Psychrometrics:Cooling Towers

Cooling Towers

Evaporate cooling is used in cooling towers for steam power plants, which require a large amount of heat rejection in the condenser. Figure 19.12 shows

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a schematic for a natural draft cooling tower, the hyperbolic shape is chosen for structural strength and low material use.

The cooling water flow m˙ cw comes from the condenser of the power plant, where it was heated to T1 while the water circulating in the steam cycle was condensed. The incoming cooling water is sprayed into the cooling tower, where some of it evaporates, which leads to cooling of the liquid. Since moist air is lighter than dry air—the low molar mass of vapor lowers the average molar mass—moist air rises and leaves the tower, while fresh environmental air at (T3, ω3) is drawn in at the bottom. Make-up water at m˙ m, T5 is added to compensate the loss of evaporated water and the mass flow m˙ cw leaves the cooling tower towards the condenser at T2. Normally, the make-up water is drawn from rivers or lakes, and that is why power plants are build close to these. As the rising moist air equilibrates with the environment, some of the added water might condense, which leads to clouds that normally can be seen above cooling towers.

The balances for air and water mass, and for energy, read

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Example: Cooling Tower

As an example we study the cooling tower for a W˙ = 300 MW power plant with a thermal efficiency η = 0.4. In the condenser, the cooling water is heated from T2 = 30 C to T1 = 40 C (the numbers refer to Fig. 19.12), thus the mass flow of cooling water is

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We ask for the required flows of air and make-up water, which both depend on the state of the incoming and exiting moist air. For further computation we assume that the incoming air is at T3 = 25 C, φ3 = 0.5, so that ω3 = 0.01, h3 kJ

1+ω = 51 kg , and that the make-up water is at T5 = 25C, so that, with (19.10), hf (T5) = 104.5 kJ , and hf (T1) = 167.2 kJ , hf (T2) = 125.4 kJ . Furthermore, we assume that the exiting air is saturated, so that ω4 = ωsat (T4).

The remaining unknowns in this problem are the air mass flow m˙ a, the make-up water flow m˙ m and the exit temperature T4. This problem differs from evaporative cooling as discussed above, due to the large amount of warm water sprayed into the air. The heat transfer between air droplets and air could only be described by a detailed heat transfer analysis. To simplify the problem, we assume T4 = 30 C which implies h4 = 100 kJ and ω4 = ωsat (T4) = 0.0272. Then we find from the conservation laws

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The river that provides the make-up water should have a sufficiently large mass flow rate, so that the removal of the make-up water will not disturb the ecological equilibrium of the river. An alternative to cooling towers is the direct use of river or lake water as cooling water. In this case, the heat rejected by the power plant is added to the river or lake. The related increase in water temperature changes the chemical environment, e.g., the amount of oxygen dissolved decreases with increasing temperature (Henry’s law, Sec. 22.10), which might disturb the ecological equilibrium more than the removal of some water for use in cooling towers.

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