Fluid and Thermodynamic Parameters - Assignment 1

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  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    1/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    2/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    SUMMARY OF RESULTS

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    3/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Question 1

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    4/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Question 2

    1

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    5/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    M = kg/s

    M

    V = L/s

    Flow Rate = L/s

    =150

    3.85 x 10

    =

    =3.896

    0.000

    ######

    3.896

    V

    ######

    2

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    6/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Using the mean temperature, calculate the required Glycol flow rates through each section of the piping system.

    Section 3 - Plate heat exchanger 2 branch line.

    Q = M C T PIPE SELECTION

    Where: From the chart (Chart 3 - Friction Loss for Water Piping

    M is unknown Systems: Copper Tube, 15.6C), I nominate 50mm

    Q = 100kW copper tube.

    C = At a flow rate of 2.5L/s, 50mm copper tube gives:

    T = 10K Velocity of: n = m/s

    QC T

    M = kg/s

    M

    V = L/s

    Flow Rate = L/s

    2.597

    3.85

    M =

    =100

    3.85 x 10

    1.4

    ######

    =2.597

    0.000

    ######

    V =

    3

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    7/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Using the mean temperature, calculate the required Glycol flow rates through each section of the piping system.

    Section 4 - Main return line, from branch lines of plate heat exchangers to chiller inlet.

    Q = M C T PIPE SELECTION

    Where: From the chart (Chart 3 - Friction Loss for Water Piping

    M is unknown Systems: Copper Tube, 15.6C), I nominate 80mm

    Q = 250kW copper tube.

    C = At a flow rate of 6.3L/s, 80mm copper tube gives:

    T = 10K Velocity of: n = m/s

    QC T

    M = kg/s

    M

    V = L/s

    Flow Rate = L/s

    3.85 x 10

    6.494

    V =

    =6.494

    0.000

    1.6

    3.85

    ######

    ######

    M =

    =250

    4

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    8/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Section 1 - Main supply line, from chiller outlet to branch lines of plate heat exchangers.

    a) Reynolds Number d) Pressure drop through pipework

    n x

    (kinematic velocity) L n2

    2

    (dynamic velocity) Where:

    f = Friction factor

    L = Length of pipe (m)

    Where: = Density of fluid (g/mL)

    = cP (0.018Pa/s) n = Velocity (m/s)

    = g/mL (1,044kg/m3) = Diameter of pipe (m)

    n = m/s pd = Pressure drop (Pa)

    = m

    pd = x x x

    =

    pd = Pa

    x P = kPa for Section 1 (pipework only).

    Re = e) Pressure drop through fittings

    Pt = Kt x Pv

    b) Flow Type Where:From the Moody Chart a Reynolds Number Pt = Total Pressure drop through fitting (kPa)

    Re =1.6 0.08

    pd = 0.028 x

    Re =

    =

    pd =

    18

    0.000

    1.6

    0.08

    =0.018

    1044

    x

    x

    x

    1044

    23,713.00

    f x

    1.28

    104450.7

    0.08x

    1.62

    2

    23.72

    0.028

    7,424

    0.0000172

    0.0000172

    633.8

    1

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    9/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Fitting

    Gate valve

    Strainer

    Check valve

    Tee (line)

    Elbow

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through pipework: kPa

    Section 1 Total Pressure Drop: kPa

    23.72

    7.85

    31.57

    2.580.43

    0.19 3 0.57

    0.15 1.34 0.21 1 0.21

    1.68 1 1.68

    7.85

    2.1 1.34 2.81 1 2.81

    0.32 1.34 6

    1.25 1.34

    0.14 1.34

    kt Pv Pt Qty Total P

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    10/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Section 2 - Plate heat exchanger 1 branch line.

    a) Reynolds Number d) Pressure drop through pipework

    n x

    (kinematic velocity) L n2

    2

    (dynamic velocity) Where:

    f = Friction factor

    L = Length of pipe (m)

    Where: = Density of fluid (g/mL)

    = cP (0.018Pa/s) n = Velocity (m/s)

    = g/mL (1,044kg/m3) = Diameter of pipe (m)

    n = m/s pd = Pressure drop (Pa)

    = m

    pd = x x x

    =

    pd = Pa

    x P = kPa for Section 2 (pipework only).

