T110D Manual - First Part

Did Di d a c ta Ita lia

T110D Air Conditioning Study Unit

User’s Manual and Exercise Guide

Did Di d a c ta Ita lia

T110D Air Conditioning Study Unit

User’s Manual and Exercise Guide

Did Did a c ta Ita lia lia Srl - Stra d a d el C a sc inott o , 139/ 139/ 30 - 10 1015 156 6 Torino Te l. +39 011 273.17.08 273.18.23 - Fa Fa x +39 011 273.30.88 http:/ / ww w.dida c ta.it/ ta.it/ - e -mail: -mail:info@dida nfo@dida c ta.it ta.it

The informa tion c onta ined in this ma nual has be en selec ted and verified with the greatest care. However, no responsibility stemming from its use can be ascribed to the Authors or to Didac ta Ita lia or any p erson or co mp any involved in its prepa rat ion. The information c onta ined in this ma nual ca n be m od ified a t any time a nd w ithout wa rning on a c c ount of tec hnic al or educ ational need s. Copyright



Didacta Italia 2004

Rep rod uc tion by any mea ns, inc luding p hotoc op ying of t his test o r pa rts thereof, or the figures co nta ined therein, is stric tly p rohibited . Printe d in Ita ly - 26/ 07/ 04 Cod e 01610E0704 —Ed itio n 01 - Revision 01

tab le of contents

Ta b le of Co ntents

1.

Gene ral .................................................................................... 1

2

Co m p osition a nd desc ription................................................. 3

2.1 Co m p o sition............................................................................................... 3 2.2 De sc rip tio n ................................................................................................. 4 2.2.1 Gene ral synop tic ..........................................................................................................4 2.2.2 Cont rol p a ne l ................................................................................................................6 2.2.3 Transd uc er and signa l c ond itioning unit ...................................................................7 2.2.4 Da ta a c q uisition a nd a na lysis softw a re .....................................................................8

3.

Sta rting ..................................................................................11

3.1 Sta rting th e unit ....................................................................................... 11 3.2 Sta rting the d a ta a c q uisition syste m .................................................... 11 3.3

Ta king the M ea surem e nts w ith Da ta A c q uisition Syste m .................. 11

3.4

Prep a ring the syste m for a p rolong ed p eriod of ina c tivity ................ 12

4.

Exp erimenta l Exe rc ises ......................................................... 13

4.1 Introd uc tio n to th e exerc ises................................................................. 13 4.2

Ba sic Exerc ises......................................................................................... 13

Exerc ise 4.1 —Air sensitive hea t va riat ion ...........................................................................13 Exerc ise 4.2 —Va ria tio n in t he lat en t hea t o f t he a ir.........................................................16 Exerc ise 4.3 —Def rosting (Eva p orat ion Co oling ) ...............................................................21 Exerc ise 4.4 —Cond ensed Co oling .....................................................................................22 Exercise 4.5 —Va luta tion of the eff icien c y o f a frig orific ma c hine .................................24

4.3 A d va nc ed Exerc ises.................................................................................. 27 Exerc ise 4.6 - Full Air Co nd itio ning Plant .............................................................................27 Exerc ise 4.7 - R22 Co m p ression Ref rig era tion Cyc le.........................................................36

T110D —User' s Ma nu a l

v

table of c ontents

5. Exe rcises with SAD/ T110D Auto m a tic d a ta a c q uisition system ...................................................................................53 5.1 Fo rmula s used fo r a ir c o nd itioning te sts (Test 01) ................................53 5.2 Fo rmu las used in co o ling t ests (Test 02) ................................................56 5.3 Experim ent a l e xerc ises............................................................................57 Exerc ise 5.1 - A ir c ond itioning te st (Test 01) ...................................................................... 57 Exerc ise 5.2 - Air c on d itioning t est (Test 01) with p re-he a te r .......................................... 61 Exerc ise 5.3 - Air c ond itioning te st (Test 01) with he a te r.................................................. 65 Exerc ise 5.4 - Air c ond itioning te st (Test 01) with c ooler .................................................. 69 Exercise 5.5 - Air c on d itioning te st (Test 01) w ith hum idifier a nd wa te r......................... 73 Exerc ise 5.6 - Air c on d itioning t est (Test 01) with ste a m hu midifier ................................ 77 Exerc ise 5.7 - Cooling test (Test 02) .................................................................................... 81 Exercise 5.8 - Co nd itioning C yc le in the Wet A ir Mo llier Diag ram (Test 03) ................. 84

6.

Anne x 1 - Ta b le for the c ollec tion of test d a ta ...................85

7.

Anne x 2 - Psyc hrom etric d iag ram s .....................................87

7.1 Ent ha lpy d ia g ram of Freo n R22 .............................................................87 7.2 Psyc hrom etric d iag ram s o f a ir ...............................................................89

8.

Anne x 3 - Ta b le for the a ir sp eed s in m / s............................95

9.

App end ix 4 - T110D/ 1 .........................................................97

vi

Dida c ta Italia

Cha pte r 1.

1.

General

The air conditioning and climate control study unit T110D has been specially designed to serve as a highly advanced teaching tool: simplicity and clarity are the basic features enabling the teacher and the students to carry out a variety of experiments without having to acquire complex utilisation procedures. The operating principle of this unit is as follows: a stream of air generated by a centrifugal fan is made to pass through a tunnel via a fluid thread rectifier. As it goes through the tunnel the air undergoes a series of treatments until it reaches a final chamber representing the environment to be conditioned. The air, in fact, is initially preheated, then humidified by means of steam diffusers, then cooled by means of the R22 evaporator and finally conveyed into the end chamber. Before this last step, the air can be heated and humidified again by means of water diffusers, enabling the students to become familiar with the heating processes and to vary the greatest possible quantity of parameters, for a thorough understanding of climate control. The study unit is self-sufficient in terms of the instrumentation necessary to perform all the tests which are described in a theoretical-practical manual. The T110D has a testing section entirely made of stainless steel and mounted on a wheeled frame that also carries part of the cooling circuit, a small steam boiler, the power supply, adjustment and control module, and a panel with a scheme of the entire circuit silk-screen printed on. The variety of tests that are possible with this study unit encompasses all the problems of air conditioning, such as: •

determination of the thermal balance

•

determination and study of the effects of removing/adding heat and humidity from/to the air, and the relative air conditioning system techniques

•

study of the influence of air velocity and flow on the temperature and humidity of the air

•

conditioning of ambient air in summer

•

conditioning of ambient air in winter

•

air conditioning for industrial processes

•

execution of psychrometric conversions on Mollier's diagrams for wet air

•

execution of enthalpy and matter balances

•

study of the principles governing the simultaneous transport of heat and matter

•

evaluation of the efficiency of the cooling cycles employed in air conditioning processes

T110D —User’s Ma nua l

1

general •

measurement of pressure drops in the air going through a channel

An automatic data acquisition and analysis system for personal computer is available. This unit, called SAD/T110D, makes it possible to perform the measurements, plot the characteristic curves on screen or print them out, file the data on a disk or print them out entirely automatically.

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2.

Com p osition a nd d esc ription

2.1

Composition

The T110D study unit (Code 954000) shown in fig. 2.1 consists of a truck mounted structure housing the following elements:

Fig. 2.1 - Overview of the system 1. Testing tunnel of stainless steel; 2. Centrifugal fan; 3. Control panel; 4. R22 sealed compressor; 5. Condenser; 6. Steam generation boiler; 7. Centrifugal pump; 8. Manometer panel; 9. Differential micro-manometer.


3

c om po sition and de sc ript ion

The automatic data acquisition system (SAD/T110D) includes: •

a kit of electronic transducers for the conversion of the quantities to be measured into the corresponding electric signals (code 914321);

•

a signal conditioning and A/D conversion unit (code 914320);

•

data acquisition and analysis software for Windows (code 914324).

2.2

Description

2.2.1 Ge nera l syno p tic

Fig. 2.2 - General synoptic scheme

Fig. 2.2 shows the overall scheme of study unit T110D. The basic components are: • •

4

Variable speed (0 to 1450 rpm) centrifugal fan No. 2 fluid threads rectifiers

•

Set of pre-heating resistors, with power continuously adjustable up to 3 kW

•

Steam generation boiler (1.3 kg/h)

•

Set of heat diffusers

•

R22 air evaporator

•

R22 air condenser with 3.8 l capacity tank

•

1.2 HP sealed compressor

Dida c ta Ita lia

Cha pte r 2. •

Drier filter

•

Thermostat controlled expansion valve

•

Set of heating resistors, with power continuously adjustable up to 3 kW

•

Set of water diffusers in a closed circuit fed by a pump from a 2 litre calibrated container

•

No. 2 chambers with inspection windows.

