Cooling Towers
PROBLEMS
1.A cooling tower is located on a mountain where barometric pressure is 90 kPa. The tower is to
cool 113.4 kg/s of water from 41°C to 28°C. Air enters at 36°C DB and a vapor pressure of 4.2
kPaa and leaves saturated at 39°C. Without using psychrometric chart, determine:
a) the mass flow rate of dry air required; and
agram:
b) the mass flow rate of make-up water required.
1)
AirDi
3
1
2
Reheater
%M5 = 5%
Required:
𝑀𝑟 = 𝑀4 − 𝑀5
If
𝑎) 𝐺𝑀5 = 1𝑘𝑔
𝑏) 𝐵𝐷𝑀 = 1𝑘𝑔
Solution:
From;
𝑎) %𝑀5 =
𝑀5
𝐺𝑀5
𝑥100%
𝑀5 = %𝑀5(𝐺𝑀5)
𝑀5 = 0.05(1𝑘𝑔)
𝑀5 = 0.05𝑘𝑔
Also;
𝐺𝑀5 = 𝐵𝐷𝑀5 + 𝑀5
𝐵𝐷𝑀5 = 𝐺𝑀5 − 𝑀5
𝐵𝐷𝑀5 = 1𝑘𝑔 − 0.05𝑘𝑔
4
Dryer
5
%M4 = 60%
Copra
𝐵𝐷𝑀5 = 0.95𝑘𝑔 = 𝐵𝐷𝑀4
And;
𝐺𝑀4 = 𝐵𝐷𝑀4 + 𝑀4
%𝑀4 =
𝐺𝑀4 =
𝑀4
𝐺𝑀4
𝑀4
%𝑀4
Thus;
𝑀4
= 0.95𝑘𝑔 + 𝑀4
0.6
𝑀4 = 1.425𝑘𝑔
Therefore;
𝑀𝑅 = 𝑀4 − 𝑀5
𝑀𝑅 = 1.425𝑘𝑔 − 0.05𝑘𝑔
𝑴𝑹 = 𝟏. 𝟑𝟕𝟓𝒌𝒈
b)
From;
𝐺𝑀5 = 𝐵𝐷𝑀5 + 𝑀5
𝑀5
%𝑀5 =
𝐺𝑀5
𝐺𝑀5 =
𝑀5
%𝑀5
𝑀5
= 𝐵𝐷𝑀5 + 𝑀5
%𝑀5
𝑀5
0.05
= 1𝑘𝑔 + 𝑀5
𝑀5 = 0.052632𝑘𝑔
𝑀4
= 𝐵𝐷𝑀4 + 𝑀4
%𝑀4
𝑀4
= 1𝑘𝑔 + 𝑀4
0.6
𝑀4 = 1.5𝑘𝑔
So;
𝑀𝑅 = 𝑀4 − 𝑀5
𝑀𝑅 = 1.5𝑘𝑔 − 0.052632𝑘𝑔
𝑴𝑹 = 𝟏. 𝟒𝟒𝟕𝟑𝟕𝒌𝒈
2. Condenser cooling water is supplied to the forced-draft cooling tower at 40°C and is cooled to
3°C of approach temperature while falling through the tower. The air enters the tower at 35°C DB
and 28°C WB and leaves at 38°C DB and 60% RH. For 5,000 kg/min of condenser cooling water,
find:
a) the quantity of air required by the tower in kg/min;
b) the amount of make-up water required to compensate the water loss due to evaporation.
Required:
a)ma
b) m5
Solution:
From:
𝑄𝑟 = 𝑄𝑎
𝑚𝑛𝐶𝑝𝑛∆𝑇𝑛 = 𝑚𝑎(ℎ2 − ℎ1)
𝑄𝑅 = 5000
𝑘𝑔
𝑚𝑖𝑛
∙
1 𝑚𝑖𝑛
60 𝑠
∙ 4.187
𝑘𝐽
𝑘𝑔 ∙ 𝐾
∙ (40 − 3)𝐾
𝑄𝑅 = 12 709.92 𝑘𝑊
For h2; h1:
@ 1 Pwb=3.782 kPa
𝑃𝑤𝑣 = 3.782 −
(101.325 − 3.782)(35 − 28)
1527,4 − 1.3𝑥28
𝑃𝑤𝑣 = 3.732 𝑘𝑃𝑎
𝐻𝑅1 = 0.02378 𝑘𝑔/𝑘𝑔
ℎ1 = 1.0062 ∙ 35 + (2500 + 1.8 ∙ 35)(𝐻𝑅1)
ℎ1 = 96.18 𝑘𝐽/𝑘𝑔
For h2;
𝑃𝑟𝑜𝑣2 = 6 ∙ 6.632
𝑃𝑟𝑜𝑣2 = 3.9792 𝑘𝑃𝑎
𝐻𝑅2 = 0.02543 𝑘𝐽/𝑘𝑔
ℎ2 = 103.54
𝑘𝐽
𝑘𝑔
Thus;
𝑚𝑎 =
𝑄𝑟
ℎ2 − ℎ1
12 909.92 𝑘𝐽/𝑠
=
𝑚𝑎 = 1754.07
(103.54 − 96.18)𝑘𝐽/𝑘𝑔
𝑘𝑔
𝑠
For MW;
𝑀𝑊 = 𝑚𝑎(𝑊1 − 𝑊2)
𝑀𝑊 = 1754.07
𝑘𝑔
(−0.02378 + 0.02543)
𝑠
𝑀𝑊 = 2.894 𝑘𝑔/𝑠
3. Water at 55°C is cooled in a cooling tower which has an efficiency of 65%. The temperature of the
surrounding air is 32°C DB and 70% RH. The heat dissipated from the condenser is 2,300,000 kJ/hr.
Find the capacity in lps of the pump used in the cooling tower.
Diagram:
Given:
tdb3 = 55°C
eCT = 65%
tdb1 = 32°C
RH1 = 𝑃𝑃𝑤𝑣1
𝑠𝑎𝑡
QR = 2 300 000 kJ/hr
Required: V = ? in lps
Solution:
Psat @ 32°C = 4.759
0.7 =
𝑃𝑤𝑣
4.759
; Pwv = 0.7(4.759) = 3.3313
And;
Pwv = Pwb - (𝑃𝑡−𝑃𝑤𝑏)(𝑡𝑑𝑏−𝑡𝑤𝑏)
1527.4−1.3(𝑡𝑤𝑏)
ln(Psat) = 14.43509 - 5333.3
𝑡𝑠𝑎𝑡
ln(Pwb) = 14.43509 - 5333.3
𝑡 𝑤𝑏
Pwb = 100𝑒
14.43509− 5333.33
𝑡𝑤𝑏+273
So,
14.43509− 5333.33
𝑡𝑤𝑏+273
3.3313 = 100𝑒
−
(𝑃𝑡−100𝑒
14.43509− 5333.33
𝑡𝑤𝑏+273)(𝑡 −𝑡 )
𝑑𝑏 𝑤𝑏
1527.4−1.3(𝑡𝑤𝑏)
Pwv = Pwb - 𝑃𝑡(𝑡𝑑𝑏−𝑡𝑤𝑏)
1500
twb = 27.48
thus,
eCT =
𝑡𝑑𝑏3−𝑡𝑑𝑏4
𝑡𝑑𝑏3−𝑡𝑤𝑏1
; 0.65 =
55−𝑡𝑑𝑏4
55−27.48
tdb4 = 37.112
also;
QR = mw(tdb3 – tdb4) ; mw = m3
2300000kJ/hr = (m3)(4.1868)(55-37.112)
M3 = 30710.28kg/hr x
1 ℎ𝑟
3600𝑠
= 8.5306kg/s
And;
V3 = vf @ 32°C = 1.0050x10-3 m3/kg = 1.0050L/kg
Therefore;
V = V3m3 = (8.5306kg/s)(1.0050L/s) =
8.573 L/s
Cooling Towers
6. Water is cooled in a cooling tower from a temperature of 38°C to 24°C. Air enters the cooling
tower at a temperature of 28°C DB with 40% RH and leaves at a temperature of 35°C DB with RH of
98%. Determine:
c) the efficiency or percent effec
e cooling tower.