    Re = e) Pressure drop through fittings

    Pt = Kt x Pv

    b) Flow Type Where:From the Moody Chart a Reynolds Number Pt = Total Pressure drop through fitting (kPa)

    5,278

    0.0000172

    2,030.50

    0.98

    1.42

    =0.018 0.065 2

    Re =1.4 0.065 2.04

    0.0000172

    1044 xpd = 0.03 x

    0.03 66.15 1044

    18

    0.000

    1.4

    0.065

    xx x

    1044

    4.3x

    =

    Re =

    pd = f

    2

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    11/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Fitting

    Gate valve

    Globe valve

    Tee (branch)

    Heat Exchange

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through pipework: kPa

    Section 2 Total Pressure Drop: kPa

    0.77 1.02

    108.27

    0.79 2 1.58

    1

    108.27

    2.04

    110.31

    100.00

    0.16 1.02 0.17 2 0.34

    6.2 1.02 6.35 1 6.35

    kt Pv Pt Qty Total P

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    12/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Section 3 - Plate heat exchanger 2 branch line.

    a) Reynolds Number d) Pressure drop through pipework

    n x

    (kinematic velocity) L n2

    2

    (dynamic velocity) Where:

    f = Friction factor

    L = Length of pipe (m)

    Where: = Density of fluid (g/mL)

    = cP (0.018Pa/s) n = Velocity (m/s)

    = g/mL (1,044kg/m3) = Diameter of pipe (m)

    n = m/s pd = Pressure drop (Pa)

    = m

    pd = x x x

    =

    pd = Pa

    x P = kPa for Section 3 (pipework only).

    Re = e) Pressure drop through fittings

    Pt = Kt x Pv

    b) Flow Type Where:From the Moody Chart a Reynolds Number Pt = Total Pressure drop through fitting (kPa)

    0.0000172

    11,934.30

    Re =1.4 0.05 11.94

    0.0000172

    4,060

    0.034 343.1 1044 0.98

    1.42

    =0.018 0.065 2

    0.05

    1044

    22.3x 1044 xpd = 0.034 x

    x

    =

    18

    0.000

    1.4

    Re =

    pd = f x x

    3

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    13/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Fitting

    Gate valve

    Globe valve

    Tee (line)

    Elbow

    Heat Exchange

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through pipework: kPa

    Section 3 Total Pressure Drop: kPa

    0.36 1.02 0.37 2 0.74

    88.46

    88.46

    11.94

    100.40

    6.96

    0.19 1.02 0.20 2

    1

    0.40

    80.00

    6.8 1.02 6.96 1

    Qty Total P

    0.17 1.02 0.18 2 0.36

    kt Pv Pt

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    14/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Section 4 - Main supply line, from chiller outlet to branch lines of plate heat exchangers.

    a) Reynolds Number d) Pressure drop through pipework

    n x

    (kinematic velocity) L n2

    2

    (dynamic velocity) Where:

    f = Friction factor

    L = Length of pipe (m)

    Where: = Density of fluid (g/mL)

    = cP (0.018Pa/s) n = Velocity (m/s)

    = g/mL (1,044kg/m3) = Diameter of pipe (m)

    n = m/s pd = Pressure drop (Pa)

    = m

    pd = x x x

    =

    pd = Pa

    x P = kPa for Section 4 (pipework only).

