An indoor pool evaporates 3 lbm / h of water which is removed by a dehumidifier to maintain 70 F

An indoor pool evaporates 3 lbm / h of water which is...

An indoor pool evaporates $3 \mathrm{lbm} / \mathrm{h}$ of water which is removed by a dehumidifier to maintain $70 \mathrm{F}, \Phi=70 \%$ in the room. The dehumidifier is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser, as shown in Fig. P12.105. For an airflow rate of $0.2 \mathrm{lbm} / \mathrm{s}$, the unit requires 1.2 $\mathrm{B}$ tu/s input to a motor driving a fan and the compressor, and it has a coefficient of performance, $\beta=\dot{Q}_{L} / \dot{W}_{c}=2.0 .$ Find the state of the air after the evaporator, $T_{2}, \omega_{2}, \Phi_{2},$ and the heat, rejected. Find the state of the air as it returns to the room and the compressor work input.

Key Concepts

Coefficient of Performance

The Coefficient of Performance (COP) is a measure of the efficiency of a refrigeration system, defined as the ratio of useful cooling (or dehumidification) provided to the work input required by the system. Understanding COP is essential in evaluating the performance of the dehumidifier, as it links the cooling effect achieved in the evaporator to the compressor work and overall energy consumption.

Heat Transfer

Heat transfer concepts are central in analyzing how energy is exchanged between the refrigerant and the air. In the context of dehumidification, heat is absorbed by the refrigerant in the evaporator and rejected in the condenser. Accounting for sensible and latent heat effects is necessary to determine the energy balance of the system and compute parameters such as the amount of heat rejected.

Energy Balance

The energy balance principle is used to account for all energy inputs and outputs in a process system like a dehumidifier. It is applied to determine various state properties of the air as it passes through the evaporator and condenser, and it helps in quantifying the work input and heat exchanges that occur during the moisture removal and heat rejection processes.

Refrigeration Cycle

The refrigeration cycle is a thermodynamic cycle used in devices such as dehumidifiers, refrigerators, and air conditioners. It involves key processes: evaporation, compression, condensation, and expansion. This cycle transfers heat from a lower-temperature reservoir to a higher-temperature reservoir, and understanding it is vital for analyzing how dehumidifiers remove moisture from indoor air and reject heat to the environment.

Psychrometrics

Psychrometrics involves the study of the thermodynamic properties of moist air and the relationships among dry-bulb temperature, wet-bulb temperature, relative humidity, humidity ratio, and enthalpy. This concept is essential for understanding how water vapor behaves within the air and for determining the state points (temperatures, humidity ratios, and relative humidities) before and after conditioning processes such as those in a dehumidifier.

Evaporator Process

The evaporator process in a dehumidifier or refrigeration cycle is where the refrigerant absorbs heat from the air, causing the refrigerant to evaporate. This cooling of the air facilitates the condensation of water vapor as the air's capacity to hold moisture decreases, and thus the evaporator plays a crucial role in determining the state of the air after moisture removal.

Transcript

0:00 Hi, ready.

00:01 So we need to find the state of aris returns to the room and the contestor works in foot.

00:07 Okay? so we need to do the find the pv one, which is going to be 0 .7 times 0 .363 equals 0 .2541 times psi -a.

00:32 Then we have absolute humidity as 0 .622 times 0 .2541.

00:42 14 .7 minus 0 .2541 equals 0 .10 .4.

00:54 Okay.

00:56 And so we can find out the w2 absolute humidity to an mass in liquid equals m.

01:04 Times one minus two.

01:08 So massive liquid is 3 over 3600 0 .2, 0 .0 .0194 minus here.

01:24 So your 2 is 0 .06773.

01:34 And now we can find the pv2.

01:39 And so we're going to do 0 .00673 equals 0 .622 times pv to, 14 .7 minus p .2.

01:54 And we're going to solve with pd2.

01:56 And you can take this guy times over here, and i'm just going to solve because it's simple algebra.

02:03 So 0 .158334.

02:09 Psia.

02:10 And if you have trouble, the reason i'm not explaining it is because you should already know algebra by the time you come to this class.

02:18 Okay? most physics is you're going to need calculus and above.

02:23 So anything below that, you should be able to decipher yourself.

02:29 So t2 equals 45.

02:31 And if i can write something, 45 plus 50 minus 45 times 0 .15.

02:44 0 .1534 minus 0 .147 divided by 0 .178, minus 0 .178, and get 46 .84f, and now we need to find the wc, which is going to be equal to m -a -2l over data, which is going to be 0 .2 times 10 .113 divided by 2, and we get 1 .01 .1.

03:42 03 b to you for a second and not done so we have to do the h3 minus h1 equals wma minus periodicities and entropy we have to be times 0 .2 minus 0 .0 .1094 minus 0 .073 times 14 .877, which equals 5 .938, b2s per ibm.

04:44 And here gets seen the v2s 3 to double 2 pv 3 equals pv2 and so we're going to look at h3 minus h1 and so this equation is going to be cpa t3 minus t1 plus 3 hv3 minus 1 on.

05:18 Okay, so i'm going to plug and chug.

05:21 0 .24, g3 minus 70 plus 0 .006773, hg3 minus 0 .0904, times 1092 .04.

05:44 And so we, we're going to, we, and so we, end up with 5 .938 equals 0 .2423 plus 0 .006773 8d3 minus 28 .746.

06:10 Okay.

06:15 And so now we're going to do trial error and we're going to assume 2b equals 1100f.

06:21 And if we plug that in to the right hand side, which is this side here, you have 0 .24 times 110, 0 .0, 0 .0, 06, 7, 3.

06:43 Getting its data from tables, 110, 9299, is 28 .746.

06:53 And we get 5 .162...