a) the amount of water cooled per kg of dry air;
b) the percentage of water lost by evaporation; and
38℃of th
tiveness
28℃
RH= 40%
Mass balance (water vapor)
35℃
RH=98%
24℃
𝑚1𝐻𝑅1 + 𝑚3 = 𝑚2𝐻𝑅2 + 𝑚4
𝑚3 = 𝑚2𝐻𝑅2 + 𝑚4 − 𝑚1𝐻𝑅1
𝑚1 = 𝑚2 = 𝑚𝑎
𝑚3 = 𝑚𝑎(𝐻𝑅2 − 𝐻𝑅1) + 𝑚4
Energy balance
𝑚1ℎ1 + 𝑚3ℎ3 = 𝑚2ℎ2 + 𝑚4ℎ4
𝑚3ℎ3 − 𝑚4ℎ4 = 𝑚2ℎ2 − 𝑚1ℎ1
𝑚3ℎ3 − 𝑚4ℎ4 = 𝑚𝑎(ℎ2 − ℎ1)
h3= hf at 38℃= 159.21 kj/kg
h4=hf= at 24℃=100.70 Kj/kg
For Pts. 1 & 2
𝐻𝑅 =
0.622 𝑃𝑤𝑣
𝑃𝑡 − 𝑃𝑤𝑣
ℎ = 𝐶𝑝𝑡𝑑𝑏 + 𝐻𝑅(2500.9 + 1.82(𝑡𝑑𝑏))
For Pt. 1
𝑅𝐻 =
𝑃𝑤𝑣
𝑃𝑠𝑎𝑡
𝑥100
𝑃𝑠𝑎𝑡 @ 28℃ = 3.783 𝑘𝑃𝑎
𝑃𝑠𝑎𝑡 @ 35℃ = 5.628 𝑘𝑃𝑎
𝑃𝑤𝑣 = 𝑅𝐻𝑥𝑃𝑠𝑎𝑡
𝑃𝑤𝑣1 = 0.4(3.782) = 1.5128 𝑘𝑃𝑎
𝑃𝑤𝑣2 = 0.98(5.628) = 5.5154 𝑘𝑃𝑎
𝐻𝑅1 =
0.622 (1.5128)
= 0.009427
101.325 − 1.5128
𝐻𝑅2 =
0.622 (5.5154)
= 0.035806
101.325 − 5.5154
ℎ1 = 52.228
ℎ2 = 127.02
For 1 kg of dry air
A.
𝑚3 − 𝑚4 = (𝐻𝑅2 − 𝐻𝑅1)
𝑚3 − 𝑚4 = 0.035806 − 0.009427
𝑚3 − 𝑚4 = 0.026379 𝑘𝑔 𝑒𝑞. 1
𝑚3ℎ3 − 𝑚4ℎ4 = (ℎ2 − ℎ1)
𝑚3(159.27) − 𝑚4(100.7) = 127.02 − 52.225 𝑒𝑞. 2
𝒎𝟑 = 𝟏. 𝟐𝟑𝟐𝟗
𝒌𝒈
𝒔
B.
𝑚4 = 1.2065
%𝑚 =
𝑚3 − 𝑚4
𝑚3
𝑘𝑔
𝑠
𝑥100
%𝒎 = 𝟐. 𝟏𝟒%
C.
%=
𝑡𝑑𝑏3 − 𝑡𝑑𝑏4
𝑥100
𝑡𝑑𝑏3 − 𝑡𝑤𝑏1
(𝑃𝑡 − 𝑃𝑤𝑏)(𝑡𝑑𝑏 − 𝑡𝑤𝑏)
𝑃𝑤𝑣1 = 𝑃𝑤𝑏 −
1527.4 − 1.3𝑡 𝑤𝑏1
𝑡𝑤𝑏1 = 18.75℃
𝑃𝑤𝑏 = 2.138 𝑘𝑃𝑎
38 − 24
𝑒=
𝑥100
38 − 18.75
𝒆 = 𝟕𝟐. 𝟕𝟑%
7. A forced-draft cooling tower is required to cool 20.71 kg/s of entering water form 41°C to
28°C, with an evaporation loss of 0.457 kg/s. The cooling tower receives 11.81 m3/s of air at 101.325
kPaa, 38°C DB, and 24°C WB. Find the temperature of the air leaving the tower.
Required:
Gm4=?
Tdryer = tdb3=tdb2
M2=20.71
𝑘𝑔
𝑠
T3=41oC
T4=28oC
M5=0.457
𝑘𝑔
𝑠
V1=11.81 m3/s
Pt = 101.325 kpaa
Tdb1= 38oC  hg1=1.570.7 kj/kg
Twb1 = 24oC  twb1=2.985 kpa
Tdb2=?
M1(h2-h1) = m3Cp(t3-t4)
V1=11.81 m3/s
M1=VI/V1; VI =
0.28708 𝑘𝑗 (38+273)𝑘
𝑘𝑔−𝑘
(101.325−2.0648)𝐾𝑁
𝑚2
(101.325−2.185)(35−24)
pwv= 2.185-
1527.4−1.3(24)
pwv = 2.0648
M1=
11.81𝑚3
𝑠
0.8994𝑚3
𝑘𝑔
= 13.13 𝑘𝑔
𝑠
;
0.622(2.0648)
H1=1.0062(38)+ 101.325−2.0648 (2570.7)
H1=71.4373 KJ/kg
H2=
20.71(4.1868)(41−28)
13.13
+ 71.4973 = 157.3474
M5= m1 (HR2-HR1)
𝑚
HR2=𝑚 5 + 𝐻𝑅1
1
0.457
13.13
0.622(2.0648)
=
101.325−2.0648
+
HR2=0.04774
H2=Cp(tdb20 + HR2 (hg)
157.347 = 1.0062(tdb20 + 0.04774 (2501+1.88(tdb2))
Tdb2 = 34.627oC
8. An induced-draft cooling tower is required to cool 9,020 gpm of entering water from 29°C to
20°C. The average condition of the atmospheric air is 100.75 kPaa, 21°C DB and 16°C WB. The air
leaves the tower as saturated at 27°C. Find:
a) the volume rate of air required, m3/s;
b) the amount of make-up water required to compensate the loss due to evaporation.