The system includes the following control and measuring instruments: •

Digital temperature indicator, connected to 4 thermal resistors selected by means of a switch

•

No. 4 PT100's T1; T2; T3; T4 from -100 to +100 °C

•

No. 2 ammeters connected to the heating sets

•

Low pressure manometer, 15 bar

•

Electronic pressure transducer Pe (Capacity 0 to 10 bar) (only in conjunction with SAD/T110D)

•

High pressure manometer, 24 bar

•

Electronic pressure transducer Pc (Capacity 0 to 25 bar) (only in conjunction with SAD/T110D)

•

No. 2 dry bulb thermometers, - 10 to 50°C, 1/5°C accuracy

•

No. 3 dry bulb PT100's Tad; Tcd; Tod, 0 to 150°C (only in conjunction with SAD/T110D)

•

No. 2 wet bulb thermometers, - 10 to 50°C, 1/5°C accuracy

•

No. 3 wet bulb PT100's Taw; Tcw; Tow, 0 to 150°C (only in conjunction with SAD/T110D)

•

No. 1 psychrometer

•

Tilting scale micro-manometer, 0 to 200 mm H 2O

•

Differential electronic pressure transducer to measure air flow rate, deltap (capacity 0 to 50 mm H 20) (only in conjunction with SAD/T110D)

•

Calibrated diaphragm, φ 100 mm

•

7 to 30 bar pressure switch, differential pressure adjustable between 2.5 and 3.8

•

Electronic thermostat with pre-selector, 0 to 99.9 °C.


5


The SAD/T110D automatic data acquisition system makes it possible to acquire automatically the following quantities: •

Taw: wet bulb ambient temperature

•

Tad: dry bulb ambient temperature

•

Tcw: wet bulb testing chamber temperature

•

Tcd: dry bulb testing chamber temperature

•

Tow: wet bulb output temperature

•

Tod: dry bulb output temperature

•

T1 Condenser input temperature

•

T2 Condenser output temperature

•

T3 evaporator input temperature

•

T4 evaporator output temperature

•

Pe low pressure (intake)

•

Pc high pressure (compression)

•

∆p differential pressure

2.2.2 Control pa nel The control panel consists of the following 6 modules, which can be recognised in the front view shown in fig. 2.3: Fan module

Makes it possible to turn on and off the fan and adjust its speed Power supply module

Makes it possible to turn on the auxiliary circuits by means of the start key or turn them off by pressing the emergency button Pre- heater mo dule

This is used to select the preheating power, 1000W, 2000W or 3000W, and to read the current absorption value on the ammeter. Tem pe rature mo dule

Makes it possible to select the operating set-point and to display the cooling circuit temperatures. 6

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Compressor module This is used to start the compressor and determine whether or not the thermostat or the pressure switch have been triggered Hea ter mod ule

This is used to select the heating power, 1000W, 2000W or 3000W, and to read the current absorption value on the ammeter.

Fig. 2.3 - Front view of the control panel

2.2.3 Transd uc er a nd signa l c ond itioning unit The transducer and signal conditioning unit (only in conjunction with SAD/T110D) consists of the following 3 modules, which can be recognised in the front view shown in fig. 2.4: Tem pe rature mo dule

Makes it possible to acquire the temperature values at the different points along the circuit on a -100 to +100°C scale by means of a selector. Differential pressure module

Makes it possible to acquire the differential pressure of the output air stream on a scale from 0 to 50 mm H 2O. T110D —User’s Ma nua l

7


Pressure module

Makes it possible to acquire the delivery pressures (0 to 25 bar scale) and intake pressures (0 to 10 bar) of the cooling compressor. The upper part of this unit (see fig. 1.4) accommodates: •

The general switch, called POWER

•

The ground socket, called GND.

Fig. 2.4 - Front view of the transducer and signal conditioning unit

2.2.4 Da ta ac q uisition a nd ana lysis softwa re The software has common characteristics that are described in detail in the "SAD Data Acquisition System - User's Manual" supplied with the system. The installation procedure is also explained step by step in this manual.

8

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Fig. 2.5 - SAD/T110D software: working environment

Fig. 2.6 - SAD/T110D software: experimental diagram


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10

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3.

Starting

3.1

Sta rting the unit

1.

Make sure that the humid bulb thermostats are filled with water and that the gauze is humid and also that the thermometer is hit by an air current and that it is not immersed in the water container;

2.

Check the availability of: •

Electrical power supply: 380 V AC 3 ~ + N 50Hz; 220V AC 60Hz

•

Maximum power 9 kW

1. Set all the selectors on zero; 3.

Fill the humidifier circuit tank and the steam generation boiler with water;

4.

That the mater input valve of the boiler is open, while the exit one is closed; and also that the seal clamps of the tubes are well closed.

5.

Connect the special plug to the mains voltage 380V AC 3 ~ + N 50Hz or 220V AC 3 ~ 60Hz

3.2

Starting the d ata a c q uisition system

1.

Connect the transducer to the signal conditioning and A/D conversion unit, by exploiting the front connectors of the unit and the cables supplied with the system.

2.

Connect the signal conditioning unit to the parallel or serial port of the Personal Computer, by means of the cable supplied as standard.

3.

Connect the signal conditioning unit to the 220V AC 50/60 Hz mains voltage.

4.

Install the SAD/T110D software in the Personal Computer according to the indications contained in the "SAD - Data Acquisition System - User's Manual".

3.3

Ta king the Me a surem ents with Da ta Ac q uisition System

1.

Power the unit by means of the special safety switch.

2.

Turn the start key in the control panel power module, so as to activate the auxiliary circuits.

3.

Start the fan and adjust fan speed by checking the flow rate on the differential micro-manometer.


11

starting

4.

Measure the values and enter them in the table shown in annex 1, or, if using the SAD/T110D, proceed with the automatic acquisition of the data.

5.

Use the heating and humidifying devices to vary the operating conditions of the study unit.

3.4

Prep a ring the system for a prolong ed p eriod of inac tivity

1.

Set all the selectors on zero.

2.

When using the SAD/T110D, turn off the signal conditioning unit.

3.

If you don't use the unit for a certain period we suggest that you empty the boiler for the vapour production, operation that must be done after having sent out all the vapour.

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Did a c ta Ita lia

Cha pte r 4.

4.

Exp erime nta l Exe rc ises

4.1

Introd uc tion to the exe rc ises

Following here will be reported a series of guide exercises that have as their aim the comprehension of the theoretical principles illustrated in the previous chapters and the valuation of the possibilities of the air conditioning machines. It is not considered in this treatment, the valuation of the problems tied with the mixing of the air treated with the fraction untreated of a given environment, as also phenomena tied with the filtration, purification and at times sterilization of air in certain environments, very important factors in air conditioning, but specific and tied to the use of air so of difficult generalization. The use of the T110D unit is devoted to the comprehension of the principles and the problems of the conditioning machines.

4.2

Basic Exe rc ises

Exe rc ise 4.1 —Air sensitive he a t va riatio n This exercise has as its aim the understading of the transformations that an air undergoes without a variation in its absolute humidity. The numerical values are examples introduced for a better comprehension of the diagram of the exercise; in reality, depending on the environment in which the group will be used, different couples of humid bulb and dry bulb temperatures will be found. Suppose you have an environmental air with the following couples of temperatures:

T s = 20°C T u = 15°C Using the psychrometric diagram (reported in the appendix) the relative umidity is found at the intersection of the humid bulb and dry bulb temperature; while the absolute humidity is read on the ordinate axis drawing a horizzontal line from the point found:

U = 0.0088 Kg/kg d.a. The temperatures Tu and Ts must be read with the hygrometer included in the group (as reported in the general treatment); therefore the fan that inputs air into the channel is turned on, choosing the work delivery, for example positioning the knob on 70 – 80%.


13

expe rimenta l exerc ises

At this point you turn on the heat of the resistance B (ref. to fig. complete diagram); operating on the knob situated on the control panel, regulate the electrical current to heat the air that flows into the channel at a temperature not higher than 50°C, to be able to use the psychrometric diagram.

U r ≅ 60% The control of the heating temperature is done manually operating on the control panel and reading the value of the temperature with the thermometers Q. Suppose you heat the air to T s ≅ 35 °C the value of T u must be of about 20 °C. Reporting these values on the psychrometric diagram, you will find that the air will have a relative humidity of:

U r = 25% and U = 0.0088 kg/kg d.a. Therefore the air will have absorbed only the sensitive heat with a measurement found once again from the psychrometric diagram through the subtraction of the enthalpy at the final point (after the heating) and the initial one: .. ∆h = h f − hi = 57.5 − 42.1 = 15.4 kJ / Kg a s Measuring the air delivery in the channel with the differential micro-gauge, according to what is reported in the general treatment; finding the value of the specific volume of the at the final conditions from the psychrometric diagram

v ≅ 0.885 m /kg a.s. 3

it is possible to find the value of the thermal power needed for the heating of the air: •

Q=

kJ ⋅ ∆h   v h

V

where V is the air delivery in m3/h. The example now shown, is that of air that acquires sensitive heat; to make this happen it is necessary to cool air to a temperature higher than the dew point, below it would also lose laten heat. For our air it would correspond to:

T r ≅ 11.5 °C that is the intersection of the horizontal pasing through the rapresentative point of the air, on the psychrometric diagram, and the curve at U r = 100%.

