which
would be maintained at 26 deg.C dry bulb at 50% relative humidity. The
unit delivers air at 15 deg.C bulb temperature and the calculated sensible
heat load is 150 KW and latent heat load is 51.3 KW. Twenty percent by
weight of extracted air is made up of outside air at 34 deg.C dry bulb and
60% relative humidity while 80% is extracted by the air conditioner’s
refrigertaion capacity in Tons and its ventilation load in KW.
 D-11 BP OCTOBER , 1981
 A continuous dryer is supplied with air at 21 deg.C dry bulb and 16 deg.C
wet bulb heated to 75 deg.C. It enters the heating coils at a velocity of 366
m/min and the size of the circular duct is 183 cm. The condition of the air
leaving the dryer is 43 deg.C dry bulb and 70% relative humiidity. The
moisture content of the entering material is 24% while the moisture
content of the leaving material is 8%. Calculate:
 [1] Amount of moisture evaporated
 [2] Capacity of dryer
 [3]Heat supplied to air
 D-12Twenty five metric tons of dessicated coconut meat containing 40%
moisture is to be dried to 10% final moisture in 24 hours. Atmospheric air
at 26 deg.C dry bulb and 40% relative humidity is first preheated to proper
temperature before reaching the product. The air leaves the dryer at 29
deg.C dewpoint temperature and 90% relative humidity. Neglecting the
heat capacity of the bone dry stock and the heat needed to heat the copra
tracks, calculate:
 [1] The proper temperature into which the air be preheated
 [2] The production rate in M.T. per hour of copra
 [3] The intake air flow rate in cu.m/sec
 [4] The mass flow rate of steam required in the preheated. If steam used
for heating is at 80 deg.C dry and saturated condition.
Solutions
Solution: A. Flow of Heated air
𝑀 𝑇𝑎
= 𝑀 𝑎 + 𝑀 𝑎 𝑆𝐻1
𝑀 𝑇𝑎
= 𝑀 𝑎 1 + 𝑆𝐻1 .. (equation 1)
Solving for the dry air flow:
𝑀 𝑎 =
𝐹 −𝑃
𝑆𝐻3 −𝑆𝐻2
BDM is constant: BDMF = BDMP
F(1 - % Mf) = P(1 - %Mp)
F =
𝑃 ( 1 −%𝑀𝑝)
1 −%𝑀𝑓
=
1000
𝑘𝑔
ℎ𝑟
(1 −0.021)
( 1 −0.20)
F = 1223.75
𝑘𝑔
ℎ𝑟
By equation 1: Humid Air Flow Rate
𝑀 𝑇𝑎
= 𝑀 𝑎 1 + 𝑆𝐻1
Solve first for 𝑀 𝑎:
𝑀 𝑎 =
𝐹 −𝑃
𝑆𝐻3 −𝑆𝐻2
𝑀 𝑎 =
1223.75
𝑘𝑔
ℎ𝑟
−1000
𝑘𝑔
ℎ𝑟
0.047 −0.028
𝑀 𝑎 = 11776.32
𝑘𝑔
ℎ𝑟
90DB
100℉ 120℉
200℉
Dry air 𝑆𝐻3
𝑆𝐻1 = 𝑆𝐻
By equation 1: Humid Air Flow Rate
𝑀 𝑇𝑎
= 𝑀 𝑎 1 + 𝑆𝐻1
