Air at 95 degree F can hold maximum of 3.52 wt% water, defined as 100% humidity

Question

Air at 95 degree F can hold maximum of 3.52 wt% water, defined as 100% humidity at 95 degree F. On a hot humid day the temperature is 95 degree F and the air contains 3.2 wt% water (90.9 % humidity). The air cooled to 68 degree F because air at 68 degree F can hold a maximum of 1.44 wt% water ( i.e 100% humidity at 68 degree F), water condenses from the air. Hot humid air (95 degree F, 90.9 % humidity) flows into a cooler at a rate of 154 .0 K.g/min. Calculate the flow rate of 2 streams leaving the cooler: Air at 68 degree F and 100% humidity ( stream 1 : air leaving the cooling unit) Water condensed from the air (stream 2: water leaving the cooling unit) What is the degree of freedom for the cooling unit

Explanation / Answer

rate at which the air enters the cooler = 154 kg/min

Now, water present in the hot air enetring per min inside the cooler = 0.032*154 = 4.928 kg

thus, mass of air enetring per min = 154 - 4.928 = 149.072 kg

Now, let the air coming out of the cooler has a flow rate of 'x' kg/min

Thus, mass of air in the hot air comin out = 0.9856*x

Now, as mass entered = mass exit

thus, 0.9856*x = 149.072

or, x = air leaving the cooling unit with 100% humidity = 151.25 kg/min

Thus, water condensing = mass of air entered - mass of air exit = 154 - 151.25 = 2.75 kg/min

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