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To carry out this other exercise, it is necessary first to turn off the heating B and then wait for a certain period of time so that the air that flows through the channel cools the resistences, to decrease this period you can increase the air delivery; then position the thermostat of the frigorific cycle at the temperature wanted higher than 11.5 °C, for example 15 °C. This operation must be done by pressing the pushbutton of the digital regulator and turning the knob to the value requested. At this point you can insert the frigorific cycle and wait until the air is cooled. Remember that the value indicated by the set-point is the value at that moment, so through it is possible to check the actual cooling of the air itself. Suppose that after a certain period the last couple of thermometers gives the following values:

T u ≅ 13.00 °C

T s = 15 °C Note

If the value read by the dry bulb is slightly different from the set-point it can depend from various factors: •

the measuring points are in different points along the channel;

•

the precision of the mercury thermometer is higher than that of the probe;

•

the two sensors can read at different heights along the section of the channel.

Differences between ± 1 °C are acceptable, if not, check if the set-point sensor is situated at the center of the channel. The values reported on the humid air diagram give:

U r ≅ 82%

U = 0.0088 Kg/kg a.s

Therefore, also in this case the air has undergone a variation in sensitive heat changing its Ur but not the absolute humidity. Similarly to before it is possible to find the value of the thermal power subtracted:

−Q =

kJ ⋅ ∆h   v h

V

The minus sign is to indicate that the heat is subtracted and not given out by the air.


15


Exe rcise 4.2 —Va riation in the late nt hea t of the a ir There is a variation in the latent heat of the air when not only its dry bulb temperature changes, but also its absolute humidity; this can happen not only with humidification heating but also with de-humidification cooling. Humidific ation heating

To do this exercise, vapour will be used, so it is better to make sure that all the contacts of the boiler with the entrance and exit of the water have been well done, checking if the electrovalve for the vapour input into the channel and the two needle valves are closed, verifying if the opening for the water in the boiler is open and that the exit is closed, turn on the boiler. At this point it is necessary to have cold water (to see better the phenomena), obtainable using the frigorific group G. Then, after having turned on the fan and set the air delivery, position the thermostat of the fridge at a low value (for ex. 5 °C) and turn on the compressor. After a certain period read the temperatures of the last couple of thermometers and, for example, you will have:

T s = 5 °C

T u = 4.5 °C

therefore, using the psychrometric diagram you have (similarly to what shown in the previous exercise):

U r = 97%

U = 0.0052 kg/kg a.s.

At this point open the steam solenoid valve (starting its own switch) to heat the air up to a given temperature. Switch off the boiler, close the solenoid valve when the temperature aims to increase too much preventing the use of the psychrometric diagram. During this heating, obviously, the frigorific group must be left in function; suppose that the air is heated and gives the following values:

T s = 35 °C;

T u = 32 °C

that correspond to:

U r = 72%;

U = 0.0256 kg/kg a.s.

Therefore the air has taken in:

0.0295 – 0.0052 = 0.0243 kg/kg a.s. 16

Did a c ta Ita lia

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of water at vapour state. If you want to calculate the latent and sensitive heat the following method is used: •

from the point representing the initial air A (fig. 4.1) draw a horizontal line, while from point B, final point draw a vertical line; the point C, intersection of the two lines, represents an imaginary point that lets us find the sensitive and latent enthalpy.

This can be done reading from the points AB and C their enthalpies on the axis slanted leftwards of the curve Ur = 100%.

h A = 18 kJ/kg;

h B = 101 kJ/kg;

hc = 49 kJ/kg a.s.

As a matter of fact, it's as if the air were heated with a constant U in the section A C and humidified in the section B C, so the sensitive heat will be:

hC – h A = 49 – 18 = 31 k J/kg a.s. while the latent one is:

h B – hC = 111 – 49 = 62 kJ/kg a.s. The thermal powers will be, knowing the delivery V and the specific volume v of the air:


17

expe rimenta l exerc ises •

Q s

•

Q1

kJ ⋅ (hC − h A )   v h

=

V

=

V

kJ ⋅ (h B − hC )   v h

The umidification heating is a typical case of air treatment in winter conditions or for industrial applications that request hot air with a certain degree of humidity. In fact, we must say that in the case of winter conditioning (quite rare) in civil environments, the air is heated and not humidified, leaving the humidity control to the hourly renewals of the air in the environment for health purposes. A very common application is, instead of changing the latent heat given by the cooling, the conditioning and control of the humidity of the air in a frigorific room for the preservation of foods; we will go back to this after having described the exercise. To subtract water from an air and therefore subtract sensitive heat it is necessary to cool the air to a temperature lower than that of the frost point. To execute this exercise operate in the following manner: turn on the fan and set the knob on a given position, suppose the air has the same values of Ts e Tu of the previous exercises:

T s = 20 °C; T u = 15 °C: U r = 60%; U = 0.0088 Kg/kg a.s. At this point the thermostat of the frigorific group must be set at a temperature lower than frost point of the given air, in this case 11.5 °C; so suppose you set the thermostat at 5 °C. At this point turn on the frigorific group; after a certain period the last couple of thermometers will show the conditions that the air has reached, suppose the following values are read:

T s = 6 °C;

T u ≅ 5.5 °C

so:

U r ≅ 95%;

U = 0.0054 kg/kg a.s.

At this point the air will have undergone a change of the type reported in fig. 4.2 exagerating to better visualize the phenomena.

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In fig. 4.2 the full line with the arrow represents the real thermodynamic transformation that the air has undergone, that is, a cooling A A', a de-humidification A' B' along the saturation line, a slight heating B' B. To find the value of the enthalpies of the case, similarly to what done in the previous exercise, draw the horizontal through B and the vertical through A, the intersection point C permits us to find the values of the enthalpies on the axis on the left of the saturation curve:

h A = 42 kJ/kg a.s.; h B = 19.5 kJ/kg a.s.; h C = 33 kJ/kg a.s. Therefore it is as if the air had undergone an imaginary transformation A C and C B; in which in the section A C it is de-humidified by subtraction of the latent heat:

h A – hC = 42 – 33 = 9 kJ/kg a.s. and in the section C B it is cooled by the subtraction of the sensitive heat:

hC – h B = 33 – 19.5 = 13.5 kJ/kg a.s. The quantity of the water that the air has lost is:

U A – U B = 0.0088 – 0.0054 = 0.0034 kg/kg a.s. It is possible to collect the condensed water and verify that the calculations are correct to certain limits: to do so it is necessary, once reached the standard conditions (for a certain air delivery, constant Ts and Tu in point B), to measure the air delivery in the channel (through the micro-gauge) V, and then in point B (with the psychometric


19


diagram) the specific volume of the air v, then if the experiment is done for a period of time T (in hours) you will have:

T ⋅

V v

⋅ (U A − U B ) = P [kg H 2 O ]

that represents the quantity of water subtracted from the air by cooling. The experiment ends by stopping at the time T, measured with a watch (or stopwatch), the air flow into the channel; at this point it is necessary to collect the water in the container M. Obviously part of the water can be in frost form at the evaporator; to collect it it's necessary to defrost the evaporator.

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Exe rc ise 4.3 —Defrosting (Eva p ora tion C ooling ) The defrosting operation is done periodically on environmental conditioning plants, when the plants are continually used; in case non-continuous functioning, it is done only in rest conditions of the plant. The evaporator is defrosted by contact with hot air, or at least hotter than the previous so also the environmental one. In the case being examined, the defrosting is done by sending hot air on the evaporator. During the defrosting operation the frigorific group must be kept still. At the end of the defrosting that must go on for a certain period of time, the water is collected during the condensing cooling. The quantity of total water thus obtained must be compared with the value of P calculated, the value measured will be lower because, not considering the sperimental errors, during the defrosting the defrosting air humidifies itself adiabatically in contact with the evaporators frost. The quantity of air acquired can be measured knowing the conditions of the air at the beginning. After having positioned the knob of the fan on a given position, and having measured the air delivery, you will read the values with the couples of thermometers before the evaporator, for example:

T s = 20 °C;

T u = 15 °C;

U r = 60%;

U 1 = 0.0088 kg/kg a.s.

suppose that at the exit of the evaporator the air has the following values:

T s = 15 °C;

T u = 14 °C;

U r = 90%;

U 2 = 0.0097 kg/kg a.s.

