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The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate at the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

The psychrometric chart below shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )

The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate at the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

The psychrometric chart below shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C. This is the limit for evaporative cooling, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )

The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate at the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

This charge shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )

The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate at the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

The psychrometric chart below shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )

The problem in your calculation is that it is not possible to evaporate the amount of water as you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

This charge shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )

The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $27\ ^\circ$C ($300$ K). But that many moles would supersaturate at the calculated $\Delta T = 24$ K. The amount of water evaporated in a parcel of air is much less.

This charge shows the evaporative cooling line from $27\ ^\circ$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $20\ ^\circ$C, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )