The occupant of an air-conditioned room sets the supply air to the room at 18°C dry bulb and 60% relative humidity. If the pressure inside the room is 100.8 kPa, what is the specific enthalpy of the air at this state? Express your answer in kJ/kg da.

Where h is the specific enthalpy of the air, c_p is the specific heat of air at constant pressure, T is the dry bulb temperature, w is the humidity ratio, h_f is the specific enthalpy of saturated air at the given temperature T_f and T_f is the saturation temperature at the given pressure P.

Now, we can substitute the given values and solve for h as follows:

Given,T = 18^oCP

= 100.8 kPa RH

= 60 \%

We know that, w = 0.622 \frac{P_v}{P-P_v} Where P_v is the partial pressure of water vapor in the air.

Now, P_v = RH \times Pws Where Pws is the saturation pressure of water vapor at the given temperature T.Using the steam tables for water, we can find that

Pws (18^oC) = 1735 Pa

Therefore,P_v = 0.6 \times 1735

= 1041 Pa

Now, we can solve for w as follows:

[tex]w = 0.622 \times \frac{1041}{100800-1041}[/tex]

= 0.0081 kg/kg_{da}

The value of c_p at 18°C is 1.005 kJ/kgK.The specific enthalpy of saturated air at 18°C is 87.85 kJ/kg.

Now, using the formula, we get:

[tex]h = c_p T + w \Bigg[ h_f + c_p (T-T_f) \Bigg]h[/tex]

= 1.005 \times (18+273) + 0.0081 \Bigg[ 87.85 + 1.005 \times (18-12.79) \Bigg]

= \boxed{44.09 \; kJ/kg_{da}}

Therefore, the specific enthalpy of the air is 44.09 kJ/kg da.

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