    Re = e) Pressure drop through fittings

    Pt = Kt x Pv

    b) Flow Type Where:From the Moody Chart a Reynolds Number Pt = Total Pressure drop through fitting (kPa)

    =

    18

    0.000

    1.6

    0.08

    Re =1.6 0.08 23.72

    0.0000172

    0.028

    0.0000172

    23,713.00

    7,424

    2

    633.8 1044 1.28

    =0.018 0.08

    1044

    pd = x41.5

    x f x

    0.028 x 1044

    xpd =

    Re =

    x1.6

    2

    4

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

    15/31

    Shawn Taylor

    Student No: 267439818

    Diploma HVAC R

    Hydronic Systems

    Refrigeration Thermodynamics

    Assignment Number 1

    Calculate the pressure loss through each section of the piping system

    Fitting

    Gate valve

    Tee (line)

    Elbow

    Chiller

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through fittings: kPa

    Total Pressure Drop through pipework: kPa

    Section 4 Total Pressure Drop: kPa

    23.72

    104.98

    0.32 1.34 0.43 2

    1 80.00

    81.26

    81.26

    0.14 1.34 0.19

    0.86

    kt Pv Pt Qty Total P

    1 0.19

    0.210.15 1.34 0.21 1

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    Shawn Taylor

    Student No 267439818

    J027B - Fluid Thermodynamic

    Parameters

    Assignment No. 1

    Fluid&ThermodynamicParameters

    Assi

    gnmentNo.1

    R22

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    Shawn Taylor

    Student No 267439818

    J027B - Fluid Thermodynamic

    Parameters

    Assignment No. 1

    You decant R22 @ -10C into a charging cylinder that has a volume of 1,000cm3.

    The cylinder is "safely" filled to 80%. The cylinder temperature rises to the air

    temperature of 20C.

    a) What will be the volume of R22 in the cylinder?

    b) Calculate if the cylinder pressure relief valve is going to be activated due

    to hydrostatic expansion if the temperature in your service van rises to 50C.

    a) v = V T

    (of R22) = x 10-4

    K-1

    =

    v = change in volume (cm3)

    V = 800cm3

    (1000 x 80%) - original volume (cm3)

    T = 30K (20C - -10C)

    v = 800cm3

    x x 30K

    v = cm3

    New volume = Original volume + Change in volume

    New volume = +

    New volume = @ 20C

    Question 1a

    0.00115K-1

    27.6cm3

    827.6cm3

    0.00115K-1

    27.6

    800cm3

    11.5

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    Shawn Taylor

    Student No 267439818

    J027B - Fluid Thermodynamic

    Parameters

    Assignment No. 1

    b) Pressure relief valve would activate when the entire volume of the tank is filled with

    liquid refrigerant. Therefore, will the initial volume of 800cm3 (@10C) expand beyond

    the total volume of 1,000cm3

    @ 50C?

    i.e. Will "Change in volume" be greater than 200cm3?

    v = V T

    (of R22) = x 10-4

    K-1

    =

    v = change in volume (cm3)

    V = 800cm3

    (1000 x 80%) Original volume (cm3)

    T = 60K (50C - -10C)

    v = 800cm3

    x x 60K

    v = cm3

    Question 1b

    0.00115K-1

    55.2

    0.00115K-1

    11.5

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    If you never paid much attention to what kind of refrigerant is circulating inside your p

    ge to Enlarge

    , R407c is already widely used in chil lers from European suppliers, like this portable unit from Frigel.

    If you never paid much attention to what kind of refrigerant is circulating inside your plastics chillers, it may be time you did. Ne

    1. ITS THE LAW

    Whats happening is not a conspiracy of chiller suppliers to make you buy new equipment. It all goes back to an international a

    2. YOU CAN KEEP YOUR OLD CHILLERS

    It is not illegal to continue to use chillers that contain R22. In fact, for a little while longer, you may be able to buy chillers built b

    3. THERE ARE NUMEROUS ALTERNATIVESThree alternative refrigerants are most commonly used in plastics chillersR407c, R410a, and R134a. They are readily availa

    4. DOES IT MATTER WHICH ONE I USE?

    Probably not very much. And in some cases, there isnt much choice anyway, because technical reasons limit the range of app