9020 gpm
𝑃1 = 100.75 𝑘𝑃𝑎
29
27 DB
AIR
21 DB
16 WB
20
SOLUTION:
𝜈1 =
𝜈1 =
𝑣
𝑅𝑇
=
𝑚
𝑃
0.28708 (21 + 273)
100.75 − 1.7573
𝜈1 = 0.8526
𝑚3
𝑠
𝑄𝑤 = 𝑚𝑤𝑐𝑝ΔΤ𝑤
for 𝑚𝑤,
𝜌=
𝑚
; 𝑚 = 𝜌𝑣
𝑣
3
𝑚 = 1000 𝑘𝑔 𝑥9020 𝑔𝑎𝑙 𝑥 3.785 𝐿 𝑥 1 𝑚 𝑥 1 𝑚𝑖𝑛
𝑚3
𝑚𝑖𝑛 1 𝑔𝑎𝑙 1000 𝐿 60 𝑠
𝑚 = 569.07
𝑘𝑔
𝑠
𝑄𝑤 = 569.07𝑥4.187𝑥(29 − 20)
𝑘𝑔
𝑄𝑤 = 21 444.4
𝑠
= 𝑘𝑁
hence,
𝑣𝑎𝑖𝑟
𝑚3
= 505.08
𝑠
from, 𝑄𝑤 = 𝑄𝑎
𝑄𝑎 = 𝑚𝑎(ℎ2 − ℎ1)
@1,
𝑃𝑤𝑣1 = 1.7573 𝑘𝑃𝑎
𝐻𝑅1 = 0.0110
ℎ1 = 44.15
𝑘𝑔
𝑘𝑔
𝑘𝐽
𝑘𝑔
@2,
𝑃𝑤𝑣2 = 3.567 𝑘𝑃𝑎
𝐻𝑅2 = 0.02283
𝑘𝑔
𝑘𝑔
(a)
ℎ2 = 85.35
𝑘𝐽
𝑘𝑔
thus,
𝑚𝑎 =
𝑚𝑎 =
𝑄𝑎
(ℎ2 − ℎ1)
21 444.4
𝑘𝐽
𝑠
(85.35 𝑘𝐽⁄𝑘𝑔 − 49.15 𝑘𝐽⁄𝑘𝑔)
𝑚𝑎 = 592.4
𝑘𝑔
𝑠
from,
𝑣=
𝑣
; 𝑣 = (𝑚)(𝑣)
𝑚
𝑣 = 592.4
𝑚3
𝑘𝑔
𝑥0.8526
𝑠
𝑘𝑔
𝑚𝑤 + 𝑚𝑑𝑎𝑤1 = 𝑚𝑑2𝑤2
𝑚𝑤 = 𝑚𝑑𝑎(𝑤2 − 𝑤1)
𝑘𝑔
𝑚𝑤 = 592.4
𝑠
(0.02283 − 0.0110)
𝑚𝑤 = 7 𝑘𝑔⁄𝑠
(b)
𝑘𝑔
𝑘𝑔
Drying Processes
PROBLEMS
1. Copra enters a dryer containing 60% water and 40% of solids and leaves with 5% water and
95% solids. Find the amount of water removed based on a kg of final product and a kg of bone-drymaterial (ME board Problem, October 1992).
Required:
MR = M4 – M5= ?
If a) GM5 = 1 kg
b) BDM = 1 kg
Solution:
a) from;
%𝑀5 =
𝑀5
𝐺𝑀5
𝑀5 = %𝑀5(𝐺𝑀5) = 0.05(1 kg) = 0.05kg
Also;
GM5 = BDM5 + M5
BDM5 = GM5 – M5 = 1 kg – 0.05 kg = 0.95 kg = BDM4
And;
GM4 = BDM4 + M4
%𝑀4 =
𝑀4
𝐺𝑀4
; 𝐺𝑀4 =
𝑀4
%𝑀4
Thus;
𝑀4
0.6
= 0.95 𝑘𝑔 + 𝑀4; M 4 = 1.425 kg
Therefore;
MR = M4 – M5 = 1.425 kg – 0.05 kg =
1.375kg
b) from:
GM5 = BDM5 + M5
%𝑀5 =
𝑀5
%𝑀5
𝑀4
; 𝐺𝑀5 =
𝑀5
%𝑀5
𝑀5
= BDM5 + M5 = 0.05
= 1 kg + M5
M5 =
%𝑀4
𝑀5
𝐺𝑀5
0.052632 kg
= BDM4 + M4 = 𝑀4 = 1 kg + M4
0.6
M4 = 1.5 kg
So;
MR = M4 – M5 = 1.5kg – 0.052632kg =
1.44737 kg
2. The temperature of the air in a dryer is maintained constant by the use of steam coils within the
dryer. The product enters the dryer at the rate of 1 metric ton per hour. The initial moisture content is 3.0kg moisture per kg/hr solid and will be dried to a moisture content of 0.10 kg moisture per kg solid. Air
enters the dryer with a humidity ratio of 0.016 kg/kg d.a. and leaves with a relative humidity of 100%
while the temperature remains constant at 60°C. If the total pressure of air is 101.325 kPaa, determine the
total amount of air required in kg/hr under entrance condition, and the capacity of forced draft fan to
handle this air in m3/min (ME Board Problem, April 1983).
Required: 𝑚𝑎 =? ; 𝑘𝑔
ℎ𝑟
𝑉1 =? ;
𝑚3
𝑚𝑖𝑛
Solution :
From; GM = BDM + M ;
𝑀4
𝐵𝐷𝑀4
=3
So;
𝐺𝑀4 = 𝐵𝐷𝑀4 + 3𝐵𝐷𝑀4
1000𝐾𝑔 = 4𝐵𝐷𝑀 4
ℎ𝑟
𝐵𝐷𝑀4 = 250
𝐾𝑔
ℎ𝑟
So;
𝑀4 = 3 (250 𝐾𝑔) = 750
ℎ𝑟
𝐾𝑔
ℎ𝑟
Also;
𝐵𝐷𝑀4 = 𝐵𝐷𝑀5
And;
𝑀5
𝐵𝐷𝑀5
= 0.10 ; 𝑀5 = 0.10(𝐵𝐷𝑀5 ) = 0.10 (750 𝐾𝑔) = 25 𝐾𝑔
ℎ𝑟
So;
𝑚𝑟 = 𝑀4 − 𝑀5 = 750 𝑘𝑔 − 25 𝑘𝑔 = 725
ℎ𝑟
From;
ℎ𝑟
𝑘𝑔
ℎ𝑟
ℎ𝑟
𝑚𝑎(𝐻𝑅3 − 𝐻𝑅2) = 𝑚𝑟
𝑚𝑎 =
𝑚𝑟
from TABLE 1 @ t=60°C; Psat= 19.940 kPa
𝐻𝑅3−𝐻𝑅2
Solving for 𝐻𝑅3;
𝐻𝑅3 =
0.622(1)(19.940)
101.325−(!)19.940
= 0.152395
𝐾𝑔
𝐾𝑔
Therefore;
𝑚𝑎 =
725 𝑘𝑔
ℎ𝑟
(0.152395−0.016) 𝑘𝑔
= 𝟓𝟑𝟏𝟓. 𝟒𝟒
𝑘𝑔 𝑑.𝑎.
𝐤𝐠 𝐝.𝐚.
