»
» »
Cooling Towers
Figure 3 . Plume abatement coil |
|
|
|
|
Hot Fluid |
t ( mm ) |
t ( in ) |
U ( J / hr sqm C ) |
U ( Btu / hr sqft F ) |
Process Water |
0.889 |
0.035 |
2,381,330 |
116.5 |
Process Water |
1.245 |
0.049 |
2,360,890 |
115.5 |
Process Water |
1.651 |
0.065 |
2,330,230 |
114.0 |
Process Water |
2.108 |
0.083 |
2,309,790 |
113.0 |
Process Water |
2.769 |
0.109 |
2,258,690 |
110.5 |
Table 1 – Overall heat transfer coefficient |
Note : This step may take several iterations unless the mass flow of the hot fluid is known . If not , starting with a 2.1 C ( 3.8 F ) difference between the inlet and outlet hot side fluid temperatures is a good first assumption .
We then calculate the heat capacity for both the hot fluid and ambient air as follows :
|
C h
= M h cp h
= Q /( T hi – T ho
)
|
( Eq . 4 ) |
|
C a
= M a cp a
= Q /( T ai
– T ao )
|
( Eq . 5 ) |
Then , the heat capacity ratio is determined as follows :
C r
= C min / C max
( Eq . 6 )
Where :
C min
= The lower value of C h or C a
C max
= The higher value of C h or C a
The maximum heat transfer rate is calculated as follows :
Q max
= C min ( T hi
– T ai ) ( Eq . 7 )
From here , heat transfer effectiveness is determined by :
E = Q / Q max
( Eq . 8 )
Now , we need to determine the value of NTU . This comes from plotting the following equation for a range of C r and
NTU :
E = 1 – exp [( 1 / C r )( NTU ) 0 . 22 { exp [ -C r
( NTU ) 0 . 78 ] -1 }] ( Eq . 9 )
Figure 4 . Plume abatement coil with louver
Figure 6 has been created from this equation for C r between 0.1 and 1.0 and NTU between 0.0 and 5.0 Calculating the value for C r and E from equations 6 and 8 above , one can determine the value of NTU from Figure 6 . Next , an estimate of the overall heat transfer coefficient ( U ) needs to be made . Based on this being process water on the tube side and air on the fin side , the range of accept-
22 Heat Exchanger World October 2022 www . heat-exchanger-world . com