    A number of suppliersincluding Advantage Engineering, AEC, Conair, Delta T, and Mokonagree that R134a is preferable f

    Two less commonly mentioned refrigerants are used for especially low-temperature chillersR507 or R404a for temperatures

    Although European equipment makers tend to rely largely on R407c, the chart shows that major domestic suppliers are leanin

    Advocates of R410asuch as AEC, Conair, and Thermal Carepoint out that its the most efficient refrigerant. That means th

    Another point of comparison between R407c and R410a is cost. Chiller makers agree that the cost of components for chillers

    Overall, it is difficult to compare efficiency, compactness, and cost of new vs. old chillers strictly on the basis of refrigerants. On

    One other point of debate concerning refrigerants is maintenance issues. Some sources note that industrial refrigeration servic

    All these various issues, whether or not of practical importance, appear to have motivated firms such as AEC, Advantage, and

    5. DO I NEED A WHOLE NEW CHILLER?

    As noted above, the higher operating pressures of most of the R22 replacements necessitate substitution of beefier compresso

    6. ITS NOT THE END OF THE STORY

    Chiller suppliers doubt that the switch from R22 to the current stable of refrigerants will be the end of the story. Several others

    One looming factor in refrigerant choices is global warming potential, or GWP. This is another atmospheric impact of refrigera

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    lastics chillers, it may be time you did.

    regulations took effect on Jan. 1, which limit the types of refrigerants that can be used in new chillers and reduce worldwide p

    greement called the Montreal Protocol for protection of the atmospheric ozone layer, signed by the U.S. in 1987 and enforced b

    efore Jan. 1 that contain R22. As of last month, a few suppliers still had some of these units in stock. For some years to come,

    le and are not really new technology, as they have been used in automotive and home air conditioning for a number of years.

    licable refrigerants. As shown in the table opposite, most suppliers offer a range of refrigerants, often segmenting them by appli

    r very small units. The reason is that it is the only commonly available refrigerant that operates at a pressure lower than that fo

    below 20 F and down to -40 F.

    toward R410a. Even some, like Mokon and Frigel, that are focusing initially on R407c, believe that R410a could be the wave

    t users may experience energy savings (kw/ton) of anywhere from 2% to 5%. Sources at Advantage Engineering and Conair a

    ith R407c is about the same as for R22, but building a chiller for R410a currently costs an average of 5% to 10% (and sometim

    e reason is that in redesigning their equipment for new refrigerants, suppliers have simultaneously implemented a number of u

    e firms (as opposed to home refrigeration specialists) may be less familiar and comfortable with R410a and the higher pressure

    hermal Care to offer both R407c and R410a for their most popular chillers, though they are encouraging customers to choose

    rs, piping, seals, etc. Whats more, the alternative refrigerants are not compatible with the mineral-type compressor oils used w

    re being actively considered, and something new could pop up. In Europe, there is even interest in propane or butane for verynts besides their ozone-depletion potential (ODP), and some chiller suppliers think it could become the next front for U.S. envir

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    roduction of the old R22 refrigerant you have been using. Chiller suppliers are now talking about a welter of replacement refrige

    y the Clean Air Act of 1990. Since then, there has been a global phase-out of the most ozone-damaging refrigerants, solvents,

    you will be able to get additional R22 to service your existing chillers. However, supplies will steadily decrease, and the cost is e

    407c is found in many European-built chillers, since the European Union phased out R22 years ago. A number of domestic su

    cation (temperature range), compressor type, or equipment size. For example, some suppliers say only R134a is compatible wi

    r R22; all the others operate at higher pressures. Lower pressures mean that lower-horsepower compressors can be used, whi

    f the future when a wider range of high-pressure components become available and component prices ease with higher volu

    re not so sure, pointing out that slightly greater compressor efficiency is only one factor in the overall efficiency of a chiller syste

    es up to 15%) more, though costs may come down in the future. Some chiller suppliers, however, say they have not raised pric

    grades in compressors, condensers, and controls.