The defrosting must be protracted until the temperature of the air starts to rise again and return to the value of 20 °C; let's call t this defrosting time. The air then has gone from a relative humidity of 60% to one of 90% and from an absolute humidity of 0.0008 to another of 0.0097 taking in

(U 2 – U 1 ) = 0.0097 – 0.0088 = 0.0009 kg/kg a.s. Therefore the

t ⋅


V v

(U 2 − U 1 ) = P s

21


The main experimental errors in this experience are that it is practically impossible to drain all the evaporator's water and that the transitions in which the air and the water are brought into equilibrium are not considered. However operating with a certain accuracy these errors can be minimised; that is, working with the same air delivery both in the frost formation stage than in the defrosting stage, and interrupt only the compressor to indicate the end of the condensing cooling from the defrosting one. Naturally the two experiences can be done separately one from the other and the effects amplified for a better comprehension. For example, you can start with an air containing more humidity and than sending to the evaporator air heated with vapour in C.

Exe rc ise 4.4 —Co nd en sed Co oling The practical execution of this exercise has already been described in the chapter dedicated to the latent heat variations in an air, here we only want to describe the application of this principle to the humidity control in a refrigerating room. Consider the variations in humidity (both absolute and relative) that the air undergoes inside the room going from 0 °C to -5 °C. Why this cooling of the air? It is evident that if you want to maintain in the room a standard temperature of 0 °C and at the same time take away the heat brought in by the foods that are introduced, we have to make the air go out from the evaporator at a lower temperature for ex. –5 °C. Consequently the temperature of the Freon must be even lower for ex. –10 °C. In this case is implemented a periodical cooling of the room's air from 0 °C to –5 °C and therefore its heating Freon –5 °C to 0 °C. This determines a cycle that presents the following phases: •

cooling without de- humidification from 0 °C to the frosting point. If hypothetically the cycle is started with the air containing 90% relative humidity, the frost points of –1.4 °C/ref fig. 4.3) but we advise to follow the tracing of the cycle on the psychrometric diagram;

•

cooling with de-humidification (along the saturation line) to the temperature wanted of –5 °C;

•

heating of the air from –5 °C to 0 °C, as a consequence of the taking away of the heat from the room. U r ≅ 65%. In this phase there can be a passage of humidity from the foods to the air of the room and therefore a drying of the foods or a loss in freshness caused by the decrease of U r .

The diagram shows that during this cycle the air has been de-humidified, because it has gone from an absolute humidity of about 0.0034 kg/kg d.a.; at to 0.0025 kg/kg d.a.; so it has lost about 0.9 gr of steam for every kg of air.

22

Did a c ta Ita lia

Cha pte r 4.

The control of humidity is then done by changing the temperature of the air at the exit of the evaporator, decreasing in the case you want to dry out, and increasing in the case you want to keep the food fresh for a short preservation. In the example given, making the air some out from the evaporator at –3 you have U r ≅ 85% so the drying effects resulting on the food will be diminished.


23


Exerc ise 4.5 —Valutation of the effic ienc y of a frigo rific ma c hine Before reading the parameters for the study of the cycle and the balances it is better to let the group run for a couple of minutes to permit the machine to go at full standard. First it is necessary to find on the diagram log P, h (in the appendix is reported the diagram for R22) the real functioning on the cycle. To do so it is necessary to find the temperatures in the various points of the cycle (t1…t4) and the drawing in and compression pressures that are indicated with P v and Pc, that are not the same of the evaporation and compression ones, because of the loss in charge that the frigorific fluid undergoes in the heat exchangers. For example, let the values found be the following for a certain air delivery at the evaporator: •

t1 = 65 °C

condenser entrance

•

t2 = 25 °C

condenser exit

•

t3 = -15 °C

after the expansion valve

•

t4 = 6 °C

exit evaporator

•

Pe = 12 bar

compressor pressure

•

Pc = 16 bar

evaporator pressure

The construction of the cycle is made up of the following phases: •

you set on the bell of the state diagram for x = 1, the points correspondent to the pressures P e and Pc;

•

prolonging the isobaric Pe until it intersects the isothermal of 65 °C, in the field of the overheated vapour, you will obtain point 4 (ref. fig.4.4); similarly for the isobaric Pc, to the intersection with the isotherm 6 °C you obtain point 3;

•

prolonging the isobaric Pc until it intersects t2 = 25 °C position point 1 on the bell for x = 0. In this case the liquid R22 go overcooling, because point 1 is in the liquid area, in this area the isotherms can be considered; to a good approximation, vertical. The state of point 1 depends on the charge losses that the R22 undergoes in the condenser, that given the specification adopted in the group, results always very limited;

•

point 2 is found going vertically from point 1 to the isotherm/isobar correspondent to t 3 = -15 °C.

Collecting the various points found (1…4) the working characteristics of the cycle are found.

24

Did a c ta Ita lia

Cha pte r 4.

The difference in pressure between point 2 and point 3 ∆P2,3 is the loss of charge that the frigorific fluid undergoes in the evaporator, this loss in an air evaporator is always big enough because of the winding paths that the fluid must follow. After having traced the cycle, the enthalpies can be measured in the various points; remember that the enthalpic values read on the diagram are for mass unit of the refrigerating fluid (kJ/kg). For example in the case considered we have: •

h1 = h2 = 230 kJ/kg

•

h3 = 415 kJ/kg

•

h4 = 440 kJ/kg

Having built the cycle on the diagram log P/h, it is possible to give a valuation to the efficiency of the frigorific machine. Referring to fig. 4.4 the values of the enthalpies in points 1, 2, 3, 4, and the quantities of heat d1 and d2 exchanged one with the other at the condenser and the evaporator, and the work of the cycle given by the compressor l ciclo are given. For the hypothesised working conditions they result: •

q1 = h4 – h1 = 215 kJ/kg

•

q2 = h3 – h2 = 185 kJ/kg

•

lcicle = h4 – h3 = 25 kJ/kg

since they derive from the measurement of the real conditions of the frigorific fluid, these values consider the various performances at the compressor (mechanic, electric), so:

ε=


q2 ciclo

=

185 25

= 7.04

25


We advise you to calculate ε with sir at different temperatures, heating with the resistance B at various humidities, heating with C and with different air deliveries in the channel as stated in general port in order to understand the mass as well as energy balances in psichrometry technique.

26

Did a c ta Ita lia

Cha pte r 4.

4.3 4.3

Ad va nc ed Exerc ises ises

Exe rc ise ise 4.6 —Full A ir Co nd itioning ition ing Pla nt a - Aim Focusing on the experimental air conditioning apparatus present in the T110D assembly, analysed in the laboratory, this chapter describes the function of the system, highlighting the thermodynamic role played by the various parts. The temperature measurements made in the laboratory are used to calculate the transformations undergone by the air, plotting them on a Mollier diagram for wet air. The following are also calculated: •

Thermal powers exchanged in the pre-heater, cooling set and post-heater;

•

The flow rate of water vaporised by the steam humidifier;

•

The flow rate of water vapour condensed in the cooling plant.

Note It is assumed: •

that water vapour from the humidifier is saturated steam at 1 bar;

•

that the water vapour present in the air at the outlet of the cooling set is in a saturated condition.

b —Instructio Instructio ns for the te a c her Conditions to b e set on the a pp aratus: aratus:

The following conditions are suitable for exercises relating to a room temperature of 16 – 20 °C. •

Fan: 50 %

•

Pre-heater: position 2 ⇒ 4-5 A

•

Post-heater: position 2 ⇒ 4 A

•

Refrigerating cycle thermostat: 12 °C

Notes •

The flow rate of steam produced by the boiler is quite high. It is therefore important to open the steam inlet valve only when readings are taken (just before,

T110D —Use Use r’s r’ s M a nua nu a l

27

expe rimenta l exerc exerc ises ises

make sure that steady state conditions have been reached), otherwise water may run out. •

If dehumidification occurs in the cold plant do NOT give the student the temperature of the air measured downstream of the plant itself (value shown on the display of the thermostat sensor sensor of the fridge cycle). For the calculations it is sufficient to assume that air has a relative humidity of 100%.

c —Ca rrying rrying out the test test c 1 —Main p oints of the summ summ er air c onditioning onditioning c yc le

This exercise consists in the realisation of a summer air conditioning cycle. This cycle did not use use the room air directly, aspirated through the fan, but pre-heated and humidified it (using steam) in order to reproduce particularly severe “summer” conditions (high temperatures, high water vapour content). A schematic diagram of the plant, highlighting the areas in which the various transformations occur, is shown in Figure 4.5.

