Consider a 10-ft^3 rigid tank containing an air-water vapor...

Consider a $10-\mathrm{ft}^{3}$ rigid tank containing an air-water vapor mixture at $14.7 \mathrm{lbf} / \mathrm{in.}^{2}, 90 \mathrm{F}$ with $70 \%$ relative humidity. The system is cooled until the water just begins to condense. Determine the final temperature in the tank and the heat transfer for the process.

Key Concepts

Energy Balance in Closed Systems

The energy balance, based on the first law of thermodynamics for a closed system, is used to relate the changes in internal energy to the net heat transfer into or out of the system. In a rigid tank with no work done, the change in internal energy directly equals the heat removed or added, making this concept essential for calculating the heat transfer during the cooling process.

Dew Point and Phase Change (Condensation)

The dew point is the temperature at which water vapor begins to condense into liquid under a constant pressure. For this problem, cooling continues until the water in the mixture reaches the saturation condition (dew point) where it just begins to condense. Understanding the dew point is critical for determining the final state temperature of the mixture.

Constant Volume Process

In a rigid tank, the volume is fixed throughout the process, meaning that any changes in the state of the system occur at constant volume. This concept is important because it simplifies the energy analysis; no work is done due to volume change, and the energy transfer is solely due to heat transfer, which affects the internal energy and temperature of the mixture.

Relative Humidity

Relative humidity is a measure of the actual amount of water vapor present in the air compared to the maximum amount that air can hold at that temperature. It is a key parameter in determining the state of the air–water vapor mixture, indicating how close the mixture is to saturation and the onset of condensation.

Transcript

00:01 Hi, everybody.

00:01 So we term at the final temperature in the tank and heat transfer for the process.

00:07 Okay? so for this, we need to find your partial pressure, which is going to be 5 .1 equal times pg1, which equals 0 .7 times 0 .699 equals 0 .489 equals 0 .489.

00:30 993 psi, psia.

00:35 And at the second, system is cooled into the water to begin to get dense.

00:41 The final state is at 100%.

00:45 And your absolute humidity 1 equals your absolute many 2.

00:50 So for you absolutely humidity 1, we have 0 .622.

00:55 And this can be 0 .4893 over 14 .7 minus 0 .4893 equals 0 .0214 .1.

01:13 And because of this, since your communities equal each other, okay, your pv1 equals pv2 as well.

01:28 Okay.

01:29 And now we can find the pg2, which is what we wanted.

01:34 So we want this guy here.

01:37 And that's going to be pv2 over five.

01:40 So we have 0 .4893 divided by 1 equals 0 .4893 psia.