Explanation Express initial partial pressure. P ν 1 = ϕ 1 P g 1 = ϕ 1 P sat@93°F (I) Here, relative humidity at state 1 is ϕ 1 , initial water vapor pressure is P g 1 and saturation pressure at temperature of 93°F is P sat@93°F . Express initial specific humidity. ω 1 = 0.622 P ν 1 P 1 − P ν 1 (II) Here, pressure at state 1 is P 1 . Express initial enthalpy. h 1 = c p T 1 + ω 1 h g 1 @ 93 °F (III) Here, initial specific enthalpy at saturated vapor and temperature of 93°F is h g 1 @ 93 °F , specific heat at constant pressure of air is c p and temperature at state 1 is T 1 . Let the liquid water is supplied at a temperature almost equal to the exit temperature of the air stream, thus the evaporative cooling process travels through line of constant wet-bulb temperature, that is almost parallel to constant enthalpy lines. Thus, the enthalpy at state 1 and 2 is almost equal. h 2 ≅ h 1 (IV) Here, enthalpy at state 2 is h 2 . Express final specific humidity. ω 2 = 0.622 P g 2 P 2 − P g 2 (V) Here, pressure at state 2 is P 2 and final water vapor pressure is P g 2 . Express the exit temperature. h 2 = c p T 2 + ω 2 ( 1060.9 + 0.435 T 2 ) (VI) Here, exit temperature is T 2 . Conclusion: Refer Table A-4E, “saturated water-temperature table”, and write saturation pressure at temperature of 93°F using an interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (VII) Here, the variables denote by x and y is temperature and saturation pressure respectively. Show the saturation pressure corresponding to temperature as in Table (1). Temperature T ( ° F ) Saturation pressure P sat ( psia ) 90 ( x 1 ) 0.6990 ( y 1 ) 93 ( x 2 ) ( y 2 = ? ) 95 ( x 3 ) 0.8164 ( y 3 ) Substitute 90°F,93°F and 95°F for x 1 , x 2 and x 3 respectively, 0 .6990 psia for y 1 and 0.8164 psia for y 3 in Equation (VII). y 2 = ( 93°F − 90°F ) ( 0.8164 psia − 0 .6990 psia ) ( 95°F − 90°F ) + 0 .6990 psia = 0 .7676 psia = P sat@93°F Substitute 0.30 for ϕ 1 and 0 .7676 psia for P sat@93°F in Equation (I). P ν 1 = ( 0.30 ) ( 0 .7676 psia ) = 0.230 psia Substitute 0 .230 psia for P ν 1 and 14

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Chapter 14.7, Problem 93P

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The exit temperature of air.

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The exit temperature of air.

Solution Summary: The author analyzes the evaporative cooling process, which travels through line of constant wet-bulb temperature parallel to constant enthalpy lines.

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