Where; 𝑀 𝑎 = 11776.32
𝑘𝑔
ℎ𝑟
𝑆𝐻1 = 𝑆𝐻2 = 0.028
𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
𝑘𝑔 𝐷𝐴
Subs the values:
𝑀 𝑇𝑎
= 11776.32
𝑘𝑔
ℎ𝑟
(1 + 0.028
𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
𝑘𝑔 𝐷𝐴
)
𝑀 𝑇𝑎
= 12106.056
𝑘𝑔 ℎ𝑢𝑚𝑖𝑑 𝑎𝑖𝑟
ℎ𝑟
Therefore, flow of heated air
𝑀 𝑎 = 11776.32
𝑘𝑔
ℎ𝑟
𝑀 𝑇𝑎
= 12106.056
𝑘𝑔 ℎ𝑢𝑚𝑖𝑑 𝑎𝑖𝑟
ℎ𝑟
B. Capacity of Forced draft (FDR) in
𝑚3
ℎ𝑟
Tower dryer
V =
𝑉𝑎
𝑀 𝑎
=
𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑠𝑢𝑝𝑝𝑙𝑦
𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑉𝑎 = volume flow rate supply ; the capacity 1 forced draft for installed before the heater
𝑉𝑎 = 𝝑 𝑀 𝑎
𝝑 = specific volume from the psychrometric chart @ condition 1 100℉ and 65% RH
𝝑 = 14.75
𝑓𝑡3
𝑙𝑏
𝑀 𝑎 = mass flow rate of dry air
V’ = for capacity in
𝑚3
ℎ𝑟
𝑉𝑎 = 𝝑𝑀 𝑎
= (14.75
𝑓𝑡3
𝑙𝑏
) (11776.32
𝑘𝑔
ℎ𝑟
)(2.2lb/kg)
𝑉𝑎 =382141.584
𝑓𝑡3
ℎ𝑟
(0.028 𝑚3
𝑓𝑡3 )
𝑉𝑎 = 10699.96
𝑚3
ℎ𝑟
C. Heat Required in the heater
𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 𝑀 𝑎 (ℎ2 − ℎ1)
ℎ1 = 100℉ DB, 65%RH
ℎ2 = 200℉ DB
Enthalpy Calculations;
h = Cp(DB) + SH (hg)
@pt 1:
h = Cp(DB) + SH (hg @ 200℉ )
= (0.24
𝐵𝑇𝑈
𝑙𝑏℉
)(100℉ ) + 0.028
𝑙𝑏 𝑚𝑜𝑖𝑠𝑡
𝑙𝑏 𝑑𝑟𝑦 𝑎𝑖𝑟
(1105.2
𝐵𝑇𝑈
𝑙𝑏
)
ℎ1 = 54.9456
𝐵𝑇𝑈
𝑙𝑏
@pt. 2 @ 200℉ DB
h = Cp(𝐷𝐵2) + 𝑆𝐻2 (hg @ 200℉ )
= (0.24
𝐵𝑇𝑈
𝑙𝑏℉
)(200℉ ) + 0.028
𝑙𝑏 𝑚𝑜𝑖𝑠𝑡
𝑙𝑏 𝑑𝑟𝑦 𝑎𝑖𝑟
(1146
𝐵𝑇𝑈
𝑙𝑏
)
ℎ2 = 80.1
𝐵𝑇𝑈
𝑙𝑏
: (80.1 – 54.9456)
𝐵𝑇𝑈
𝑙𝑏
= 25.2
𝐵𝑇𝑈
𝑙𝑏
Subs the values:
𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 𝑀 𝑎 (ℎ2 − ℎ1)
= 11776.32
𝑘𝑔
ℎ𝑟
(25.2
𝐵𝑇𝑈
𝑙𝑏
)(
𝐾𝑐𝑎𝑙
3.96 𝑏𝑡𝑢
) (
2.2𝑙𝑏
𝑘𝑔
)
𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 164868.48
𝐾𝑐𝑎𝑙
ℎ𝑟
D. Fresh air (%)
𝑀 𝑓
𝑀 𝑎
Energy balance: Mixing point
Mass Balance
𝑀𝑓 𝐷𝐵𝑓 + 𝑀 𝑅 (𝐷𝐵3) = 𝑀 𝑎 (𝐷𝐵1)
𝑀𝑓 𝐷𝐵𝑓 + (𝑀 𝑎 - 𝑀𝑓) (𝐷𝐵3) = 𝑀 𝑎 (𝐷𝐵1)
𝑀 𝑓
𝑀 𝑎
𝐷𝐵𝑓 + (
𝑀 𝑎
𝑀 𝑎
−
𝑀 𝑓
𝑀 𝑎
) (𝐷𝐵3) =
𝑀 𝑎
𝑀 𝑎
(𝐷𝐵1)
𝑀 𝑓
𝑀 𝑎
𝐷𝐵𝑓 + (1-
𝑀 𝑓
𝑀 𝑎
) (𝐷𝐵3) = (𝐷𝐵1)
𝑀 𝑓
𝑀 𝑎
𝐷𝐵𝑓 + (𝐷𝐵3) -
𝑀 𝑓
𝑀 𝑎
(𝐷𝐵3) = (𝐷𝐵1)
𝑀 𝑓
𝑀 𝑎
𝐷𝐵𝑓 -
𝑀 𝑓
𝑀 𝑎
(𝐷𝐵3) = (𝐷𝐵1) (𝐷𝐵3)
Factor out
𝑀 𝑓
𝑀 𝑎
:
𝑀 𝑓
𝑀 𝑎
𝐷𝐵𝑓 − (𝐷𝐵3) = (𝐷𝐵1)- (𝐷𝐵3)
𝑀 𝑓
𝑀 𝑎
=
( 𝐷𝐵1)− ( 𝐷𝐵3)
𝐷𝐵 𝑓 − ( 𝐷𝐵3)
=
100 −120
92−120
=
−20
−28
𝑀 𝑓
𝑀 𝑎
= 71.4%
𝑀 𝑎
𝑀𝑓
𝑀𝑓 + 𝑀𝑟 = 𝑀 𝑎
𝑀 𝑅
From high temperature psychrometry:
W1 = W2 = 0.0695
𝑙𝑏
𝑙𝑏
W3 = 0.0885
W3’ = 0.126
V2 = 0.0695
𝑓𝑡3
𝑙𝑏
a) Total lbs. of water in air before heating.
