Evaporative cooling, enthalpy

The specific enthalpies ia equation 9 can be determined as described earUer, provided the temperatures of the product streams are known. Evaporative cooling crystallizers operate at reduced pressure and may be considered adiabatic (Q = 0). As with of many problems involving equiUbrium relationships and mass and energy balances, trial-and-error computations are often iavolved ia solving equations 7 through 9. 

Example 4 Evaporative Cooling Air at 95 F dry-bulb temperature and 70 F wet-bulb temperature contacts a water spray, where its relative humidity is increased to 90 percent. The spray water is recirculated makeup water enters at 70 F. Determine exit dry-bulb temperature, wet-bulb temperature, change in enthalpy of the air, and quantity of moisture added per pound of dry air. 

When the dryer is seen as a heat exchanger, the obvious perspective is to cut down on the enthalpy of the air purged with the evaporated water. Minimum enthalpy is achieved by using the minimum amount of air and cooling as low as possible. A simple heat balance shows that for a given heat input, minimum air means a high inlet temperature. However, this often presents problems with heat-sensitive material and sometimes with materials of constmction, heat source, or other process needs. AH can be countered somewhat by exhaust-air recirculation. 

In hot, dry climates an inexpensive alternative to air conditioning is the swamp cooler. In this device water continuously wets porous pads through which fans blow the hot air. The air is cooled as the water evaporates. Use the information in Tables 6.2 and 6.3 to determine how much water must be evaporated to cool the air in a room of dimensions 4.0 m X 5.0 m X 3.0 m by 20.°C. Assume that the enthalpy of vaporization of water is the same as it is at 25°C. 

The cooling cycle starts when all parts of the refrigerator are at about 1.3K. At this temperature, the 3He is completely adsorbed by the pump. The pump temperature is now raised to about 25 K by means of an heater. At 25 K, the 3He is desorbed, and its pressure increases over the saturation pressure at 1.3 K. Consequently, 3He condenses in the part of the tube T internal to the copper support C and drops down into the evaporator E. In this phase, the latent heat of condensation and the enthalpy variation are delivered to the 4He bath. The cooling phase starts when all the 3He is condensed in E and the power on the pump heater is switched off. The pump starts cooling towards the bath temperature, reducing the pressure on liquid 3He in E. The adsorption heat of the 3He vapour is delivered to the 4He bath by L. 

Example 5.1. If the inner wall of a transfer tube is stainless steel of 4 mm outer diameter, 0.25 mm wall thickness and 1.5 m length, the room temperature enthalpy of the tube is about 2700 J. Should only the latent heat be used to precool the tube, about 11 of liquid 4He would be needed. However, the heat capacity of the evaporated gas contributes to the cooling. The room temperature enthalpy of the gas produced by 1 cc of liquid is about 190 J. If half of this enthalpy is used (slow initial transfer), then only about 30 cc of liquid is needed for precooling the transfer line. 

One important component of any evaporator installation is the equipment for condensing the vapour leaving the last effect of a multiple-effect unit, achieved either by direct contact with a jet of water, or in a normal tubular exchanger. If M is the mass of cooling water used per unit mass of vapour in a jet condenser, and H is the enthalpy per unit mass of vapour, then a heat balance gives ... 

Thus evaporation is a cooling process, and air moving through the tower will gain both moisture and enthalpy until, if possible, equilibrium with the water is reached. 

For the cycle, the total enthalpy change is zero. At the compressor, outside energy Wm is needed, and at the evaporator, heat transfer qm from the matter to be cooled is used to evaporate the refrigerant R-134a. 

Many processes other than chemical reactions absorb or release heat. For example, think about what happens when you step out of a hot shower. You shiver as water evaporates from your skin. That s because your skin provides the heat needed to vaporize the water. As heat is taken from your skin to vaporize the water, you cool down. The heat required to vaporize one mole of a liquid is called its molar enthalpy (heat) of vaporization Similarly,... 

The humid air takes water vapour from the film, by which the film of water and the air are cooled, until a time and position constant temperature is reached. It will be constant over the whole film because the adjoining wall is adiabatic and therefore no heat can be added to it. This adiabatic permanent temperature is called the wet bulb temperature The resistance to mass transfer is only on the gas side. Once the permanent temperature has been reached water still evaporates in the unsaturated air flowing over it. As the temperature of the water film is constant, the enthalpy of vaporization required for the evaporation will be removed as heat from the air. Fig. 1.52 indicates how the temperature and partial pressure of the water vapour in the air changes at this permanent state. The wet bulb temperature is lower than the temperature of the humid air flowing over the water surface. Therefore a wet substance can be cooled down to its wet bulb temperature by evaporation. 

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