Batteria di refrigerazione Batteria di post riscaldamento Nebulizzatore Batteria di preriscaldamento

As mentioned earlier, in order to simulate a case of summer conditioning, air is taken in from the room at the following conditions T abs = 16.5 °C and T abu = 11 °C (point a) and is treated in the pre-heating set where it is heated (transformation a-a’) and then humidified (trasformation a’-1) using a steam diffuser. The output consists of a flow of hot and wet air (conditions in point 1) to be used by the “summer” transformation cycle. 28

Did Did a c ta Ita lia lia

Cha pte r 4.

The summer conditioning cycle includes: •

Cooling and dehumidification in the cooling set (transformation 1-1’), supplied with an appropriate refrigerant (see refrigerating cycle exercise). This transformation brings the air to ϕ = 100 % in point 1’.

•

Heating of the air using the post-heating set (transformation 1’-2).

The data measured during the course of a typical exercise are summarised in Table 4.1:

°

°

a

16.5

11

1

42

28

2

28

18

. Using these values it is possible to assess the representative state of thermodynamic conditions of wet air. In fact, if we remember that the wet-bulb temperature corresponds to the adiabatic saturation point temperature, it is possible to identify points a, 1 and 2 using the Mollier wet air diagram. To do this, we must identify the isoenthalpic curves passing through the intersection between the respective wet bulb isotherms and the saturation curve ϕ = 100 %. Having drawn these straight lines, the points representing the thermodynamic state of the air are identified in correspondence with the intersection between the isoenthalpic lines and the respective dry-bulb isotherms. The following can then be measured: relative humidity, titres and enthalpy. These data are set out in Table 4.2

ϕ a

7.63

31.92

50

6.0

1

21.50

90.00

35

18.5

2

12.10

50.65

38

9

.


29

expe rimenta l exerc exerc ises ises

In order to identify the remaining points (a’ and 1’) it is necessary to calculate energy and mass balances for the various parts making up the plant.

c 2 —Ba lanc es for pre- hea ter + hum idifier (sec (sec tions a -1)

& Energy balance: Q 1

& Q 1

= Q& prpre− ri scsc

− L& i = ∑

& m

⋅ ∆h =>

q1

& Q 1

& a ⋅ (h1 − h a ) − m & v ⋅ hv =m

= h1 − ha − h v ⋅

& v m & a m

The enthalpy hv of steam corresponds to saturated steam at 1 bar: From the table we find: h v = 2676 kJ/kg. Mass balance for steam: & a ⋅ x1 m

& a ⋅ xa − m & v =0 −m

⇒

& v m & a m

= x1 − x a

By replacing the values in Table 4.2, we obtain: & v m

= 18.5 - 6 = 12.5 g/kg

& a m

q1

= 90 − 31.92 − 2676 ⋅ 12.5 ⋅ 10 −3 = 24.63 kJ / kg q1

30

= 24.63 kJ / kg

Did Did a c ta Ita lia lia

Cha pte r 4.

In order to identify point a’, the end of pre-heating, it is necessary to define the mass and energy balance of the thermodynamic system included between sections a and a’ (pre-heater) (the same results could also be obtained by writing the balances the humidifier alone, system a’-1) (see Figure 4.5).

c 3 - Balanc Balanc e for pre-hea ter (sec (sec tions tions a- a’)

Energy balance: & Q 1

& a ⋅ (h a' − ha ) =m

Mass balance of steam: & a ⋅ x a' m

& a⋅ −m

xa

= 0 ⇒ xa = xa’

by replacing the numeric values you obtain:

x a ' &a Q& 1 / m

= 6 g / kg

= q1 = (ha ' − ha ) ⇒ ha '

= q1 + h a

and therefore

ha’ = 24.63 + 31.92 = 56.55 kJ/kg = 13.51 kcal/kg Having calculated the enthalpy for point a’, it is identified on the Mollier diagram as the intersection between isoenthalpic ha’ and the isotitre straight line xa’ = xa corresponding to point a. In order to calculate heat q2 removed by the refrigeration set and the heat q 3 provided by the post-heater, a balance of mass and energy must first be calculated for the system consisting of the refrigeration set plus post-heating set. In this way, you calculate the overall thermal flow exchange by the system.


31


Subsequently, using a balance for the post-heater alone, the separate contributions of q2 and q3 can be calculated.

c 4 - Bala nc e for the refrige rator system + po st-he ate r (sec tions 1-2):

Energy balance: & Q

− L& i = ∑

& m

⋅∆h => + Q& 3 −

& Q 2

& a ⋅ (h 2 − h1 ) + m & l ⋅ hl =m

If we consider the enthalpic flow issued with the flow of condensed steam to be & l ⋅ h l ≈ 0, this relation can be written as: negligible, m

+ Q& 3 −

& Q 2

= + q3 −

& a m

by replacing the numeric values:

+ q3 −

q2

≈ (h 2 − h1 )

q 2 = 50.65 - 90 = -39.35 kJ/kg

Balance of water vapour mass: & a ⋅ x2 m

& a ⋅ x1 + m & l = 0 −m

from which: & l m & a m

= x1 − x 2

by replacing the numeric values:

32

Did a c ta Ita lia

Cha pte r 4. & l m

= 18.5 - 9 = 9.5 g/kg

& a m

In order to make a separate calculation of the contribution provided by q 2 and that by q3, and to find the thermodynamic state of point 1’, we must lastly define the energy and steam mass balance for the post-heater alone (or, equivalently, the cooling set alone).

c 5 - Bala nc e for po st-he ate r (sec tions 1’-2):

Energy balance: & Q 3

& a ⋅ ( h 2 − h 1' ) =m

Steam mass balance: & a ⋅ x2 m

& a ⋅ x 1' = 0 −m

⇒

x2 = x1’

point 1’ is the most representative point for the thermodynamic conditions of air output from the cooling set. As indicated in the text, this point will lie on the curve ϕ = 100%. Moreover, the balance of the steam mass is used to calculate that the titre of 1’ is equal to the title of 2. It is therefore possible to determine, univocally, point 1’ on the Mollier diagram as the intersection of the isotitre passing through 2 and the curve ϕ = 100%. Having fixed 1’, it is possible to read the corresponding enthalpy from the Mollier diagram:

h1’ ≈ 8.3 kcal/kg = 34.75 kJ/kg


33


by developing the calculations and replacing the numeric values, you obtain:

x1’ = 9 g/kg & / m & a Q 3

= q 3 = ( h 2 − h 1' )

⇒ q3 = 50.65 - 34.75 = 15.90 kJ/kg and therefore: q 2 = q3 + 39.35 = 15.90 + 39.35 = 55.25 kJ/kg

q3 = 15.90 kJ/kg e q 2 = -55.25 kJ/kg Using these data, it is possible to report the fixed points (a, a’, 1, 1’, 2) and the relative transformations on the Mollier wet air diagram. The trend from a typical experience is shown in the diagram below (Fig. 3).

c 6 —Ca lc ulating the air flow rate The air flow rate is measured using a calibrated diaphragm that detects a fall in pressure. Using this finding and the characteristic tables for the diaphragm (enclosed) it is possible, using linear interpolation, to determine the speed of air flow. Values measured:

∆ p = 6.35 mmH 2O T air = T 2 = 28 °C Diameter of diaphragm = 0.1 m The following is shown by the table:

vair = 10.5 m/s and therefore:

&a m

= ρ air ⋅ v air ⋅ Sez = ρ ⋅ vI a ⋅

π ⋅ ∅2 4

= 1.29⋅(273/308)⋅10.5⋅7.85⋅10-3 = 0.0942

kg/s ≈ 339 kg/h

34

Did a c ta Ita lia

Cha pte r 4.

Once the flow of treated air is known, it is then possible to calculate the thermal power exchanged, the flow rate of vaporised water and condensed water.

c 7 —Calc ulation of the thermal p owe rs exc hang ed and the flow rates of c ondensed and vapo rised water & Q pre − risc

& a = 24.63⋅0.0942 = 2.32 kW = Q& 1 = q 1 ⋅ m

Q& refrig = Q& 2

= q 2 ⋅ m& a = (-55.25)⋅0.0942 = -5.20 kW

Q& post − heat = Q& 3

&l = m

&v= m

&l m &a m

&v m &a m

&a ⋅m

= q3 ⋅ m& a = 15.90⋅0.0942 = 1.50 kW

9.5⋅0.0942 = 0.895 g/s

≈ 3.22 kg/h

& a = 12.5⋅0.0942 = 1.18 g/s ≈ 4.25 kg/h ⋅m

To sum up: & Q = 2.32 kW pre − ri sc

& Q raff

= -5.20 kW

& Q = 1.50 kW pos t− r is c

& l = 0.895 g/s m

& v = 1.18 g/s m

≈ 3.22 kg/h ≈ 4.25 kg/h

Exe rc ise 4.7 —R-22 Com p ression Refrig eratio n Cyc le a - Aim Focusing on the compression refrigerating machine, operating with R22 in the T110D assembly, this chapter describes the function of the refrigerating assembly, highlighting the thermodynamic function of the various parts. T110D —User’s Ma nua l

35


By measuring pressure and temperature in the laboratory, it is also possible to calculate: 1. The main points of the thermodynamic cycle, plotting the cycle itself on the equilibrium diagram (h, log p) and (T,s). (To carry out this part, the following hypotheses are used to make the calculations: slight falls in pressure in exchangers, condensation end points to give a saturated liquid), 2. The energy exchanged by the system during the various transformations (heat and work), 3. The refrigerating efficiency of the cycle, 4. The flow rate of R22 in circulation. For this calculation, it is supposed that all the heat removed from the air during cooling is transmitted to R22 running inside the evaporator, 5. The electrical power absorbed by the electric motor (assuming that the compressor has a mechanical output of 0.8 and the motor an electrical output of 0.9). Lastly, calculate the efficiency of the refrigerating cycle in the event that this consists of an inverse Carnot cycle between: •

The same evaporation and condensation pressures of the real cycle.

