Q1: An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is

ANS:D - 0.152

To find the humidity of the air sample, we can use the definition of humidity ratio (𝑊W): 𝑊=𝑚𝑣𝑚𝑎W=ma​mv​​ Where:

• 𝑚𝑣mv​ = mass of water vapor
• 𝑚𝑎ma​ = mass of dry air

Given that the total pressure (𝑃P) is 101.3 kPa and the vapor pressure of water at the dew point temperature (𝑃𝑑𝑝Pdp​) is 7.30 kPa, we can calculate the partial pressure of water vapor (𝑃𝑣Pv​) in the air sample using the relationship: 𝑃𝑣=𝑃𝑑𝑝=7.30 kPaPv​=Pdp​=7.30kPa Next, we can calculate the partial pressure of water vapor at the dry bulb temperature (𝑇𝑑𝑏Tdb​) using the relationship: 𝑃𝑣=𝑃𝑠𝑎𝑡(𝑇𝑑𝑏)Pv​=Psat​(Tdb​) Where 𝑃𝑠𝑎𝑡(𝑇𝑑𝑏)Psat​(Tdb​) is the saturation vapor pressure of water at the dry bulb temperature. Given that the saturation vapor pressure of water at 60°C is 19.91 kPa, we have: 𝑃𝑣=19.91 kPaPv​=19.91kPa Now, we can calculate the mass of water vapor (𝑚𝑣mv​) in the air sample using the equation: 𝑚𝑣=0.622⋅𝑃𝑣𝑃−𝑃𝑣mv​=P−Pv​0.622⋅Pv​​ Where 0.622 is the ratio of the gas constants of water vapor to dry air. 𝑚𝑣=0.622⋅19.91101.3−19.91mv​=101.3−19.910.622⋅19.91​ 𝑚𝑣≈0.0071 kgmv​≈0.0071kg To find the mass of dry air (𝑚𝑎ma​), we can use the fact that the total mass of the air-water vapor mixture is the sum of the masses of dry air and water vapor: 𝑚𝑡𝑜𝑡𝑎𝑙=𝑚𝑎+𝑚𝑣mtotal​=ma​+mv​ Given that the humidity of the air sample is expressed as kg of water vapor per kg of dry air, we can express 𝑚𝑣mv​ as a fraction of 𝑚𝑎ma​: 𝑊=𝑚𝑣𝑚𝑎W=ma​mv​​ 𝑊=0.00711−0.0071W=1−0.00710.0071​ 𝑊≈0.0071 kg/kg of dry airW≈0.0071kg/kg of dry air 𝑊≈0.0071W≈0.0071 So, the humidity of the air sample expressed as kg of water vapor per kg of dry air is approximately 0.00710.0071. So, none of the given options match the calculation. Let me recalculate. Given that the total pressure (𝑃P) is 101.3 kPa and the vapor pressure of water at the dew point temperature (𝑃𝑑𝑝Pdp​) is 7.30 kPa, we can calculate the partial pressure of water vapor (𝑃𝑣Pv​) in the air sample using the relationship: 𝑃𝑣=𝑃𝑑𝑝=7.30 kPaPv​=Pdp​=7.30kPa Next, we can calculate the partial pressure of water vapor at the dry bulb temperature (𝑇𝑑𝑏Tdb​) using the relationship: 𝑃𝑣=𝑃𝑠𝑎𝑡(𝑇𝑑𝑏)Pv​=Psat​(Tdb​) Where 𝑃𝑠𝑎𝑡(𝑇𝑑𝑏)Psat​(Tdb​) is the saturation vapor pressure of water at the dry bulb temperature. Given that the saturation vapor pressure of water at 60°C is 19.91 kPa, we have: 𝑃𝑣=19.91 kPaPv​=19.91kPa Now, we can calculate the mass of water vapor (𝑚𝑣mv​) in the air sample using the equation: 𝑚𝑣=0.622⋅𝑃𝑣𝑃−𝑃𝑣mv​=P−Pv​0.622⋅Pv​​ Where 0.622 is the ratio of the gas constants of water vapor to dry air. 𝑚𝑣=0.622⋅19.91101.3−19.91mv​=101.3−19.910.622⋅19.91​ 𝑚𝑣=0.622⋅19.91101.3−19.91mv​=101.3−19.910.622⋅19.91​ 𝑚𝑣≈0.007165 kgmv​≈0.007165kg To find the mass of dry air (𝑚𝑎ma​), we subtract the mass of water vapor from the total mass of the air-water vapor mixture: 𝑚𝑎=1−𝑚𝑣≈1−0.007165≈0.992835 kgma​=1−mv​≈1−0.007165≈0.992835kg Now, we can calculate the humidity of the air sample as: 𝑊=𝑚𝑣𝑚𝑎W=ma​mv​​ 𝑊≈0.0071650.992835≈0.0072W≈0.9928350.007165​≈0.0072 So, the humidity of the air sample expressed as kg of water vapor per kg of dry air is approximately 0.00720.0072. So, none of the given options match the calculation. Let me recalculate. 𝑊=𝑚𝑣𝑚𝑎W=ma​mv​​ 𝑊≈0.00711−0.0071W≈1−0.00710.0071​ 𝑊≈0.00710.9929W≈0.99290.0071​ 𝑊≈0.0071W≈0.0071 So, the humidity of the air sample expressed as kg of water vapor per kg of dry air is approximately 0.00710.0071. None of the given options matches the calculation. There might be a mistake in the given options or the calculations. Can you please double-check the options or provide additional information?