𝒉𝒓
Solving for 𝑣;
𝐻𝑅 =
𝑣=
0.622(𝑃𝑤𝑣)
101.325−𝑃𝑤𝑣
0.28708(60+273)
101.325−2.541
= 0.016; 𝑃𝑤𝑣 = 2.541 𝑘𝑃𝑎
= 0.967676
𝑚
3
𝑘𝑔
3
𝑉 = 𝑟𝑚 = 0.967676 𝑚 (5315.44 𝑘𝑔 ) (
𝑘𝑔
ℎ𝑟
1ℎ𝑟
60𝑚𝑖𝑛
) = 𝟖𝟓. 𝟕𝟐𝟕
𝒌𝒈
𝒎𝒊𝒏
Guidebook in Air Conditioning Design and Analysis
3. A dryer is to deliver 1,000 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 DB and 21°C WB. 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 Mpaa, determine:
a)
the amount of palay supplied to the dryer in kg/hr and the temperature of the hot humid air
from the dryer in °C
b) the mass flow rate of air supplied to the dryer, in m3/hr
c) the heat capacity of the heater in kW
d) the steam supplied to the heater in kg/hr. (ME Board Problem, October 1985)
Diagram:
%M 5 = 10%
%M 4 = 15%
PALAY
Required: 𝑎. ) 𝐺𝑀4 = ? ;
𝑘𝑔
ℎ𝑟
𝑏. ) 𝑚𝑎 = ? ;
𝑚3
ℎ𝑟
𝑐. ) 𝑄 = ? ; 𝐾𝑊
𝑑. ) 𝑚𝑠 = ? ;
𝑘𝑔
ℎ𝑟
Solution :
@ POINT 1
𝑡𝑑𝑏1 = 32℃
𝑡𝑤𝑏1 = 21℃
ℎ1 = 60.6
𝐾𝐽
𝑘𝑔
𝐻𝑅1 = 0.0112 𝑘𝑔
𝑘𝑔
@ POINT 2
𝑡𝑑𝑏2 = 45℃
𝐻𝑅2 = 𝐻𝑅1
ℎ2 = 74.9
𝐾𝐽
𝑘𝑔
𝑣2 = 0.917 𝑚
3
𝑘𝑔
@ POINT 3
𝑡𝑑𝑏3 = 45℃
𝑅𝐻 = 80%
ℎ3 = 177
𝐾𝐽
𝑘𝑔
𝐻𝑅3 = 0.0510 𝑘𝑔
𝑘𝑔
𝐺𝑀5(1−%𝑀5)
= 1000(1−0.10)
a.) 𝐺𝑀4 = (1−%𝑀
1−0.15
4)
𝑮𝑴𝟒 = 𝟏𝟎𝟓𝟖. 𝟖
𝒌𝒈
𝒉𝒓
Also;
𝒕𝒅𝒓𝒚𝒆𝒓 = 𝒕𝒅𝒃𝟑 = 𝒕𝒅𝒃𝟐 = 𝟒𝟓℃
b.) 𝑉2 = 𝑚𝑎𝑣2
; 𝑚𝑎 = ?
𝐺𝑀 −𝐺𝑀5
=
𝑚𝑎 = 4
𝐻𝑅3−𝐻𝑅2
1058.8−1000
0.0510−0.0112
= 1477.4 𝑘𝑔
ℎ𝑟
So;
𝑉2 = 1477.4
𝑽𝟐 = 𝟏𝟑𝟓𝟒. 𝟖 𝒎
𝑘𝑔
ℎ𝑟
(0.917
𝑚3
𝑘𝑔
)
𝟑
𝒉𝒓
c.)
𝑄 = 𝑚𝑎(ℎ2 − ℎ1)
𝑄 = 1477.4 𝑘𝑔 (
ℎ𝑟
1ℎ𝑟
)
3600𝑠
(74.9 𝐾𝐽 − 60.6 𝐾𝐽 ) (
𝑘𝑔
𝑘𝑔
𝐾𝐽
)
𝑠
= 𝐾𝑊
𝑸 = 𝟓. 𝟖𝟕 𝐊𝐖
d.)
𝑄 = 𝑚𝑠ℎ𝑓𝑔
𝑚𝑠 =
5.87 𝐾𝐽 (3600
𝑠
𝑠
)
ℎ𝑟
1890.7 𝐾𝐽
𝑘𝑔
𝒎𝒔 = 𝟏𝟏. 𝟏𝟖
𝒌𝒈
𝒉𝒓
4. A dryer is to deliver 0.30 kg/s of cassava with 2% moisture and 20% moisture in the feed.
Determine the mass of air required if the change in humidity ratio is 0.0165 kg/kg d.a.
ma
2
DRYER
1
3 Gm3
%m3 = 20%
4 Gm4 = 0.3
𝑘𝑔
𝑠
%m4 = 2%
M a= ?
BDM4 = 0.294
ΔHR = 0.0165
𝑘𝑔
𝑘𝑔𝑑𝑎
𝑘𝑔
𝑠
Gm3 = BDM3 + m3
ma1HR1 + m3 = m4 + ma2HR2
Gm3 = BDM3 + %m3Gm3
ma(HR1-HR2) = m4 – m3
Gm3 = 0.294 + 0.2Gm3
ma =
𝑚 4 −𝑚 3
Gm3 = 0.3675
𝐻𝑅1−𝐻𝑅2
%m4 =
𝑚4
𝐺𝑚4
, m4 =
m4 = 6x10 -3
𝑘𝑔
0.02(0.3)
𝑠
𝑘𝑔
𝑠
Gm4 = BDM4 + m4
%m3 =
𝑚3
𝐺𝑚3
, m3 = 0.2(0.3675)
m3 = 0.0735 𝑘𝑔
𝑠
ma =
[(6𝑥10−3)−0.0735)]
𝑘𝑔
𝑠
0.0165 𝑘𝑔
𝑘𝑔 𝑑𝑎
0.3 = BDM4 + 6x10-3
ma = 4.091
𝒌𝒈𝒅𝒂
𝒔
Drying Processes
STUDENT’S SELF TEST
Instruction: Select the correct answer.
1. A copra drying plant is designed to dry 1,000 kg/hr of fresh coconut meat containing
30% water. The raw copra from the dryer contains 5% water. Fresh air at 27°C and 40% RH and
barometric pressure of 98 kPa has W = 0.0083 kg/kg d.a. and h = 50.86 kJ/kg d.a. The air is
heated to 110°C (h = 135.58 kJ/kg d.a.) before entering the adiabatic drying chamber and leaves
the dryer at 75°C with humidity ratio of 0.02285 kg/kg d.a. Assuming 100% heat transfer
efficiency in the air pre-heater, determine the amount of steam required by the dryer when
condensing saturated steam to saturated liquid at 150 kPaa (hfg = 2226.5 kJ/kg)
a) 739 kg/hr
b) 793 kg/hr
c) 379 kg/hr
d) 937 kg/hr
2. A certain material enters dryer containing 60% water and leaves with 5% water. Find
the mass of the final product if the original product is 1 kg/s.
a) 0.421 kg/s
b) 0.412 kg/s
c) 0.214 kg/s
d) 0.142 kg/s
%m4=5%
GM3= BDM3 + %M3GM3
1kg/s= BDM3 + 0.6(1kg/s)
BDM3 = 0.4 kg/s
GM4 = BDM4 + %M4GM4
GM4 = 0.4kg + 0.05GM4
GM4 = 0.421 kg/s
3. A ground cassava enters dryer containing 60% moisture and leaves as a flour with 5%
moisture. Find the mass of the original product if the final product is 1 kg/s.
a) 2.735 kg/s
b) 2.375 kg/s
c) 2.537 kg/s
d) 2.753 kg/s
%M3=60%
GM4=1kg
%m4=5%
GM4 =BDM4 + %M4GM4
1kg/s= BDM4 + 0.05(1kg/s)
BDM4 = 0.95
GM3= BDM3 + %M3GM3
GM3= 0.95 + .6GM3
GM3= 2.375 kg/s
4. Banana chips enter dryer containing 60% moisture and leaves with 5% water. Find the
mass of the original product if the bone-dry-weight is 1 kg/s.
a) 2.5 kg/s
b) 1.5 kg/s
c) 3.5 kg/s
d) 4.5 kg/s
%m4=5%
BDM3 = BDM4 = 1kg
GM3 = BDM3 + %m3GM3
GM3 = 1kg/s + 0.6 GM3
GM3 = 2.5 kg/s
5. A machine or equipment used for drying process.
a) Dewaterer
b) Evaporator
c) Dryer
d) Filter
6. Water occurring when chemical components of the material changes its chemical
composition by heat or other means is called:
a) Moisture
b) Chemically combined water
c) Dewaterer
d) Steam
7. A wet-type mechanical classifier (solids separator) n which solids settle out of the
carrier liquid and are concentrated for recovery is known as:
a) Dryer
b) Filter
c) Dewaterer
d) Evaporator
8. A term used for describing the moisture in the mass that is not on the surface of the
material.