0.0695
𝑙𝑏
𝑙𝑏
( 800
𝑙𝑏
ℎ𝑟
)
= 55.6
𝑙𝑏
ℎ𝑟
b) Final lbs of water in air when discharged
= 0.0885
𝑙𝑏
𝑙𝑏
(800
𝑙𝑏
ℎ𝑟
)
= 70.8
𝑙𝑏
ℎ𝑟
c) Water evaporated from material.
= 800
𝑙𝑏
ℎ𝑟
(W3 - W2)
= 800
𝑙𝑏
ℎ𝑟
(0.0885 – 0.0695)
= 15.2
𝑙𝑏
ℎ𝑟
D) Relative humidity of air leaving Dryer (psychrometric chart)
= 59%
c) Percentage of air leaving the dryer
W3
W3’
=
0.0885
0.126
= 70.24%
D-3
For Total Air Requirement
𝑀𝑡𝑎 = 𝑀 𝑎 + 𝑀 𝑎 𝑆ℎ1
𝑀𝑡𝑎 = 𝑀 𝑎 (1 + 𝑆ℎ1)
Solve for 𝑴 𝒂
𝑀 𝑎 =
𝐹−𝑃
𝑆ℎ2−𝑆ℎ1
=
𝑘𝑔
ℎ𝑟
𝑘𝑔
𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟
Eq 1
Feed Product Calculation
𝐵𝐷𝑀 𝑝 = 𝐵𝐷𝑀 𝐹
For 𝑺𝒉 𝟐 (use chart)
𝑅𝐻2 = 100%
𝑡2 = 60 𝑑𝑒𝑔. 𝐶
𝑆ℎ1 = 0.016
𝑘𝑔 𝑜𝑓 𝐻2 𝑂
𝑘𝑔 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑉2 = 0.966
𝑚3
𝑘𝑔
Chart is out of bounds
𝑆ℎ2 =
0.622 𝑃𝑉2
𝑃 𝑎𝑡𝑚 𝑃𝑉2
Eq. 2
𝑅𝐻2 =
𝑃𝑉2
𝑃𝑠𝑎𝑡 𝑎𝑡 60 𝑑𝑒𝑔. 𝐶
where:
𝑃𝑠𝑎𝑡 𝑎𝑡 60 𝑑𝑒𝑔. 𝐶 = 19.941 𝑘𝑃𝑎
𝑆ℎ2 =
0.622 𝑃𝑉2
𝑃𝑎𝑡𝑚 − 𝑃𝑉2
=
0.622 (19.941𝑘𝑃𝑎)
101.325−19.941 𝑘𝑃𝑎
𝑺𝒉 𝟐 = 𝟎. 𝟏𝟓𝟐
Using eq. 1
𝑀 𝑎 =
𝐹−𝑃
𝑆ℎ2−𝑆ℎ1
=
𝑘𝑔
ℎ𝑟
𝑘𝑔
𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑀 𝑎 =
(1000 − 275)
𝑘𝑔
ℎ𝑟
(0.152 − 0.016)
𝑘𝑔
𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑴 𝒂 = 𝟓𝟑𝟑𝟏
𝒌𝒈
𝒉𝒓
dry air
For Humid air
𝑀𝑡𝑎 = 𝑀 𝑎 (1 + 𝑆ℎ1)
𝑀𝑡𝑎 = 5331
𝑘𝑔
ℎ𝑟
dry air 1 + 0.016
𝑘𝑔 𝑜𝑓 𝐻2 𝑂
𝑘𝑔 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑴 𝒕𝒂 = 𝟓𝟒𝟏𝟔
𝒌𝒈
𝒉𝒓
𝒉𝒖𝒎𝒊𝒅 𝒂𝒊𝒓
D - 4.
