Cooling Tower (2)
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Results Heater (kW) 0 0.5 1.0 1.5 Fully Opene d Blowe r Half Opene d Blowe r Fully Opene d Blowe r Half Opene d Blowe r Fully Opene d Blowe r Half Opene d Blowe r Air Inlet dry bulb, T 1 (C) 31.5 30.4 30.2 30.4 30.5 30.6 30.6 Air Inlet wet bulb, T 2 (C) 29.9 25.8 25.6 25.6 25.8 26 25.8 Air Outlet dry bulb T3 (C) 29.2 26.9 26.4 27 27.4 28 28.7 Ar Outlet bulb, T 4 (C) 26.5 27.0 26.0 26.8 27.4 28.3 29.1 Water Inlet Temperature, T 5 (C) 45.5 37.3 35.6 40 39.1 43.4 44.5 Water Outlet Temperature, T 6 (C) 51.1 25.8 25.4 25.7 26.2 26.5 26.9 Heater Power (W) 0 402 408 793 755 1190 1149 Dp Orifice 0 96 82 96 80 95 82 Dp Column 0 10 9 11 9 11 10 SAMPLE CALCULATION EXPERIMENT 1: EFFECT OF HEATING LOAD Fixed variables; 1. Air flow = 100% (Damper fully open) 2. Water flow rate = 1.0 LPM a) Heating load = 0.5 kW Range of cooling tower;
description
calculation of cooling tower
Transcript of Cooling Tower (2)
Results
Heater (kW) 0 0.5 1.0 1.5Fully
Opened Blower
Half Opened Blower
Fully Opened Blower
Half Opened Blower
Fully Opened Blower
Half Opened Blower
Air Inlet dry bulb, T1 (C)
31.5 30.4 30.2 30.4 30.5 30.6 30.6
Air Inlet wet bulb, T2 (C)
29.9 25.8 25.6 25.6 25.8 26 25.8
Air Outlet dry bulb T3 (C)
29.2 26.9 26.4 27 27.4 28 28.7
Ar Outlet bulb, T4
(C)26.5 27.0 26.0 26.8 27.4 28.3 29.1
Water Inlet Temperature,T5
(C)
45.5 37.3 35.6 40 39.1 43.4 44.5
Water Outlet Temperature,T6
(C)
51.1 25.8 25.4 25.7 26.2 26.5 26.9
Heater Power (W) 0 402 408 793 755 1190 1149Dp Orifice 0 96 82 96 80 95 82Dp Column 0 10 9 11 9 11 10
SAMPLE CALCULATION
EXPERIMENT 1: EFFECT OF HEATING LOAD
Fixed variables;
1. Air flow = 100% (Damper fully open)
2. Water flow rate = 1.0 LPM
a) Heating load = 0.5 kW
Range of cooling tower;
Range = Water inlet temperature, T5 - water outlet temperature, T6
= 37.3⁰C – 25.8⁰C
=11.5⁰C
Approach of cooling tower;
Approach = Water outlet temperature, T6 – Air outlet wet bulb, T2
= 25.8⁰C – 25.8 ⁰C
= 0⁰ C
Efficiency of cooling towers;
η = range
range+approach ×100
=11.511.5+0 × 100
= 100 %
Total cooling load;
Cooling load = pump input, Q1 + heating load
= 402 W + (0.5 kW × 1000W1kW ) = 902 W
b) Heating load = 1.0 kW
Range of cooling tower;
Range = Water inlet temperature, T5 - water outlet temperature, T6
= 40.0⁰C – 25.7⁰C
=14.3⁰C
Approach of cooling tower;
Approach = Water outlet temperature, T6 – Air outlet wet bulb, T2
= 25.7⁰C – 25.6 ⁰C
= 0.1⁰ C
Efficiency of cooling towers;
η = range
range+approach ×100
=14.3
14.3+0 .1 × 100
= 99.31 %
Total cooling load;
Cooling load = pump input, Q1 + heating load
= 793 W + (1.0kW × 1000W1kW ) = 1793 W
c) Heating load = 1.5 kW
Range of cooling tower;
Range = Water inlet temperature, T5 - water outlet temperature, T6
= 43.4⁰C – 26.5⁰C
=16.9⁰C
Approach of cooling tower;
Approach = Water outlet temperature, T6 – Air outlet wet bulb, T2
= 26.5⁰C – 26.0 ⁰C
= 0.5⁰ C
Efficiency of cooling towers;
η = range
range+approach ×100
=16.9
16.9+0.5 × 100
= 97.13 %
Total cooling load;
Cooling load = pump input, Q1 + heating load
= 1190 W + (1.5kW × 1000W1kW ) = 2690 W
Heating load (kW) 0.5 1.0 1.5
Range (⁰C) 11.5 14.3 16.9
Approach of cooling water (⁰C) 0 0.1 0.5
Efficiency of cooling water, η
(%)
100 99.31 97.13
Total cooling load (W) 902 1793 2690
EXPERIMENT 2: BLOWER DAMPER
Fixed variables;
1. Heating load = 0.5 kW
2. Water flow rate = 1.0 LPM
a) Air flow = 100 % (Blower fully open)
Range of cooling tower;
Range = Water inlet temperature, T5 - water outlet temperature, T6
= 37.3⁰C – 25.8⁰C
=11.5⁰C
Approach of cooling tower;
Approach = Water outlet temperature, T6 – Air outlet wet bulb, T2
= 25.8⁰C – 25.8 ⁰C
= 0⁰ C
Efficiency of cooling towers;
η = range
range+approach ×100
=11.511.5+0 × 100
= 100 %
Total cooling load;
Cooling load = pump input, Q1 + heating load
= 402 W + (0.5 kW × 1000W1kW ) = 902 W
a) Air flow = 50 % (Blower half open)
Range of cooling tower;
Range = Water inlet temperature, T5 - water outlet temperature, T6
= 35.6⁰C – 25.4⁰C
=10.2⁰C
Approach of cooling tower;
Approach = Water outlet temperature, T6 – Air outlet wet bulb, T2
= 25.4⁰C – 25.6 ⁰C
= -0.2⁰ C
Efficiency of cooling towers;
η = range
range+approach ×100
=10.2
10.2−0.2 × 100
= 102 %
Total cooling load;
Cooling load = pump input, Q1 + heating load
= 408 W + (0.5 kW × 1000W1kW ) = 908 W
Air Flow 100% 50%
Range (⁰C) 11.5 10.2
Approach of cooling water (⁰C) 0.7 0.5
Efficiency of cooling water, η
(%)
100 102
Total cooling load (W) 902 908