•

The pressures of condensation and evaporation corresponding, respectively to room temperature and the temperature of refrigerated air.

In both cases determine the mechanical powers and the flow rates of refrigerating fluid required to remove a head from the refrigerated environment equivalent to that measured experimentally.

b —Instructio ns for the te a c her The T110D unit is used to adjust the refrigerating cycle automatically. This creates a few problems during the test given that the assembly tends to make adjustments both through the laminar valve (controlled by bellows by the end evaporation temperature) (continuous fine regulation) and the on-off function of the compressor (approximate regulation), controlled by a thermostat that detects air temperature after the evaporator (extent of regulation, set-point set by user). If the set-points are not chosen appropriately, the system makes continuous adjustments, preventing stationary operating conditions from being obtained. This means that the equipment has to be put “in crisis” by setting parameter values that cannot be met even if the system works at full head (in this way the cycle will operate in stable conditions at full power). In order to achieve this, the following adjustments are valid for room temperatures of approximatley 20°C (small adjustments may be required for other values): •

36

fan ON with potentiometer at 75%; Did a c ta Ita lia

Cha pte r 4. •

set-point of compressor thermostat (in air channel downstream of the evaporator) set at about 12 °C;

•

electrical air preheater (upstream of evaporator) ON 2 with approx. 7 ÷8 amperes of absorbed power (adjust using potentiometer).

Using these measurements, the value of the parameters will be as follows, in order of size: •

air temperature upstream of the evaporator ≈ 43° C

•

air temperature downstream of the evaporator …≈ 16° C

•

pressure at start of compression

•

pressure at end of compression

≈ 3 bar ≈ 10.4 bar (10.5 atm)

•

temperature at end of compression = T 1 ≈ 70.6° ÷ 75.4° C

•

temperature at start of compression (evaporator outlet) = T4 ≈ 11.8° ÷ 12.4° C

•

temperature at end of condensation (condenser outlet) = T2 ≈ 28.3° C

•

temperature at end of lamination (evaporator input) = T3 ≈ 5.7° C

•

∆ p of air astride the diaphragm = 11 ÷ 20 Pa

It is important to observe that the pressure at the start and end of compression does not coincide with the corresponding evaporation and condensation pressures of the cycle owing to the loss of head. These are relatively small for the condenser, but are much greater for the evaporator. This means that if all the above values were to be detected and plotted on the R22 equilibrium diagram, they would produce a strongly distorted cycle (see Figure 4.6). For teaching purposes it is not appropriate to use this real cycle, but it is clearer and more instructive to use an ideal cycle and overlook the losses of head in the condenser and evaporator, also given that the various energie involved (q 2, q1, li,comp, efficiency) do not change as they pass from one cycle to another (they are differences of enthalpy and they stay the same if point 3 is set in the real cycle, or it is set as 3’ as in the ideal cycle). Errore. Non è stato specificato un argomento.

It follows that students must solely be given the following measured parameters: •

pressure at start of compression coinciding with evaporation pressure (supposed constant along the entire evaporator);

•

pressure at end of compression coinciding with condensation (supposed constant along the entire evaporator);


37

expe rimenta l exerc ises •

temperature at start of compression (T4);

•

temperatura at end of compression (T 1).

It is therefore necessary to hypothesise that the end point of condensation provides a saturated liquid (lower curve). Using these data students can plot the complete ideal cycle on the equilibrium diagram (NOT provide values for T 2 and, above all, T 3 because they are not compatible with the ideal cycle).

c - Performing the test c 1 —Ma in po ints of the refrige ration c yc le During the course of the experimental exercise, analysis is focused on the refrigerating cycle used for the cooling set in an air conditioning plant, for laboratory purposes, of the full air type. Figure 2.8 shows a detail of the refrigerating system operating on R22.

38

Did a c ta Ita lia

Cha pte r 4.

Valvola di laminazione Condenser Compressore Evaporatore

This is a continuous flow system that operates in a steady condition. Each of the transformations take place in a given device.


39


These are:

4-1

Irreversible adiabatic compression

Compressor

1-2

Reversible isobar condensation (until saturated liquid condition)

Condenser

2-3

Isenthalpic leak

3-4

Reversible isobar evaporation (until superheated steam conditions)

Laminar valve Evaporator

A thermodynamic diagram is required which is used to display the properties of the refrigerant fluid in conditions of equilibrium: for example, the diagram (h, log p) known as a Mollier diagram, shown in Figure 4.9. The following values measured experimentally using laboratory aparatus are also known:

Evaporation pressure (p3=p4)

3 bar

Condensation pressure (p 1=p2)

10.5 atm

Compressor aspiration temperature (or end of evaporation) (T4)

11.8° C

Compressor delivery temperature (or start of condensation) (T1)

70.6° C

Air temperature above the evaporator

43° C

Air temperature below the evaporator

16° C

∆ p on calibrated diaphragm

25.6 mm H2O

Point 1

Point 1 at the end of compression occurs in conditions of superheated steam. Given that it has a known temperature T 1 = 70.6° C = 343.6 K and pressure p 1 = p condensation = 10.5 atm = 10.4 bar, it is possible to make a full definition of its thermodynamic state using the Mollier diagram. Pressure, enthalpy and entropy are measured (see Table 4.3). 40

Did a c ta Ita lia

Cha pte r 4.

Tra nsforma tion 1-2 a nd p oint 2

This consists of an isobar transformation consisting of two parts: one takes the form of overheating from point 1 to the condensation temperature corresponding to the pressure of 1, and the other, inside the limit curve, of isotherm condensation. Given that the final conditions are saturated liquid, point 2 is competely determined and both enthalpy and entropy can be determined from the Mollier diagram. The transformation will obviously be a straight horizontal line in the diagram (h, log p) whereas in Gibbs’ diagram it will be a straight horizontal line in the isotherm tract (inside the limit curve) and in the superheated steam part the intermediate points of the transformation can again be determined using the Mollier diagram.

Trasfo rm a tion 2-3 a nd p oint 3 The 2-3 transformation take the form of a leak, in other words an isenthalpic transformation (without producing work or thermal exchange); it is irreversible and takes place to reduce the pressure of refrigerant fluid to the pressure of evaporation. Point 3, marking the end of expansion, can be determined as the intersection between the isenthalpic line passing through 2 and the isobar/isotherm passing through point 4 (evaporation isobar-isotherm at 3 bar). The other thermodynamic values for point 3 can be read on the Mollier diagram. Given that this is an irreversible transformation, the intermediate stages of the transformation are not known: strictly speaking, it is only known that final enthalpy is equal to the initial stage, but it does not necessarily remain constant throughout the transformation. However, it is usual to consider enthalpy as remaining constant throughout the course of transformation in order to plot the quality of transformation. This will result in a vertical segment in the diagram (h, log p), whereas it will create a decreasing curve in the diagram (T, s).


41


Point 4

Point 4 marking the start of compression occurs in conditions of superheated steam. Given that its temperature T4 = 11.8° C = 284.8 K and pressure p 4 = p evaporation = 3 bar are known, it is possible to make a complete definition of its thermodynamic state using the Mollier diagram. In particular, pressure, enthalpy and entropy are determined (see Table 4.3). Tra nsforma tion 3-4 a nd p oint 4

Transformation 3-4 is the transformation involving the evaporation of fluid. It will lie, partly (up to point 3’), inside the limit curve (where it coincides with an isobarisotherm) and the last part consists of a tract of isobar inside the zone of superheated steam. The main points of 4 have already been determined earlier and the plotting of the transformation 3-4 does not present any difficulty either on the diagram (h, log p) or the diagram (T, s) because the two ends need only be joined by a straight segment.

42

Did a c ta Ita lia

Cha pte r 4.