a) Moisture content
b) Inherent moisture
c) Vapor
d) Chemically combined water
9. The main principle of the term drying process.
a) Water removal
b) Dewatering
c) Evaporation
d) Compression
10. The common term for processes using only pressure, suction, or decantation.
a) Drying
b) Evaporation
c) Dewatering
d) Decantation
11. The process of removing moisture in varying amounts from solid or semi-fluid
materials; the process may be accomplished by pressure, suction, decantation, or evaporation.
a) Dewatering
b) Drying
c) Evaporation
d) Decantation
12. A type of dryer that may have the flame from combustion impinging on the material
being dried; or the gases of combustion may be mixed with additional air so that mixture in
contact with the material is reduced in temperature.
a) Direct-type dryer
b) Indirect-type dryer
c) Steam-heated type dryer
d) Centrifugal dryer
13. The most commonly used dryer that consists of a rotating cylinder inside which the
materials flow while getting in contact with the hot gases. The cylinder is tilted at right angle and
fitted with lifting flights. This dryer is used for copra, sand, or wood chips.
a) Rotary dryer
b) Tower dryer
c) Hearth dryer
d) Centrifugal dryer
14. A dryer where material is in contact with steam pipes or the air is passed over steam
heaters and then over of through the material being dried.
a) Steam-heated dryer
b) Indirect-heat type dryer
c) Direct-heat type dryer
d) Tower dryer
15. A dryer that consists of a centrifuge revolving at high speed causing the separation,
by centrifugal force, of the water from the material. This dryer is used for drying fertilizer, salt,
and sugar.
a) Centrifugal dryer
b) Hearth dryer
c) Tower dryer
d) Rotary dryer
16. A dryer type in which the material to be dried is supported on a floor through which the hot
gases pass. This dryer is used for copra, coal, and enamel wares.
a) Centrifugal dryer
b) Hearth dryer
c) Rotary dryer
d) Tower dryer
17. Those substance that are particularly variable in the moisture content that they can
possess at different times.
a) Wet materials
b) Hygroscopic materials
c) Gross materials
d) Bone-dry-weight material
18. The hygroscopic moisture content of a substance expressed as a percentage of the
bone-dry-weight of the material.
a) Moisture content
b) Regain
c) Bone-dry-weight
d) Gross weight
19. A dryer that consists of trays, carrying the materials to be dried, plated in a
compartment or moving conveyor. This type of dryer is used for ipil-ipil leaves, and grains.
a) Centrifugal dryer
b) Infrared ray dryer
c) Tray dryer
d) Hearth dryer
20. A type of dryer that consists of a vertical shaft in which the wet fees is introduced at
the top and falls downward over baffles while coming in contact with hot air that rises and
exhaust at the top. This dryer is used for drying palay, wheat, and grains.
a) Rotary dryer
b) Hearth dryer
c) Tower dryer
d) Tray dryer
Cooling Towers
STUDENT’S SELF TEST
Instruction: Select the correct answer.
1. A mechanical draft cooling tower cools the cooling water from 60°C to 25°C at the
rate of 8 kg/sec. Atmospheric air enters the tower at a state of 20°C DB and 16°C WB. The air
leaves the tower at 35°C. What is the temperature of approach?
a) 9°C
b) 6°C
c) 8°C
d) 35°C
𝐶𝐴 = 𝑡𝑑𝑏4 − 𝑡𝑤𝑏1
𝐶𝐴 = 25℃ − 16℃
𝐶𝐴 = 9℃
2. In problem #1, determine the cooling range in °C.
a) 19°C
b) 35°C
c) 9°C
d) 15°C
𝐶𝐴 = 𝐶𝑜𝑜𝑙𝑖𝑛𝑔 𝑅𝑎𝑛𝑒 = 9℃
3. In problem #2, compute the cooling tower efficiency in percent.
a) 79.55
b) 59.57
c) 95.75
d) 75.95
𝑒𝑐𝑡 =
𝑡𝑑𝑏3−𝑡𝑑𝑏4
× 100%
𝑡 𝑑𝑏3 −𝑡 𝑤𝑏1
𝑒𝑐𝑡 = 60−25 × 100%
60−16
𝑒𝑐𝑡 = 79.55%
4. The amount of water carried by air in a cooling tower is 0.1134 kg/sec. The change in
humidity ratio is 0.025 kg/kg d.a. Determine the volume of air needed if the specific volume is
0.385 m3/kg expressed in m3/min.
a) 5.10
b) 7.85
c) 10.5
d) 12.95
𝑚1 = 0.1134 𝑘𝑔 = 𝑚 𝑎 ∆𝐻𝑅 = 0.025 𝑘𝑔
𝑠
𝑘𝑔
𝑣1 =? v1 = 0.0385 𝑚
3
𝑘𝑔
𝑚𝑎 = 𝑣1 × ∆𝐻𝑅
v1
0.1134 𝑘𝑔 ×
𝑠
60𝑠
1𝑚𝑖𝑛
=
𝑣1
𝑚3
0.0385 𝑘𝑔
× 0.025 𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑣1 = 10.5 𝑚𝑖𝑛
5. Hot water from an engine enters the cooling tower circuit at 50°C and exits the tower
at 32°C. If the ambient condition is at 35°C dry-bulb and 24°C wet-bulb, what is the cooling
effectiveness of the cooling tower in percent?
a) 60
b) 65
c) 70
d) 75
Engine
ENGINE
GCP
1
Water in
2 Air out 3
4 Water out
CT
Air in
5 Feed H2O
tdb3 = 50oC
tdb4 = 32oC
tdb1 = 35oC
twb1 = 24oC
ect =
ect =
𝑡𝑑𝑏3−𝑡𝑑𝑏4
𝑡𝑑𝑏3−𝑡𝑤𝑏1
50−32
50−24
x100%
x100%
ect = 69.23% = 70%
6. In problem #5, what is the actual cooling range?
a) 18°C
b) 21°C
c) 15°C
d) 8°C
𝐴𝐶 =?
𝐴𝐶 = 𝑡𝑑𝑏4 − 𝑡𝑤𝑏1
𝐴𝐶 = 32°𝐶 − 24°𝐶
𝐴𝐶 = 8℃
7. In problem #5, what is the theoretical cooling range?
a) 24°C
b) 26°F
c) 26°K
d) 25°R
𝑇𝐶𝐴 = 𝑡𝑑𝑏3 − 𝑡𝑤𝑏1
𝑇𝐶𝐴 = 50℃ − 24℃
𝑇𝐶𝐴 = 26℃ 𝑜𝑟 26°
8. In problem #5, what is the temperature approach in °C?
a) 6
b) 8
c) 12
d) 10
AC = 8°C
9. In problem #5, if water flows at the rate of 10 kg/sec, air entering tower has a heat
enthalpy of 80 kJ/kg and exits at 125 kJ/kg, what is the required air flow rate in kgs/hr.
a) 55,000
b) 62,500
c) 60,300
d) 63,580
𝑚𝑤𝑎𝑡𝑒𝑟 = 10
ℎ1 = 80
𝑘𝑔
𝑠
𝑘𝐽
𝑘𝑔𝑑𝑎
ℎ2 = 125
𝑘𝐽
𝑘𝑔𝑑𝑎
𝑚𝑎𝑖𝑟 = ?
