Ambient moist air enters a steady-flow airconditioning unit at 102 kPa and 30^∘ C with a 60 % re

Ambient moist air enters a steady-flow airconditioning unit...

Ambient moist air enters a steady-flow airconditioning unit at $102 \mathrm{kPa}$ and $30^{\circ} \mathrm{C}$ with a $60 \%$ relative humidity. The volume flow rate entering the unit is 100 L/s. The moist air leaves the unit at $95 \mathrm{kPa}$ and $15^{\circ} \mathrm{C}$ with a relative humidity of $100 \% .$ Liquid condensate also leaves the unit at $15^{\circ} \mathrm{C}$. Determine the rate of heat transfer for this process.

Key Concepts

Mass and Energy Balances Involving Phase Change

In air conditioning systems, phase changes often occur when the air is cooled to its dew point, resulting in condensation. Incorporating mass balances for both the dry components and the condensed liquid, along with the associated energy effects, is essential for accurate analysis of the system’s performance.

Moist Air Enthalpy Calculation

Calculating the enthalpy of moist air requires accounting for both the sensible heat of the dry air and the latent heat of vaporization of the water vapor. This concept is crucial when changes involve condensation or evaporation, as it directly affects the energy balance of air conditioning processes.

Psychrometrics

Psychrometrics is the study of the thermodynamic properties of moist air. It involves understanding variables like temperature, pressure, relative humidity, and the moisture content (humidity ratio), which are essential for predicting condensation and the energy changes associated with it.

Steady?Flow Energy Analysis

This concept involves applying the steady-state energy balance to a control volume. In such systems, the rate of energy entering the system (via mass and heat transfer) is balanced by the rate of energy leaving. It is critical for determining heat transfer rates in processes where conditions change between inlet and outlet.

Transcript

00:01 Hi, everybody.

00:02 So, we need to determine the rate of heat transfer for the process.

00:11 Okay, so we have a temperature at t1 equals 30 degrees celsius and a saturation pressure.

00:24 Actually, i'll just write it as, if i can write, as p -set equals equals, the p of the gas one as 4 .246k pascels.

00:43 And so we can set the phi equals pv1 over pg1.

00:51 So here we got 0 .6 equals pv1 equals 4 .26.

00:58 And you just move this guy over here and we get a pv of 2 .54.

01:07 5476k pascal's.

01:10 And now i'm going to find a humidity ratio of one.

01:14 Okay, and that ratio equation is this guy right here, p minus pv1.

01:23 And for this, we get 0 .622, 2 .5476.

01:32 And we get 102 minus 2 .5476 equals 0 .0159.

01:44 And now we need to do the mass of air, which is pavratt 1 equals 1102 minus 2 .5476.

02:03 100 times 10 to negative 1, divided by 0 .287.

02:14 Let me get that 7 to look normal, 7 times 303 .15.

02:24 And this is going to be 0 .1143 kilograms per second.

02:32 Okay.

02:34 And now we have to do an at exit condition.

02:39 So at exit condition, so we can say exit.

02:44 Okay, exit condition is equal to 15 degrees celsius, and we have a saturation pressure, so p set equals p of gas, 2 equals 1 .705 kilopascals, okay? and we have a phi 2 equals 100%.

03:09 So, and from this case, we can do, we have pv2 is going to equal to pg2, which is going to be, so we actually have this guy right here, and i'm going to move the saturation, is actually going to equal to also the pv2.

03:39 So now we can do the humidity ratio for the exit and we get 0 .622 times 1 .705 divided by 95 minus 1 .75 and we get 0 .01367.

04:03 And these are zeros.

04:08 And now so we have energy to this equation.

04:11 Conditioning.

04:13 So for this, we can find the q, which is going to be m -a -c -p -a for the air temperature 2 minus temperature 1 plus the humidity 2 times m -a -h -v -2 minus humidity 1 of m -a -h -v -1 plus m -v -1 minus m -v -1.

04:43 V2 times hv3...

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