Transfo rma tion 4-1

Transformation 4-1 is an irreversible adiabatic compression. Strictly speaking, only the initial and final stages of balance are known. It therefore cannot be plotted on the equilibrium diagrams; however, by convention it is indicated using a “fictitious” transformation (using a dashed line to recall that it is irreversible).

1

10.4

70.6

450

1.830

2

10.4

25.5

231

1.105

3

3

-14.7

h 3=h2=231

1.123

4

3

11.8

417

1.830

Main

Based on these values, it is possible to trace the cycle on the enclosed equilibrium diagram (Figure 4.9 shows the general trade of the cycle on the plane (h,log p).

– T110D —User’s Ma nua l

43


In order to plot the trend of the cycle on the diagram (T, s), it is important to measure the assumed values for temperature and entropy in correspondence with some points along the curve representing the cycle on the diagram (h, log p) plotted earlier. It is also necessary to plot the limit curve, using the values of entropy and temperature shown in Table form. Fig. 4.10 shows the cycle on the plane (T, s) in qualitative terms. A realistic example of cycle on (T,s), obtained using MS-Excel, is shown in Figure 4-11.

44

Did a c ta Ita lia

Cha pte r 4.

-

Temperatura = Entropia

Note for the teac her only In the same experimental conditions used in this exercise, if the temperatures T 2 and T3 had been measured, the following results would have been obtained: •

T2 = 28.3° C

•

T3 = 5.7° C

It can be seen that the value of T 2 is very close to the value obtained by supposing point 2 on the C.L.I., namely approximately 26°C. What changes most is the temperature of point 3. The real point 3 is at 5.7°C, which corresponds to an evaporation pressure of approximately 6 bar. Given that the pressure at the outlet of the evaporator is 3 bar (corresponding to an evaporation temperature of –14.7°C), this means that the loss of pressure while crossing the evaporator is approximately 3 bar. The trend of the real cycle is shown in red on the enclosed equilibrium diagram.


45


c 2 —Calc ulating the energy e xc hange d The energy exchanged is calculated by resorting to the first principle of thermodynamics in the open-system form, by adopting the following hypotheses: •

Steady condition

•

A single input and output

•

Negligible changes in kinetic and potential energy

The following formula can be proposed using these hypotheses:

q - l i = dh Hea t yielde d t o the outsid e e nvironm ent (q 1 )

Heat is only yielded to the outside in isobar transformation 1-2 in which there is no exchange of work. This can be expressed as:

q1 = h2 - h1 = 231 - 450 = - 219 kJ/kg Hea t substa c ted from outsid e e nvironm ent (q 2 )

Heat is only absorbed from the outside in isobar transformation 3-4 in which there is no exchange of work. This can be expressed as:

q2 = h4 - h3 = 417 - 231 = 186 kJ/kg Inside w ork ab sorbe d b y the syste m (l ic )

Work is only exchanged in transformation 4-1: this is work carried out on the system and given that transformation 4-1 is adiabatic, this produces

l ic = - (h1 - h4 ) = - (450 - 417) = - 33 kJ/kg Having calculated the energy exchanged, it is important to check that the first principle of thermodynamics has been complied with, namely that:

∫ δq = ∫ δl

i

or also:

q1 + q2 = l ic (quantity taken with sign)

46

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Cha pte r 4.

In fact, given that: - 219 + 186 = - 33 the identity will be checked.

c 3 —Ca lc ulating the effic ienc y of the refrige rating c yc le The following is obtained on the basis of the above calculations:

ε = q2 / |l ic| = 186/33 = 5.64 c 4 —Ca lc ulating the p ow ers exc hang ed and the flow rate o f R22 The flow rate of the refrigerating fluid carrying out the cycle is not known. However, the sensitive heat removed from the air as it passes through the evaporator can be determined If the channel through which air is ducted is supposed to be adiabatic towards the outside, the following formula can be used to express the first principle of thermodynamics:

Q& 2

= Q& sensitive, air

from which:

& ⋅ ρ ⋅ c ⋅ ∆T & air ⋅ c p ⋅ ∆T = V Q& sensitive, air = m air p

The volumetric displacement of air can be calculated based on the characteristics of the calibrated diaphragm. The volumetric displacement of air can be determined using the characteristics of the calibrated diaphragm. If the ∆ p between upstream and downstream is known, using the enclosed tables, air speed can be calculated:

T aria = 16° C, ∆ p = 14.4 mmH 2O Given that the diameter of the diaphragm is:

⇒ da tabelle v ≈ 15.2 m/s ∅ = 10 cm the following is obtained:

A = π⋅ D /4 = 7.854 ⋅10 m 2


-3

2

47

expe rimenta l exerc ises & = v ⋅ A = 0.119 m3 /s = 428.4 m 3 /h V air

and therefore & air = 0.119⋅1.29⋅273/289 = 0.145 kg/s m

& air ⋅ c p ⋅ ∆T = 0.145 ⋅ 1004 ⋅ (43 − 16 ) = 3931 W Q& sensitive, air = m

and lastly: & Q 2

= 3931 W

the maximum deplacement of R22 can therefore be calculated as:

& R − 22 m

=

& Q 2

=

q2

3931 186 ⋅ 1000

= 0.0211 kg / s ≈ 76 kg / h

It is now possible to calculate the mechanical power supplied by the fluid and the electric power absorbed (assuming the value of 0.8 as the mechanical output of the compressor and 0.9 as the electrical output of the motor): & L i ,c

& L

& R − 22 = 696 W = l i ,c ⋅ m

e

=

& L i ,c

η ⋅ ηel

= 967 W

c 5 —Ca rnot’s inverse c yc le The refrigerating efficiency, mechanical power and flow rate of the refrigerant in Carnot’s inverse cycle can now be calculated, considering the thermal load to be removed from the air as constant. 1st Case

On the basis of a cycle with an evaporation and condensation cycle equal to those calculated experimentally ( pev = 10.4 bar; p cond = 3 bar. These values correspond to a condensation temperature of 25.5° C and an evaporation temperature T evap. equal to 14.7 °C. The corresponding ideal Carnot’s cycle plotted on the diagram (h,log p) is shown in the following figure.

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The following efficiency and energies are exchanged:

ε = T evap. / (T cond. - T evap. ) = 258.3 / (298.5

- 258.3) = 6.4

q1 = h2 - h1 = ( 231 - 412.3 ) = - 181.3 kJ/kg q2 = h4 - h3 = (385 - 228) = 157 kJ/kg m R-22 = [ma

× c p × (T ai -T ao )] / q2 = 0,025 kg/s a

& / ε = 3931/ 6.4 = 614 W & = Q L ic 2

2nd Ca se

The second case considers: •

condensation temperature Tcond equivalent to room temperature, 20° C,

•

evaporation temperature Tevap. equal to the cold air temperature,16° C.


49


The following efficiency and energies are exchanged:

ε = T evap. / (T cond. - T evap. ) = 289 / (293

- 289) = 72.3

q1 = h2 - h1 = ( 224.1 - 411.4 ) = - 187.3 kJ/kg q2 = h4 - h3 = (407.5 - 223 ) = 184.5 kJ/kg m R-22 = [ma

× c p × (T ai -T ao )] / q2 = 0,0213 kg/s a

& / ε = 3931/ 6.4 = 54.4 W & = Q L ic 2

c 6 ana lysis (only theo retic al, not nume ric ) It is possible to demonstrate that there is a relationship between minimum power (or work), power effectively requested and changes in entropy of the universe during each cycle: & = L & L min

+ T0 ⋅ ∆ S& tot

The change in the entropy of the universe is only caused by an increase in the overall entropy of the two heat sources (cold – namely the air treated – and hot – namely the outside environment), given that the system operates cyclically we can suppose:

∆

50

S&

tot

=∫

∂Q& 2 Taria

∂Q& 1

+ ∫

Tamb

Did a c ta Ita lia

Cha pte r 4.

Air treated by the system changes its temperature during the course of transformation passing from the value of T a1 upstream of the evaporator (in our case 43° C) to the value Ta2 downstream (16° C). In the same way, even in the condenser, the cooling air passes from room temperature to a higher temperature. For this reason, it can be said (supposing the isobar of the air cooling/heating process):

∆

S&

tot

= c p

=∫

∂Q& 2 Taria

+∫

& Q 1

′ Taria

=

2

∫

& aria ⋅ c p ⋅ dT m aria

Taria

Ta1

Ta

+

2

∫

T′ a1

m ′aria ⋅ c p aria ⋅ dT

′ Taria

=

 Ta   Ta′    ⋅ ln + c p ⋅ ln  T   T′   a   a  2

aria

Ta

2

aria

1

1

It is not possible to proceed with the numeric calculation because we have no data relating to flow rate and the temperature of air in the condenser.


51


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5.