Solution:
𝐵𝐷𝑀 𝐹 = 𝐵𝐷𝑀 𝑃
𝐹 1 − %𝑚 𝐹 = 𝑃(1 − %𝑚 𝑃)
𝐹 1 − 0.07 = 20 000 1 − 0.005
𝑭 = 𝟐𝟏 𝟑𝟗𝟕. 𝟖𝟓
𝒌𝒈
𝒉𝒓
(𝒂)
𝑚 𝐸 = 𝐹 − 𝑃 = 21 397.85 − 20 000 = 𝟏𝟑𝟗𝟕. 𝟖𝟓
𝒌𝒈
𝒉𝒓
(𝒃)
𝐵𝐷𝑀 = 𝑃 1 − %𝑚 𝑃 = 20 000
𝑘𝑔
ℎ𝑟
1 − 0.005
2.205 𝑙𝑏
𝑘𝑔
= 43 879.5
𝑙𝑏
ℎ𝑟
𝑄 = 𝐵𝐷𝑀𝑐 𝑃 𝑡 𝐹 − 𝑡 𝑃 = 43 879.5
𝑙𝑏
ℎ𝑟
0.21
𝐵𝑇𝑈
𝑙𝑏 ⋅ ℉
239 − 86 ℉ = 𝟏 𝟒𝟎𝟗 𝟖𝟒𝟖. 𝟑𝟒
𝑩𝑻𝑼
𝒉𝒓
𝒄
𝜂 𝐷𝑅𝑌𝐸𝑅 =
𝐻𝑒𝑎𝑡 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝐹𝑢𝑒𝑙
0.60 =
1 409 848.34
𝑚 𝑂𝐼𝐿(17 987.96)
𝑚 𝑂𝐼𝐿 = 130.63
𝑙𝑏
ℎ𝑟
𝑘𝑔
2.205 𝑙𝑏
= 59.24
𝑘𝑔
ℎ𝑟
𝑆𝐺 𝑂𝐼𝐿 =
𝜌 𝑂𝐼𝐿
𝜌 𝐻2 𝑂
0.90 =
𝜌 𝑂𝐼𝐿
1000
𝜌 𝑂𝐼𝐿 = 900
𝑘𝑔
𝑚3
𝑚3
1000 𝐿
= 0.9
𝑘𝑔
𝐿
𝑉𝑂𝐼𝐿 =
𝑚 𝑂𝐼𝐿
𝜌 𝑂𝐼𝐿
=
59.24
𝑘𝑔
ℎ𝑟
0.9
𝑘𝑔
𝐿
= 65.82
𝐿
ℎ𝑟

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Board exam on druyers

  • 2. DRYER Board exam problems  Drier 5 booklet BP Oct 181 A tower-type, moist air dryer is to deliver 1000 kg/hr of cassava flour with 2% residual moisture of 20% in the feed. The air to be heated in the heating chamber is a mixture of fresh air at 92 deg.F dry bulb and 65% relative humidity is heated by steam coil to 200 deg.F. The dryer is properly insulated so that moisture absorption can be adiabatic. Compute: (1) Required flow of heated air mixture to dryer in pounds per minute. ; (2) Capacity forced draft fan on dryer in cu.mhr ; (3) Heat in kcal/hr for heating the the air mixture in heating chamber. ; (4) Percentage of weight of fresh air in mixture.
  • 3.  D-2 BP APRIL , 1982  Eight hundred pounds of dry air per hour is saturated at 115 deg.F, then heated to 224 deg.F, passed through an adiabatic drier and discharged at 142 deg.F. Determine each of the following: (1) Total pounds of water in air before heating, per hour. ; (2) Final pounds of water in air when discharged, per hour. ; Water evaporated from material in the drier per hour ; (4) Relative humidity of air leaving the drier ; (5) Dewpoint of air leaving the drier ; (6) Percentage of saturation of air leaving the drier ; (7) Number of cubic feet of air per min. before drying
  • 4.  D-3 BP APRIL , 1983  The temperature of air in the dryer is maintained constant by the use of steam coils within the dryer. The product enters the dryer at the rate of one metric ton per hour. The initial moisture content is 3 kilograms moisture per kilogram of dry solid. Air enters the dryer with a humidity ratio of 0.016 kilogram moisture per kilogram dry air and leaves with a relative humidity 100% while the temperature remains constant at 60 deg.C. If the total pressure of the air is 101.3 Kpa, determine: (a) Total required amount of air in kh/hr under entrance condition. ; (b) Capacity of the forced draft fan to handle this air in cu.m/min.
