The cooling water from the condenser of a power plant enters...

The cooling water from the condenser of a power plant enters a wet cooling tower at $40^{\circ} \mathrm{C}$ at a rate of $90 \mathrm{~kg} / \mathrm{s}$. The water is cooled to $25^{\circ} \mathrm{C}$ in the cooling tower by air that enters the tower at $1 \mathrm{~atm}, 23^{\circ} \mathrm{C},$ and 60 percent relative humidity and leaves saturated at $32^{\circ} \mathrm{C}$. Neglecting the power input to the fan, determine $(a)$ the volume flow rate of air into the cooling tower and $(b)$ the mass flow rate of the required makeup water.

Key Concepts

Conservation of Energy

This principle is fundamental in analyzing any thermal process. In the context of cooling towers, it involves accounting for the energy removed from the water stream as it cools and matching that energy with the energy absorbed by the air and the latent energy involved in evaporating water. By applying this principle, one can determine the heat transfer rates between the water and air, which is essential for evaluating the cooling tower performance.

Mass and Energy Balances

Mass and energy balances are used to track the flow of substances and their associated energy content in a system. In a cooling tower, these balances involve accounting for the cooled water, the evaporated water, and the air that interacts with the water. This approach is crucial to determine both the volume flow rate of air required for effective cooling and the makeup water needed to replace what is lost through evaporation.

Psychrometrics

Psychrometrics is the study of the thermodynamic properties of moist air. It deals with parameters such as temperature, humidity, and saturation, which are key when analyzing air-water interactions in a cooling tower. By understanding psychrometric principles, one can assess how much water vapor the air can hold at various temperatures and humidity levels, which directly influences the evaporative cooling process.

Evaporative Cooling

Evaporative cooling is a process where water absorbs heat from a fluid (usually air) during the phase change from liquid to vapor. In wet cooling towers, a portion of the water is intentionally evaporated to remove heat from the remaining water stream, thus lowering its temperature. This mechanism is central to the operation of cooling towers and is key to determining both the required air flow for sufficient evaporation and the makeup water needed to replenish the lost moisture.

Transcript

00:01 All right guys, to solve this problem, we first need to use the psychometric chart to determine the enthalpy's h1 specific volume v1 and absolute humidity that is w1 by using the given temperature that is t1, which is 23 degrees centigrade, and the relative humidity that is 51 is equal to 60%.

00:22 So we've got h1 is equal to 49 .9 kilojoules per kg.

00:30 We've got v1 is equal to 0 .85 cubic meters per kg.

00:43 Then we've got w1 is equal to 0 .01053.

00:48 So we will also need the same properties for the exit stream and the enthalpy is h3 and h4.

00:55 For the calculation, we will use the given temperature, t2 is equal to 32 degrees centigrade, t3 is equal to 40 degrees centigrade, t4 is equal to 25 degrees centigrade and the relative humidity 5 .2 that is 100%.

01:17 So we have h2 is equal to 110 .6 kilojoules per kg.

01:24 Then we've got w2 is equal to 0 .36065.

01:30 Then we've got h3 is equal to 168 kilojoules per kche.

01:39 And we have h4 is equal to 105.

01:44 Kilojoules per kg.

01:48 So we will now use the mass flow rate balances for water and air, rearrange them to simplify the upcoming calculations.

01:56 So we have ma1 is equal to ma2 is equal to ma.

02:05 So m3 minus m4 is equal to ma into w2 minus w1 is equal to m8 into w2 minus w1 is equal to m so for part a guys, we can take the energy balance equation and solve it for the needed mass flow rate ma.

02:30 For the calculations, we will need the given mass flow rate of water, that is m3 is equal to 90 kg per second...

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