Exe rc ises with SAD/ T110D Autom a tic Da ta Acquisition System

5.1

Formulas used for a ir c ond itioning te sts (Test 01)

Enthalpy The automatic data acquisition system enables the enthalpy values at 3 characteristic system points to be determined automatically on the basis of the temperatures and the wet air titre. The formulas for the determination of enthalpy at the various points are:

h = 2501 ⋅ X bu

+ (1 + 1.9 ⋅ X bu ) T W

where: Tw = tw + 273.16 (K) tw = wet bulb temperature

X bu

= 0.622 ⋅

P ws 101325 − P ws

wet air titre

P ws

=e

 C 1   + C 2 +C 3 ⋅T + C 4 ⋅T 2 + C 5 ⋅T 3 +C 6 ⋅ln (T )  T  

C1 = -5800.2206 C2 = 1.3914993 C3 = -0.04860239 C4 = 0.000041764768 C5 = -0.000000014452093 C6 = 6.5459673 T = Tw

Wet a ir titre The air titre at the 3 system points is determined on the basis of temperature of dry bulb temperature. The air titre is calculated through the following formula: T110D —User' s Ma nua l

53

Exercises with SAD/ T110D Au to ma tic Da ta Ac q uisition Syste m

X =

h − T d 2501 + 1.9 ⋅ T d

where: h = enthalpy Td = td + 273.16 (K) td = dry bulb temperature

Relative humidity The relative humidity at the 3 system points is determined through the following formula:

X ⋅ 101325 Hr = X ⋅ 0.622 ⋅ P ws where: X = air titre

P ws

=e

 C 1   +C 2 +C 3 ⋅T +C 4 ⋅T 2 +C 5 ⋅T 3 +C 6 ⋅ln (T )  T  

T = Td = temperature of the dry bulb

Air flow- rate The flow-rate of the output air is calculated on the basis of the differential pressure before and after the special calibrated diaphragm. Flow-rate (Q) may therefore be obtained through the following formula:

Q

=

C 1− β 4

ε1

π 4

d 2 2 ∆pρ 1

Where: Q is the air flow-rate expressed in m3/s C is the discharge coefficient

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Cha pte r 5.

C

= 0.5959 + 0.0312β

2.1

8

− 0.1840β +

2.5 10 0.0029β ( Re

6 )

0.75

D

ReD is Reynolds number relative to diameter D

ε1 is the expansion factor ε 1 = 1 − (0.41 + 0.35β 4 )

∆ p k ρ 1

k is the iso-entropic exponent; for air it is 1.4

β is the ratio between the diameter of the diaphragm (d) and the pipe diameter (D) expressed in m.

β=

d D

d = 0.1 m; D = 0.3 m

∆p is the differential pressure before and after the diaphragm 1 mmH2O = 9.797 Pa

ρ is the specific weight of air in kg/m3 therefore we can say that:

Q = K ∆ pρ1 Where: K = 0.007853

T110D —User' s Ma nua l

55


5.2

Formula s used in c ooling tests (Test 02)

Efficiency of C.O.P. cooler

ε=

− h1 h4 − h3 h3

where: h1, h3 and h4 are the enthalpy values at the relative points in the cycle on the diagram of Freon R22. For the air flow-rate which is cooled, use the same formulas as in Test 01. Experimental procedure for Test 02

56

•

adjust the SP of the cooler and the air flow-rate

•

acquire the temperature and the pressure

•

from the Freon 22 diagram, determine the enthalpy values at points 1,3,4; enter then via the keyboard and save the tests.

•

modify air flow-rate and following the same procedure.

Did a c ta Ita lia

Cha pte r 5.

5.3

Exp erime ntal ex erc ises

Exe rc ise 5.1 - Air co nd ition ing test (Test 01) Exe c ution o f the test 1. Set a Set Point of 18°C 2. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 3. Wait for the flow-rate to become stable. 4. Acquire the data automatically by means of the Read and Input commands. 5. Gradually increase the flow-rate and repeat steps 3 and 4. 6. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199801: Test 01, followed by the following diagrams: •

Sensible heat - air flow-rate

•

Dry temperature - air flow-rate

•

Wet temperature - air flow-rate

•

Relative humidity - air flow-rate


57


Table 5.1

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Fig. 5.1 – Enthalpy vs air flow rate

Fig. 5.2 – Dry temperature trend vs air flow rate


59


Fig. 5.3 – Wet temperature trend vs air flow rate

Fig. 5.4 – Relative humidity vs air flow rate

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Exe rc ise 5.2 - A ir co nd itioning test (Test 01) with p re- hea ter Exe c ution o f the test 1. Set a Set Point of 18°C 2. Turn on the pre-heater and position it on the value at which you wish to perform the test. 3. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 4. Wait for the flow-rate to become stable. 5. Acquire the data automatically by means of the Read and Input commands. 6. Gradually increase the flow-rate and repeat steps 4 and 5. 7. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199802: Test 01 with pre-heater, followed by the following diagrams: •


•


•


•



61


Table 5.2

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63




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Exe rc ise 5.3 - Air c ond itioning test (Test 01) with he a ter Exe c ution o f the test 1. Set a Set Point of 18°C 2. Turn on the heater and position it on the value at which you wish to perform the test. 3. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 4. Wait for the flow-rate to become stable. 5. Acquire the data automatically by means of the Read and Input commands. 6. Gradually increase the flow-rate and repeat steps 4 and 5. 7. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199803: Test 01 with heater, followed by the following diagrams: •


•


•


•



65


Table 5.3

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67




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Exe rc ise 5.4 - A ir co nd itioning test (Test 01) with c oo ler Exe c ution o f the test 1. Set a Set Point lower than ambient temperature in the “Keyboard Value” window, from the keyboard. 2. Turn on the cooler and adjust the Set Point of the instrument installed in the electric panel located under the channel to the same value as set in the software. 3. Wait for the coolant temperature to approach the SP. 4. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 5. Wait for the flow-rate to become stable. 6. Acquire the data automatically by means of the Read and Input commands. 7. Gradually increase the flow-rate and repeat steps 3, 5 and 6. 8. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199804: Test 01 with coolant set point of 10°C, followed by the following diagrams: •


•


•


•



69


Table 5.4

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71




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Exe rc ise 5.5 - Air c ond itioning test (Test 01) with hum idifier a nd wa ter Exe c ution o f the test 1. Set a Set Point of 18°C. 2. Start the pump located under the channel by means of the relative button. 3. Adjust the flow of the water coming out from the nozzle located inside the chamber by working on the valve fitted to the pump. 4. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 5. Wait for the flow-rate to become stable. 6. Acquire the data automatically by means of the Read and Input commands. 7. Gradually increase the flow-rate and repeat steps 5 and 6. 8. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199805: Test 01 with pump delivery valve fully open, followed by the following diagrams: •


•


•


•



73


Table 5.5

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75




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Exe rc ise 5.6 - A ir co nd itioning test (Test 01) with stea m hum idifier Exe c ution o f the test 1. Set a Set Point of 18°C. 2. Make sure that the boiler located under the channel is full of water. 3. Turn on the boiler by means of the button located on top of it and wait for the steam to form. 4. Press the steam delivery button. 5. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 6. Wait for the flow-rate to become stable. 7. Acquire the data automatically by means of the Read and Input commands. 8. Gradually increase the flow-rate and repeat steps 6 and 7. 9. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 1809199806: Test 01, followed by the following diagrams: •


•


•


•



77


Table 5.6

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79




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Exe rc ise 5.7 - Cooling test (Test 02) Exe c ution o f the test 1. Make sure that option “Calculate on command” inside “Option - Input mode” of SAD/T110D (test 02) is active. 2. Set a Set Point lower than ambient temperature in the “Keyboard Value” window, from the keyboard. 3. Turn on the cooler and adjust the Set Point of the instrument installed in the electric panel located under the channel to the same value as set in the software. 4. Wait for the coolant temperature to approach the SP. 5. Set an air flow-rate corresponding to a pressure difference at the diaphragm of 1.5 2 mm H2O. 6. Wait for the flow-rate to become stable. 7. Acquire the data automatically by means of the Read command. 8. Based on the pressure and temperature data for the 3 points, determine the enthalpy value from the enthalpy diagram for Freon R22. 9. Enter from the keyboard the enthalpy values measured on the diagram and work on the Calculate and Input commands. 10. Gradually increase the flow-rate and repeat steps 4, 5, 6, 7, 8 and 9. 11. After a certain number of acquisition processes, examine the working file and/or the experimental diagrams; print them out if you wish to.

Test results The following page shows a table with the results of test 2109199802: Test 02 with coolant set point of 10°C, followed by a diagram: •

Efficiency - air flow-rate


81


Table 5.7

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Fig. 5.25 – Efficiency vs flow-rate


83

T110D Manual - First Part

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