  • 5.  D-4 BP NOVEMBER , 1983  A rotary dryer fired with bunker oil of 10,000 kcal/kg HHV is to produce 20 metric tons per hour of dried sand with 0.5% moisture from wet feed containing 7% moisture, specific heat of sand is 0.21 BTU/lb per deg.F, temperature of wet feed is 30 deg.C and temperature of dried product is 115 deg.C.  QUESTION:  1. Calculate the weight in kg. of wet feed per hour  2. Calculate the weight of water to be removed per hour in kilograms.  3. Calculate heat in BTU/hour required  4. Calculate liters of Bunker oil per hour if specific gravity of bunker oil is 0.90 and dryer efficiency is 60%
  • 6.  D-6 BP APRIL , 1985  A rotary dryer produces 20 metric tons per hour of dried sand containing 0.5% moisture from a wet feed containing 10% moisture. Temperature of wet sand is 30 deg.C and that of dried sand is 114 deg.C. Fuel used per hour of bunker oil is 275 liters, with a HHV of 39,000 BTU/liter. Specific heat of sand is 0.21 BTU/lb-deg.Centrigrade Neglecting radiation loss, calculate the efficiency of the sand dryer. Specific gravity of oil is 0.9.  Required:  Feed rate in kg/hr  Volume flow rate in the dryer  Heat Supplied to the dryer  Efficiency of the sand dryer
  • 7.  D-7 BP OCTOBER , 1985  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 deg.C dry bulb and 21 deg.C wet bulb. The dryer is maintained at 45 deg.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.abs., determine the following:  1] Palay supplied to the dryer in kg/hr  2] Temperature of the hot humid air from the dryer in deg.C  3] Air supplied to dryer in cu.m/hr  4] Heat supplied by the heater in KW  5] Steam supplied to heater in kg/hr
  • 8.  D-9 BP OCTOBER , 1990  Wet material containing 215% moisture (dry basis) is to be dried at the rate of 1.5 kg/sec in a continuous dryer to give a product containing 5% moisture (wet basis_. The drying medium consist of air heated to 373 K and containing water vapor equivalent to a partial pressure of 1.40 kPa. The air leaved the dryer at 310 K and 70% saturated. Calculate how much air will be required to remove the moisture.
  • 9.  D-10 BP APRIL , 1988  An assembly hall was to have an air conditionaing unit installed which would be maintained at 26 deg.C dry bulb at 50% relative humidity. The unit delivers air at 15 deg.C bulb temperature and the calculated sensible heat load is 150 KW and latent heat load is 51.3 KW. Twenty percent by weight of extracted air is made up of outside air at 34 deg.C dry bulb and 60% relative humidity while 80% is extracted by the air conditioner’s refrigertaion capacity in Tons and its ventilation load in KW.
  • 10.  D-11 BP OCTOBER , 1981  A continuous dryer is supplied with air at 21 deg.C dry bulb and 16 deg.C wet bulb heated to 75 deg.C. It enters the heating coils at a velocity of 366 m/min and the size of the circular duct is 183 cm. The condition of the air leaving the dryer is 43 deg.C dry bulb and 70% relative humiidity. The moisture content of the entering material is 24% while the moisture content of the leaving material is 8%. Calculate:  [1] Amount of moisture evaporated  [2] Capacity of dryer  [3]Heat supplied to air
  • 11.  D-12Twenty five metric tons of dessicated coconut meat containing 40% moisture is to be dried to 10% final moisture in 24 hours. Atmospheric air at 26 deg.C dry bulb and 40% relative humidity is first preheated to proper temperature before reaching the product. The air leaves the dryer at 29 deg.C dewpoint temperature and 90% relative humidity. Neglecting the heat capacity of the bone dry stock and the heat needed to heat the copra tracks, calculate:  [1] The proper temperature into which the air be preheated  [2] The production rate in M.T. per hour of copra  [3] The intake air flow rate in cu.m/sec  [4] The mass flow rate of steam required in the preheated. If steam used for heating is at 80 deg.C dry and saturated condition.