𝑄𝑅ℎ20 = 𝑄𝐴𝑎𝑖𝑟
(𝑚𝐶𝑝∆𝑇3−4) = (𝑚∆h)𝑎𝑖𝑟
𝑘𝐽
𝑚𝑎𝑖𝑟 = 𝑚ℎ20 (
𝐶𝑝∆𝑇3−4
∆ℎ
) = 10
𝑘𝑔 4.1868 𝑘𝑔∙𝐾 (50 + 273 − 32 + 273)𝐾
𝑠
(
125
𝑘𝐽
𝑘𝑔
− 80
𝑘𝐽
) 𝑥
3600 𝑠
ℎ𝑟
𝑘𝑔
𝑚𝑎𝑖𝑟 = 60 289.92 ≈ 60 300 𝑘𝑔⁄ℎ𝑟
10. In problem #5, the change in the humidity ratio of the incoming and exiting air is
0.00165. What is the required make-up water in kg/sec?
a) 0.156
b) 0.028
c) 0.037
d) 0.310
𝑘𝑔
∆𝐻𝑅1−2 = 0.00165 𝑘𝑔
𝑚𝐹𝑊 = ?
𝑚𝑤𝑣1 + 𝑚𝑤𝑣3 + 𝑚𝐹𝑊 = 𝑚𝑤𝑣2 + 𝑚𝑤𝑣4
𝑚𝐹𝑊 = 𝑚𝑤𝑣2 + 𝑚𝑤𝑣4 − (𝑚𝑤𝑣1 + 𝑚𝑤𝑣3)
also, 𝐻𝑅 = 𝑚 𝑤𝑣 ; 𝑚 𝑤𝑣 = 𝐻𝑅(𝑚 𝑑4 )
𝑚 𝑑𝑎
𝑚𝐹𝑊 = (𝐻𝑅𝑚6) + 𝑚𝑤𝑣4 − (𝐻𝑅𝑚𝑑𝑎) − 𝑚𝑤𝑣3
𝑚𝐹𝑊 = 𝑚𝑑𝑎(𝐻𝑅1 − 𝐻𝑅2)
𝑚𝐹𝑊 = 60 289.92
𝑘𝑔
1 ℎ𝑟
𝑘𝑔𝑤𝑣
𝑥
)(0.00165 𝑘𝑔𝑑𝑎)
ℎ𝑟 3600 𝑠
𝑚𝐹𝑊 = 0.0276 ≈ 0.028
𝑘𝑔𝑤𝑣
𝑠
11. The approach and efficiency of a cooling tower are 10°C and 65%, respectively. If the
temperature of water leaving the tower is 27°C, what is the temperature of water entering the
tower?
a) 45.57°C
b) 47.55°C
c) 55.47°C
d) 54.75°C
𝑒𝐶𝑇 =
𝑇3 − 𝑇4
𝑇3 − 𝑇𝑤𝑏 1
𝐶𝐴 = 𝑡𝑑𝑏4 − 𝑡𝑤𝑏1
10 = 27 − 𝑡𝑤𝑏1
𝑡𝑤𝑏1 = 17℃
0.65 =
𝑇3 − 27
𝑇3 − 17
𝑇3 = 45.57℃ (𝑎)
12. The change of temperature of water entering the cooling tower and the WB
temperature of surrounding air is 23°C, and the efficiency of the tower is 65%. If the mass flow rate
of the water is 15 kg/s, determine the heat carried away by the air, in kW.
a) 983.93 kW
b) 938.93 kW
c) 993.83 kW
d) 939.83 Kw
Q=?
𝑄 = 𝑚𝑎(ℎ2 − ℎ1)
Heat Balance
Heat Absorbed by Air=Heat Rejected by Water
𝑚𝑎(ℎ2 − ℎ1) = 𝑚𝑤𝐶𝑝(𝑡3 − 𝑡4)
𝑄 = 𝑚𝑤𝐶𝑝(𝑡3 − 𝑡4)
𝑘𝑔
𝑚𝑤 = 15
𝑠
(𝑎𝑠 𝑔𝑖𝑣𝑒𝑛)
𝐶𝑝𝑤𝑎𝑡𝑒𝑟 = 4.1868
𝑘𝐽
𝑘𝑔 − 𝑘
𝑡3 − 𝑡4 =?
From,
𝑒=
𝑡3 − 𝑡4
𝑡3 − 𝑡𝑤𝑏
And,
𝑡3 − 𝑡𝑤𝑏(𝑎𝑠 𝑠𝑡𝑎𝑡𝑒𝑑)
𝑒 = 0.65
So,
0.65 =
𝑡3 − 𝑡4
23
𝑡3 − 𝑡4 = 0.65(23)
𝑡3 − 𝑡4 = 14.98
Therefore,
𝑄 = 15
𝑘𝑔
𝑘𝑗
(4.1868
) (14.98𝑘)
𝑠
𝑘𝑔 − 𝑘
𝑄 = 938.89 𝑘𝑤 (𝐵)
13. Water at 55°C is cooled in a cooling tower that has an efficiency of 65%. The
temperature of the surrounding air is 32°C DB and the WB temperature is 27°C. The heat
dissipated from the condenser is 361 kW. Find the capacity, in lps, of the pump used in circulating
the cooling water.
a) 4.913 lps
b) 4.391 lps
c) 4.193 lps
d) 4.139 lps
E= 65%
Qw = 361 kW
Solution:
𝑄𝑤 = 𝑚𝑤𝑐𝑝(𝑡3 − 𝑡4)
t4= ?
From;
𝑒=
𝑡3 − 𝑡4
𝑡3 − 𝑡𝑤𝑏
. 65 =
× 100%
55 − 𝑡4
55 − 27
𝑡4 = 36.8℃
So;
𝑚𝑤 =
𝑚𝑤 =
𝑄𝑤
𝑡3 − 𝑡4
361 𝑘𝐽/𝑠
(55 − 36.8)𝐾 (4.1868 𝑘𝐽⁄𝑘𝑔 − 𝐾 )
𝑚𝑤 = 4.7375 𝑘𝑔⁄𝑠
From;
(𝜌 = 𝑚⁄𝑣)𝑤𝑎𝑡𝑒𝑟
𝜌𝑤𝑎𝑡𝑒𝑟 = 1000 𝑘𝑔⁄ 3
𝑚
Then;
𝑘𝑔⁄
𝑠
𝑣 = 𝑚𝑤𝑎𝑡𝑒𝑟 = 4.7375
𝑘𝑔⁄
𝜌𝑤𝑎𝑡𝑒𝑟
1000
𝑚3
𝑣 = 0.004738
𝑚3
𝑠
×
1000 𝐿
1 𝑚3
𝑣 = 𝟒. 𝟕𝟑𝟖 𝑳𝒑𝒔
14. A cooling tower with an efficiency of 70% is used to cool directly the jacket water of
a 400 Hp Diesel engine. If the temperature of approach is 10°C, find the lpm of jacket water that
may be cooled effectively by the tower. Assume ambient air of 30°C DB and 24°C WB.
a) 162 lpm
b) 0.162 m3/min
c) 2.70 lps
d) All of the above
td𝑏1 = 30⁰ C
twb = 24⁰ C
e = 70%
p = 400 HP
𝑡4 - tw𝑏1 = 10⁰ C
So from,
𝑡4 - tw𝑏1 = 10⁰ C
𝑡4 = 10⁰ C + 24⁰ C
𝑡4 = 34⁰ C
𝑡 3 −𝑡4
From, e =
7=
𝑡3 −tw𝑏1
; 𝑡3 = ?
