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There are a lot of parameters which effects to the cooling
tower design and operation. Some will be discussed here
through the examples below.
Example 4-1.
The water circulating rate is one of most important
primary variables. Obviously it is a key number in the
original design. A problem frequently encountered is
the prediction of the effects of changes in water circulation
rate on the temperatures of the water entering and leaving
an existing cooling tower. Assume an existing mechanical
draft cooling tower is operating at the following conditions,
and estimate the cold & hot water temperature when
the water flow rate is increased to 20,000 GPM, assuming
no change in the entering air mass flow rate, wet bulb
temperature, and heat load. (Actually, the air mass
is decreased due to the increase of pressure drop at
the fill with the increase of water.)
Given,
- Water Flow Rate (L1):
16000
- Entering Air Flow Rate
(G1): 80848
- Ambient Wet Bulb Temperature:
80.0
- Site Altitude: sea level
- Hot Water Temperature
(HWT, tw2): 104.0
- Cold Water Temperature
(CWT, tw1): 89.0
- Characteristic Curve Slope
(m): -0.800
- Alternative Water Flow
Rate (L2): 20000
(Solution)
Range, R1 = HWT - CWT = tw2 -
tw1 = 104 - 89 = 15oF
Water Flow Rate in Pound, L1 = Water Flow
Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3
lb/min
Heat Load, D1 = L1 x R1
= 133,333.3 x 15 = 2,000,000 BTU/min
Air Mass Flow Rate, G1 = 80,848 lb/min
Liquid to Gas Ratio, L/G1 = L1
/ G1 = 133,333.3 / 80,848 = 1.6492
Water Flow Rate in Pound, L2 = Water Flow
Rate x (500 / 60) = 20,000 x (500 / 60) = 166,666.7
lb/min
Heat Load, D2 = D1 = 2,000,000
BTU/min
Air Mass Flow Rate, G2 = G1 =
80,848 lb/min
Liquid to Gas Ratio, L/G2 = L2
/ G2 = 166,666.7 / 80,848 = 2.0615
Range, R2 = D2 / L2
= 2,000,000 / 166,666.7 or = R1 x (L1
/ L2) = 12oF
(The range must be calculated since the heat load is
same as the design condition but water flow rate was
changed.)
In estimating the cold water
temperature with the new water flow rate, there are
two methods. One is to find a new approach by means
of the computer. Another is to find it using the CTI
performance curves. Two methods shall be discussed.
With the use of the computer, the iteration is required
until the value of the new tower characteristic is exactly
equal to the new KaV/L (NTU) varying the approach by
means of computer.
First Step: Calculate
NTU at the design conditions as follows;
|
WATER SIDE |
AIR SIDE |
ENTH
DIFF. |
| Descriptions |
tw
(oF) |
hw
(Btu/Lb) |
Description |
ha
(Btu/Lb) |
1/(hw-ha) |
| tw1
+ 0.1 x R |
90.50 |
56.6478 |
ha1
+ 0.1 x L/G x R |
46.1645 |
0.0954 |
| tw1
+ 0.4 x R |
95.00 |
63.3426 |
ha1
+ 0.4 x L/G x R |
53.5858 |
0.1025 |
| tw1
+ 0.6 x R |
98.00 |
68.2591 |
ha1
+ 0.6 x L/G x R |
58.5334 |
0.1028 |
| tw1
+ 0.9 x R |
102.50 |
76.4013 |
ha1
+ 0.9 x L/G x R |
65.9547 |
0.0957 |
|
Sum
of 1 / (hw - ha) |
0.3964 |
|
Total Tower Demand
(NTU) = Cooling Range x Sum of 1 / (hw - ha) |
1.4866 |
Second Step: Calculate
a value of "C" of tower characteristic for
the design conditions as follows;
C = KaV/L / (L/G1)-m
= KaV/L x (L/G1)m = 1.4866 x (1.6492)0.8
= 2.21825
Third Step: Calculate
a new tower characteristic for the increased water flow
as follows;
New Tower Characteristic
= C x (L/G2)-m = 2.21825 x (2.0615)-
0.8 = 1.2436
(Note that "C" value is a constant value regardless
the change of water flow rate in finding the approach
at the alternative temperature conditions.) The new
tower characteristic for the increased water flow rate
can be calculated as above.
Forth Step: Iterate
until the value of new characteristic is equal to the
new NTU varying the value of approach.
To be continued. Please press the next button.... |