  • 13. Solution: A. Flow of Heated air 𝑀 𝑇𝑎 = 𝑀 𝑎 + 𝑀 𝑎 𝑆𝐻1 𝑀 𝑇𝑎 = 𝑀 𝑎 1 + 𝑆𝐻1 .. (equation 1) Solving for the dry air flow: 𝑀 𝑎 = 𝐹 −𝑃 𝑆𝐻3 −𝑆𝐻2 BDM is constant: BDMF = BDMP F(1 - % Mf) = P(1 - %Mp) F = 𝑃 ( 1 −%𝑀𝑝) 1 −%𝑀𝑓 = 1000 𝑘𝑔 ℎ𝑟 (1 −0.021) ( 1 −0.20) F = 1223.75 𝑘𝑔 ℎ𝑟 By equation 1: Humid Air Flow Rate 𝑀 𝑇𝑎 = 𝑀 𝑎 1 + 𝑆𝐻1 Solve first for 𝑀 𝑎: 𝑀 𝑎 = 𝐹 −𝑃 𝑆𝐻3 −𝑆𝐻2 𝑀 𝑎 = 1223.75 𝑘𝑔 ℎ𝑟 −1000 𝑘𝑔 ℎ𝑟 0.047 −0.028 𝑀 𝑎 = 11776.32 𝑘𝑔 ℎ𝑟 90DB 100℉ 120℉ 200℉ Dry air 𝑆𝐻3 𝑆𝐻1 = 𝑆𝐻
  • 14. By equation 1: Humid Air Flow Rate 𝑀 𝑇𝑎 = 𝑀 𝑎 1 + 𝑆𝐻1 Where; 𝑀 𝑎 = 11776.32 𝑘𝑔 ℎ𝑟 𝑆𝐻1 = 𝑆𝐻2 = 0.028 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 𝑘𝑔 𝐷𝐴 Subs the values: 𝑀 𝑇𝑎 = 11776.32 𝑘𝑔 ℎ𝑟 (1 + 0.028 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 𝑘𝑔 𝐷𝐴 ) 𝑀 𝑇𝑎 = 12106.056 𝑘𝑔 ℎ𝑢𝑚𝑖𝑑 𝑎𝑖𝑟 ℎ𝑟 Therefore, flow of heated air 𝑀 𝑎 = 11776.32 𝑘𝑔 ℎ𝑟 𝑀 𝑇𝑎 = 12106.056 𝑘𝑔 ℎ𝑢𝑚𝑖𝑑 𝑎𝑖𝑟 ℎ𝑟
  • 15. B. Capacity of Forced draft (FDR) in 𝑚3 ℎ𝑟 Tower dryer V = 𝑉𝑎 𝑀 𝑎 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑠𝑢𝑝𝑝𝑙𝑦 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑉𝑎 = volume flow rate supply ; the capacity 1 forced draft for installed before the heater 𝑉𝑎 = 𝝑 𝑀 𝑎 𝝑 = specific volume from the psychrometric chart @ condition 1 100℉ and 65% RH 𝝑 = 14.75 𝑓𝑡3 𝑙𝑏 𝑀 𝑎 = mass flow rate of dry air V’ = for capacity in 𝑚3 ℎ𝑟 𝑉𝑎 = 𝝑𝑀 𝑎 = (14.75 𝑓𝑡3 𝑙𝑏 ) (11776.32 𝑘𝑔 ℎ𝑟 )(2.2lb/kg) 𝑉𝑎 =382141.584 𝑓𝑡3 ℎ𝑟 (0.028 𝑚3 𝑓𝑡3 ) 𝑉𝑎 = 10699.96 𝑚3 ℎ𝑟
  • 16. C. Heat Required in the heater 𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 𝑀 𝑎 (ℎ2 − ℎ1) ℎ1 = 100℉ DB, 65%RH ℎ2 = 200℉ DB Enthalpy Calculations; h = Cp(DB) + SH (hg) @pt 1: h = Cp(DB) + SH (hg @ 200℉ ) = (0.24 𝐵𝑇𝑈 𝑙𝑏℉ )(100℉ ) + 0.028 𝑙𝑏 𝑚𝑜𝑖𝑠𝑡 𝑙𝑏 𝑑𝑟𝑦 𝑎𝑖𝑟 (1105.2 𝐵𝑇𝑈 𝑙𝑏 ) ℎ1 = 54.9456 𝐵𝑇𝑈 𝑙𝑏 @pt. 2 @ 200℉ DB h = Cp(𝐷𝐵2) + 𝑆𝐻2 (hg @ 200℉ ) = (0.24 𝐵𝑇𝑈 𝑙𝑏℉ )(200℉ ) + 0.028 𝑙𝑏 𝑚𝑜𝑖𝑠𝑡 𝑙𝑏 𝑑𝑟𝑦 𝑎𝑖𝑟 (1146 𝐵𝑇𝑈 𝑙𝑏 ) ℎ2 = 80.1 𝐵𝑇𝑈 𝑙𝑏 : (80.1 – 54.9456) 𝐵𝑇𝑈 𝑙𝑏 = 25.2 𝐵𝑇𝑈 𝑙𝑏 Subs the values: 𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 𝑀 𝑎 (ℎ2 − ℎ1) = 11776.32 𝑘𝑔 ℎ𝑟 (25.2 𝐵𝑇𝑈 𝑙𝑏 )( 𝐾𝑐𝑎𝑙 3.96 𝑏𝑡𝑢 ) ( 2.2𝑙𝑏 𝑘𝑔 ) 𝑄 𝐻𝑒𝑎𝑡𝑒𝑟 = 164868.48 𝐾𝑐𝑎𝑙 ℎ𝑟
  • 17. D. Fresh air (%) 𝑀 𝑓 𝑀 𝑎 Energy balance: Mixing point Mass Balance 𝑀𝑓 𝐷𝐵𝑓 + 𝑀 𝑅 (𝐷𝐵3) = 𝑀 𝑎 (𝐷𝐵1) 𝑀𝑓 𝐷𝐵𝑓 + (𝑀 𝑎 - 𝑀𝑓) (𝐷𝐵3) = 𝑀 𝑎 (𝐷𝐵1) 𝑀 𝑓 𝑀 𝑎 𝐷𝐵𝑓 + ( 𝑀 𝑎 𝑀 𝑎 − 𝑀 𝑓 𝑀 𝑎 ) (𝐷𝐵3) = 𝑀 𝑎 𝑀 𝑎 (𝐷𝐵1) 𝑀 𝑓 𝑀 𝑎 𝐷𝐵𝑓 + (1- 𝑀 𝑓 𝑀 𝑎 ) (𝐷𝐵3) = (𝐷𝐵1) 𝑀 𝑓 𝑀 𝑎 𝐷𝐵𝑓 + (𝐷𝐵3) - 𝑀 𝑓 𝑀 𝑎 (𝐷𝐵3) = (𝐷𝐵1) 𝑀 𝑓 𝑀 𝑎 𝐷𝐵𝑓 - 𝑀 𝑓 𝑀 𝑎 (𝐷𝐵3) = (𝐷𝐵1) (𝐷𝐵3) Factor out 𝑀 𝑓 𝑀 𝑎 : 𝑀 𝑓 𝑀 𝑎 𝐷𝐵𝑓 − (𝐷𝐵3) = (𝐷𝐵1)- (𝐷𝐵3) 𝑀 𝑓 𝑀 𝑎 = ( 𝐷𝐵1)− ( 𝐷𝐵3) 𝐷𝐵 𝑓 − ( 𝐷𝐵3) = 100 −120 92−120 = −20 −28 𝑀 𝑓 𝑀 𝑎 = 71.4% 𝑀 𝑎 𝑀𝑓 𝑀𝑓 + 𝑀𝑟 = 𝑀 𝑎 𝑀 𝑅
  • 18.
  • 19. From high temperature psychrometry: W1 = W2 = 0.0695 𝑙𝑏 𝑙𝑏 W3 = 0.0885 W3’ = 0.126 V2 = 0.0695 𝑓𝑡3 𝑙𝑏 a) Total lbs. of water in air before heating. 0.0695 𝑙𝑏 𝑙𝑏 ( 800 𝑙𝑏 ℎ𝑟 ) = 55.6 𝑙𝑏 ℎ𝑟 b) Final lbs of water in air when discharged = 0.0885 𝑙𝑏 𝑙𝑏 (800 𝑙𝑏 ℎ𝑟 ) = 70.8 𝑙𝑏 ℎ𝑟 c) Water evaporated from material. = 800 𝑙𝑏 ℎ𝑟 (W3 - W2) = 800 𝑙𝑏 ℎ𝑟 (0.0885 – 0.0695) = 15.2 𝑙𝑏 ℎ𝑟
  • 20. D) Relative humidity of air leaving Dryer (psychrometric chart) = 59% c) Percentage of air leaving the dryer W3 W3’ = 0.0885 0.126 = 70.24%