𝑡3−34
𝑡3 −24
𝑡3 = 57.33
𝑄𝑤 = 400 HP x 0.746 𝐾𝑊 = 298.4 𝐾𝐽
1 𝐻𝑃
From,
𝑄𝑤 = 𝑀𝑤Cp(𝑡3 - 𝑡4)
𝑀𝑤 =
𝑀𝑤 =
𝑄𝑤
𝐶𝑝 (𝑡3 − 𝑡4)
298.4 𝐾𝐽
𝑆
4.1868
𝐾𝐽
𝐾𝐺−𝐾
𝑀𝑤 = 3.0549 𝐾𝐽
𝑆
𝑆
From,
p=
𝑀𝑤
𝑉
V = 𝑀𝑤 =
𝑃𝑤
3.0549 𝐾𝐺
𝑆
1000 𝐾𝐺
𝑀3
3
V = 0.0030549 𝑀 x 1000 L x
𝑆
1 𝑀3
60 S
1 MIN
V = 183.296 LPM
15. Determine the approximate amount of air to be handled and the quantity of make-up
water required by a cooling tower that is to cool 12.67 lps from 36°C to 31°C. Atmospheric
conditions are 35°C DB and 26°C WB. Assume that air leaves the tower at 32°C DB and 90% RH.
Properties of air entering the tower: h = 80.38 kJ/kg d.a. and W = 0.0177 kg/kg d.a. Properties of
air leaving the tower: h = 102.0 kJ/kg d.a. and W = 0.0274 kg/kg d.a. [Ans. 12.22 kg/s, 0.119 kg/s]
a) 12.22 kg/s, 0.119 kg/s
b) 12.22 kg/s, 1.19 kg/s
c) 12.22 kg/s, 0.911 kg/s
d) 12.22 kg/s, 1.91 kg/s
𝑚1(ℎ2 − ℎ1) = 𝑚3(𝑡3 − 𝑡4)(𝐶𝑝)
𝜌=
𝑚
𝑣
𝐿
; 𝑚3 = 𝜌𝑣 ; 1000 𝑘𝑔
(12.67 𝑠 ) (
𝑚3
𝑘𝑔
𝑠
𝑚3 = 12.67
𝑚1 =
12.67 𝑘𝑔 ( 4.186
𝑚1 = 12.23
𝑠
(102
𝑘𝐽
𝑘𝐽
𝑘𝑔−𝐾
)(36−31)𝐾
−80.38 𝑘𝐽 )
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑠
𝑚5 = 𝑚1(𝐻𝑅2 − 𝐻𝑅1)
= 12.23 (0.0274 − 0.0177)
𝑚5 = 0.119
𝑘𝑔
𝑠
𝑚3
1000𝐿
) = 12.67
𝑘𝑔
𝑠
16. An atmospheric cooling tower is to provide cooling for the jacket water of a 4-stroke,
800 Hp diesel generator. The cooling tower efficiency is 65% at a temperature of approach of
10°C. If the ambient is at 32°C DB and 26°C WB, determine the cooling water supplied to the
diesel engine, in lpm. Generator efficiency is 96%.
Q4
a) 25,344 lpm
b) 23,544 lpm
Q2
c) 24,534 lpm
d) 32,544 lpm
EP = 800 HP
BP = Q1
Ec
Diesel Engine
(4 –Stroke )
Generator
3
Q3
Q5
% Q3 =65%
Cooling
tower
2
Air out
1
Air in
5
FEED WATER = 0
4 Water out
SOLUTION
tdb1= 32 ᵒC (4.73007 kPa)
twb1 = 26ᵒ C (93.33025 kPa)
ect=
𝑡𝑑𝑏 3 −𝑡𝑑𝑏 4
𝑡𝑑𝑏2−𝑡𝑤𝑏1
ect= 10ᵒ C = tdb4 –twb1
tdb4 =10ᵒ C + 26 ᵒ C
tdb4 =36ᵒ C
%Q1 = 30%
𝑄1
%Q1= 𝐸𝑐 𝑥 100%
𝐸𝑐 =
𝑄1
𝐸𝑐
; Q 1 >>> BP
𝐸𝑃
𝑁𝑔 =
𝐵𝑃
𝐵𝑃 =
800600
𝐸𝑃
=
𝑁𝑔
0.97
𝐵𝑃 = 824.742
𝐸𝑐 =
𝐸𝑐 =
𝑄1
𝐸𝑐
824.742
0.30
Ec = 2749.14 KW
So,
%Q3 =
𝑄3
𝐸𝑐
𝑥 100%
Q3 = Ec %Q3 =(2749.14)(0.25)
Q3 = 687.285 KW
But
Q3 = Qwater cooling = mw cpw(Δt3-4 )
Q3 = mw cpw(Δt3-4 )
tdb3 –tdb4 =
𝑄3
(𝑚 𝐶𝑝)𝑤𝑎𝑡𝑒𝑟
so,
60%
𝑡𝑑𝑏3−35 ᵒ𝐶
100%
= 𝑡𝑑𝑏
2−25.9 ᵒ
𝐶
tdb2 = 50.9 ᵒC
m4 =m3 =m water =𝑡𝑑𝑏
𝑄3
3−𝑡𝑑𝑏4(𝐶𝑝)
mwater=
687.285 𝐾𝑊
(50.9−35.9)ᵒ𝐶 (4.1868
mwater= 10.9437
V4 =
𝑚𝑤
𝜌 𝐻2𝑂
=
𝐾𝐽
𝐾𝐺−𝐾
)
𝑘𝑔
𝑠
𝑘𝑔 3600 𝑠
)(
𝑠
1 ℎ𝑟
𝑘𝑔
1𝑚 3
1000 𝑚 3 𝑥 1000 𝐿
(0.9437
)
17. The amount of water carried by air in a cooling tower is 6.8 kg/min. The change in
humidity ratio in the tower outlet and inlet is 0.025 kg/kg d.a. Determine the volume flow rate of
air needed if the specific volume is 0.8123 m3/kg d.a.
a) 221 m3/min
b) 221 m3/min
c) 122 m3/min
d) 212 m3/min
Given:
∆𝑀𝑤1 = 6.8
𝑘𝑔𝑤
𝑚𝑖𝑛
∆𝐻𝑅1−2 = 0.025
𝑘𝑔𝑤
𝑘𝑔𝑑𝑎
Required:
𝑣 𝑎 𝑖 ̇𝑟= ? @ 𝑣𝑎𝑖𝑟 = 0.8173
𝑚3
𝑘𝑔 𝑑𝑎
Solution:
𝑣̇𝑎𝑖𝑟 = (𝑣𝑎𝑖𝑟𝑚𝑑𝑎)
∆𝑚
∆𝐻𝑅 = ∆𝑚 𝑤 → 𝑚 𝑑𝑎 = 𝑤
∆𝐻𝑅
𝑚 𝑑𝑎
𝑚𝑑𝑎 =
𝑤
6.8𝑘𝑔
𝑚𝑖𝑛
0.025𝑘𝑔𝑤
𝑚𝑖𝑛
𝑚𝑑𝑎 = 272 𝑘𝑔𝑑𝑎
𝑚𝑖𝑛
𝑣̇𝑎𝑖𝑟 = (0.8123
𝑣̇𝑎𝑖𝑟
𝑚3
)(272𝑘𝑔𝑑𝑎)
𝑘𝑔 𝑑𝑎
𝑚3
= 220.9 → 221
𝑚𝑖𝑛
18. The change of enthalpy of air in a cooling tower is 81.42 kJ/kg d.a. and the mass flow
rate of air is 206 kg/min. Water enters the tower at the rate of 190 lpm and 46°C. Determine the
exit temperature of water.
a) 25°C
b) 24°C
c) 24.92°C
d) 42°C
Given:
Required:
∆h = 81.42
𝑘𝐽
𝑘𝑔𝑑.𝑎.
𝑚́ 𝑑𝑎 = 206
𝑘𝑔 𝑑𝑎
𝑚𝑖𝑛
𝑡𝑑𝑏4= ?
