Answer: The local atmospheric pressure in that location is 86.1 kPa.
Explanation:
Using the rate of evaporation and rate of heat transfer the heat of evaporation is calculated as follows.
[tex]h_{evap} = \frac{Q}{m}\\= \frac{nW}{m}\\= \frac{0.75 \times 2 kJ/kg}{\frac{1.19}{30 \times 60}}\\= 2268.9 kJ/kg[/tex]
Now, using enthalpy of vaporization the local atmospheric pressure is determined from data in A-5 using interpolation:
[tex]P = P^{*}_{1} + \frac{P^{*}_{2} - P^{*}_{1}}{h^{*}_{2} - h^{*}_{1}} (h_{evap} - h^{*}_{1})\\= 75 kPa + \frac{(100 - 75)}{(2257.5 - 2278)} \times (2268.9 - 2278) kPa\\= 86.1 kPa[/tex]
Thus, we can conclude that the local atmospheric pressure in that location is 86.1 kPa.