  • 21. D-3 For Total Air Requirement 𝑀𝑡𝑎 = 𝑀 𝑎 + 𝑀 𝑎 𝑆ℎ1 𝑀𝑡𝑎 = 𝑀 𝑎 (1 + 𝑆ℎ1) Solve for 𝑴 𝒂 𝑀 𝑎 = 𝐹−𝑃 𝑆ℎ2−𝑆ℎ1 = 𝑘𝑔 ℎ𝑟 𝑘𝑔 𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟 Eq 1 Feed Product Calculation 𝐵𝐷𝑀 𝑝 = 𝐵𝐷𝑀 𝐹
  • 22. For 𝑺𝒉 𝟐 (use chart) 𝑅𝐻2 = 100% 𝑡2 = 60 𝑑𝑒𝑔. 𝐶 𝑆ℎ1 = 0.016 𝑘𝑔 𝑜𝑓 𝐻2 𝑂 𝑘𝑔 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑉2 = 0.966 𝑚3 𝑘𝑔 Chart is out of bounds 𝑆ℎ2 = 0.622 𝑃𝑉2 𝑃 𝑎𝑡𝑚 𝑃𝑉2 Eq. 2 𝑅𝐻2 = 𝑃𝑉2 𝑃𝑠𝑎𝑡 𝑎𝑡 60 𝑑𝑒𝑔. 𝐶 where: 𝑃𝑠𝑎𝑡 𝑎𝑡 60 𝑑𝑒𝑔. 𝐶 = 19.941 𝑘𝑃𝑎 𝑆ℎ2 = 0.622 𝑃𝑉2 𝑃𝑎𝑡𝑚 − 𝑃𝑉2 = 0.622 (19.941𝑘𝑃𝑎) 101.325−19.941 𝑘𝑃𝑎 𝑺𝒉 𝟐 = 𝟎. 𝟏𝟓𝟐
  • 23. Using eq. 1 𝑀 𝑎 = 𝐹−𝑃 𝑆ℎ2−𝑆ℎ1 = 𝑘𝑔 ℎ𝑟 𝑘𝑔 𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑀 𝑎 = (1000 − 275) 𝑘𝑔 ℎ𝑟 (0.152 − 0.016) 𝑘𝑔 𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑴 𝒂 = 𝟓𝟑𝟑𝟏 𝒌𝒈 𝒉𝒓 dry air For Humid air 𝑀𝑡𝑎 = 𝑀 𝑎 (1 + 𝑆ℎ1) 𝑀𝑡𝑎 = 5331 𝑘𝑔 ℎ𝑟 dry air 1 + 0.016 𝑘𝑔 𝑜𝑓 𝐻2 𝑂 𝑘𝑔 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑴 𝒕𝒂 = 𝟓𝟒𝟏𝟔 𝒌𝒈 𝒉𝒓 𝒉𝒖𝒎𝒊𝒅 𝒂𝒊𝒓
  • 24. D - 4. Solution: 𝐵𝐷𝑀 𝐹 = 𝐵𝐷𝑀 𝑃 𝐹 1 − %𝑚 𝐹 = 𝑃(1 − %𝑚 𝑃) 𝐹 1 − 0.07 = 20 000 1 − 0.005 𝑭 = 𝟐𝟏 𝟑𝟗𝟕. 𝟖𝟓 𝒌𝒈 𝒉𝒓 (𝒂) 𝑚 𝐸 = 𝐹 − 𝑃 = 21 397.85 − 20 000 = 𝟏𝟑𝟗𝟕. 𝟖𝟓 𝒌𝒈 𝒉𝒓 (𝒃) 𝐵𝐷𝑀 = 𝑃 1 − %𝑚 𝑃 = 20 000 𝑘𝑔 ℎ𝑟 1 − 0.005 2.205 𝑙𝑏 𝑘𝑔 = 43 879.5 𝑙𝑏 ℎ𝑟 𝑄 = 𝐵𝐷𝑀𝑐 𝑃 𝑡 𝐹 − 𝑡 𝑃 = 43 879.5 𝑙𝑏 ℎ𝑟 0.21 𝐵𝑇𝑈 𝑙𝑏 ⋅ ℉ 239 − 86 ℉ = 𝟏 𝟒𝟎𝟗 𝟖𝟒𝟖. 𝟑𝟒 𝑩𝑻𝑼 𝒉𝒓 𝒄 𝜂 𝐷𝑅𝑌𝐸𝑅 = 𝐻𝑒𝑎𝑡 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝐹𝑢𝑒𝑙 0.60 = 1 409 848.34 𝑚 𝑂𝐼𝐿(17 987.96) 𝑚 𝑂𝐼𝐿 = 130.63 𝑙𝑏 ℎ𝑟 𝑘𝑔 2.205 𝑙𝑏 = 59.24 𝑘𝑔 ℎ𝑟
  • 25. 𝑆𝐺 𝑂𝐼𝐿 = 𝜌 𝑂𝐼𝐿 𝜌 𝐻2 𝑂 0.90 = 𝜌 𝑂𝐼𝐿 1000 𝜌 𝑂𝐼𝐿 = 900 𝑘𝑔 𝑚3 𝑚3 1000 𝐿 = 0.9 𝑘𝑔 𝐿 𝑉𝑂𝐼𝐿 = 𝑚 𝑂𝐼𝐿 𝜌 𝑂𝐼𝐿 = 59.24 𝑘𝑔 ℎ𝑟 0.9 𝑘𝑔 𝐿 = 65.82 𝐿 ℎ𝑟