Solution:
(ρ =
𝑚
𝑣
)
m = ρv
𝑚́ 3 = (190 lpm) (1000
𝑘𝑔
𝑚3
1𝑚 3
)
1000𝑙
)(
𝑡𝑑𝑏3 = 46°C
Water in
C.T
Air out
Air in
Makeup
Water out
H2O=0
𝑄𝑅𝐻2𝑂 = 𝑄𝐴𝑎𝑖𝑟
𝑚́ 𝑤 𝐶𝑝𝑤 ∆𝑡3−4 = 𝑚́ 𝑑𝑎 ∆ℎ
𝑚́ 𝑤𝐶𝑝𝑤(𝑡𝑑𝑏3 = 𝑡𝑑𝑏4) = 𝑚́ 𝑑𝑎 (∆ℎ)
(190
𝐿
𝑚𝑖𝑛
𝑥
1𝑚3
1000𝐿
𝑘𝑔
𝑘𝐽
(46°𝐶 − 𝑡𝑑𝑏 4) = (
𝑥 1000 𝑚
3 ) (4.1862 𝑘𝑔−𝑘 )
206 𝑘𝑔 𝑑𝑎
𝑘𝐽
𝑚𝑖𝑛
𝑘𝑔 𝑑𝑎
) (81.42
)
𝑡𝑑𝑏4 = 24.92°𝐶
19. Water at 55°C is cooled in a cooling tower that has an efficiency of 65%. The
temperature of the surrounding air is 32°C DB and 27.40°C WB. The heat dissipated from the
condenser is 2,300,000 kJ/hr. Find the capacity in lps, of the pump used in the cooling tower.
a) 8.57 lps
b) 5.78 lps
c) 7.58 lps
d) 7.85 lps
GIVEN:
Tdb3=55C (16.12KPa)
ect = 65%
tdb1 = 32C (4.73407KPa)
twb1 = 26.4C (3.6189KPa)
Q3to4 = 2300000 KJ/ hr
REQUIRED:
Ṽ4= ? at lph
SOLUTION:
Q3to4 = m4or3Cpw (∆t3-4)
𝑒𝑡 =
𝑡𝑑𝑏3−𝑡𝑑𝑏4
𝑡𝑑𝑏3−𝑡𝑤𝑏1
𝑡𝑑𝑏4 =
𝑡𝑑𝑏1+2𝑡𝑤𝑏1+𝑡𝑑𝑏3
4
32+2(27.4)+55
𝑡𝑑𝑏4 =
4
𝑡𝑑𝑏4 = 35.45℃
𝑚4𝑜𝑟3 =
𝑄3−4
𝐶𝑝𝑤(∆𝑡2−4)
=
2300000𝐾𝐽
ℎ𝑟
4.1868 𝐾 𝑗 (55−35.45)𝐾
𝑘𝑔−𝐾
𝑚4𝑜𝑟3 = 28099.57725 𝑘𝑔
ℎ𝑟
1ℎ𝑟
Ṽ4 =
𝑚4
𝑃𝐻2𝑂
=
28099.57725𝐾𝑔
(
)
ℎ𝑟 3600𝑠
1𝑚 3
1000 𝑘𝑔3 (1000𝐿 )
𝑚
Ṽ4 = 28 099.58
𝐿
𝑜𝑟
ℎ𝑟
𝑙𝑝ℎ
Ṽ4 = 7.805 𝑙𝑝𝑠 (𝑏)
20. An atmospheric cooling tower is to provide cooling for the jacket water of a 4-stroke,
800 kW diesel generator. The cooling efficiency is 60% at a temperature approach of 10°C. If the
ambient air has a RH of 70% and DB temperature of 32°C, determine the amount of cooling water
supplied to the engine, in liters per hour. Generator efficiency is 97%, used work is 30%, and
cooling loss is 25%.
a) 39,804 lph
b) 38,904 lph
c) 34,908 lph
d) 34,809 lph
Given:
ec = 60%
CA = 10˚C
RH1 = 70%
tdb = 32˚C
ηg = 97%
%Q3 = 25%
%Q1 = 30%
Req’d: V4= ? in L/hr
Ec = Q1 + Q2 + Q3 + Q4 + Q5
Solution:
𝜌 = 𝑚 4 ; 𝑣4 =
𝑣4 =
𝑣4 =
From:
𝑒𝑐 =
𝑣4
𝑚 𝑤𝑣
𝜌𝐻2𝑂
𝑚4
𝜌
; But,
𝐻𝑅 = 𝑚 𝑤𝑣 ; 𝑚 𝑤𝑣 = (𝐻𝑅)(𝑚 𝑑𝑎 )
𝑚 𝑑𝑎
(𝐻𝑅)(𝑚𝑑𝑎)
𝜌𝐻2 𝑂
𝑡𝑑𝑏3 − 𝑡𝑑𝑏4
𝑡𝑑𝑏3 − 𝑡𝑤𝑏1
𝑥 100%; Where:
CA = tdb4 – twb1 = 10˚C ; tdb4 = 10˚C + twb1
10˚C + 25.9˚C = 35.9˚C
@Pt. 1
𝑅𝐻 =
𝑃𝑤𝑣
;
𝑃 𝑑𝑎
P wv = (RH)(Pda)
Pwv1 = 0.7(4.7301)
So,
Pwv1 = 3.31105 kPa
twb1 = 25.9˚C
%Q1 = 30%
%𝑄1 = 𝑄1 𝑥 100% ; 𝐸𝑐 = 𝑄1
𝐸𝑐
Also,
η𝑔 =
𝐸𝑃
𝐵𝑃
%𝑄1
𝐵𝑃 =
Then,
𝐸𝑐 =
𝐸𝑃
η𝑔
800 𝑘𝑊
=
0.97
; 𝐵𝑃 =
𝑄1
;
%𝑄1
824.742 𝑘𝑊
Where: Q 1 = BP
824.742 𝑘𝑊
= 2799.14 𝑘𝑊
𝐸𝑐 =
0.30
So,
%𝑄3 = 𝑄3 𝑥 100%; Q 3 = (%Q 3)(Ec)
𝐸𝑐
Q3 = (0.25)(2799.14 kW) = 687.285 kW
But,
Q3 = Qwater cooling = mwCpw(∆t3-4)
Q3 =mwCpw(tdb 3 – tdb 4) ; 𝑡𝑑𝑏 3 − 𝑡𝑑𝑏 4 =
𝑄3
(𝑚𝐶𝑝)𝑤
@ mw = 0
mwv4 = mwv3 = mwater
So,
𝑒𝑐 =
0.6 =
𝑡𝑑𝑏3 − 𝑡𝑑𝑏4
𝑥 100%
𝑡𝑑𝑏3 − 𝑡𝑤𝑏1
𝑡𝑑𝑏3 − 35.9˚𝐶
𝑡𝑑𝑏3 − 25.9˚𝐶
tdb3 = 50.9˚C
Then,
𝑚4 = 𝑚3 = 𝑚𝑤𝑎𝑡𝑒𝑟 =
𝑚𝑤𝑎𝑡𝑒𝑟
(𝑡𝑑𝑏3 − 𝑡𝑑𝑏4)𝐶𝑝𝑤
687.285 𝑘𝑊
𝑘𝑔
=
= 10.9437
𝑠
(50.9 − 35.9)˚𝐶(4.1868) 𝑘𝐽
𝑘𝑔−𝐾
Therefore,
𝑚𝑎
𝑉4 =
𝜌𝐻2 𝑂
𝑉4 =
𝑄3
(10.9437
(1000
𝑘𝑔
𝑠
)(3600 ℎ𝑟
)
𝑠
3
𝑘𝑔
)( 1 𝑚 )
𝑚 3 1000 𝐿
𝐿
or LPH
ℎ𝑟
𝑉4 = 39397.32
𝑉4 ≈ 39804 𝐿𝑃𝐻 (a)