    s involved. R407c is more similar in pressure to R22, so more service firms may be equipped to handle it, at least initially. On t

    R410a.

    ith R22 and require synthetic oils instead. While it is theoretically possible to replace the oil, compressor, and other component

    small chillers up to 1 ton.onmental regulators. GWP is already said to be the primary driver in European refrigerant regulation. R22 and most of the curre

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    rants with designations like R410a, R407c, R134a, R404a, and R507. What does all this mean for you?

    nd foam blowing agentsfirst R12 in the 1980s and now R22, which has been the standard in plastics chillers. Under U.S. la

    xpected to rise accordingly. January 1 saw a 60% drop in the global cap on R22 production mandated by the Montreal Protocol

    ppliers have been using R134a in the very smallest chillers for years. Some of the major U.S. suppliers began introducing mod

    th central chillers based on oilless, centrifugal compressorswhich are sold by Thermal Care and Frigel.

    h reduces operating costs. Lower pressure and compressor horsepower with R134a are also the reasons why some suppliers

    es.

    m. Advocates also note that the efficiency of R410a allows use of smaller internal components and piping, which could enable

    es on chillers with R410a.

    e other side of the coin, it is sometimes pointed out that R407c is a mixture of three gases, while R410a is a more compatible

    especially if the switch is to R407cit may or may not be cost-effective to do so. That would depend on the size and techno

    nt substitutes have GWPs within a fairly similar range.

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    , no new chillers could be built to use R22 after Jan. 1, 2010.

    , and prices simultaneously jumped more than 20% higher. The global production cap will drop another 60% in 2015, and all R

    ls with R410a or R407c more than a year ago. They report that there have been no issues with these units in field.

    avor it for the largest central chillers as well. Components for building very large chillers with the highest-pressure refrigerant, R

    slightly more compact chiller design. Advantage sources again believe that overall chiller design is a bigger influence on footp

    ix of two gases. One may encounter arguments that if a chiller with R407c leaks some refrigerant, the lighter components esc

    logical age of the chiller. Mokon sources said it could be cost-effective to switch a smaller chiller to R407c. However, no suppli

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    2 production is scheduled to cease in 2020. After that, R22 will be available only from recovered and recycled stock.

    410a, are not yet readily available. R410a operates at 340 to 390 psi or higher, vs. 215 to 245 psi for R22.

    rint than slight differences in component sizes.

    pe first, thereby changing the composition of the remaining mix and its refrigeration performance. That allegedly means the lea

    er interviewed has received any requests to retrofit a unit for the new refrigerants.

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    king chiller could not simply be topped up with more R407cinstead, the whole refrigerant charge would have to be removed

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    and replaced, at higher cost. Because of the more compatible mix of ingredients, R410a reportedly is not susceptible to these i

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    ssues. Bu

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    3500

    2500

    4,7008,5007,000

    7,500

    1,500

    1

    Bladder Type

    Expansion Tank

    Glycol Chiller 250kW

    Entering Temp: 0C

    Leaving Temp: -10C R22

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    Plate Heat Exchanger 1

    Cooling Load 150kW;

    2,800

    1,500

    9,0008,500

    9,000

    2

    3

    Plate Heat Exchanger 2

    Cooling Load 100kW;

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    REFRIGERATION SCHEMATIC 001

    DWG

    Student No.

    267439818 REFRIGERATION PIPEWORK SCHEMATIC

    SHAWN TAYLOR 8 OCT 2012

    DRAWN BY DATE

    PROJECT

    UPRIGHT BOTTLE CABINEThawn Taylo

    DRAWING

    UNIT

    UEENEEJ032B

    COMMERCIAL

    SYSTEM DESIGN

    REFRIGERATION

    SIZE

    A4 ARS-001

    DWG NO. REVSCALE

    NTS

  • 7/27/2019 Fluid and Thermodynamic Parameters - Assignment 1

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    hawn TayloPROJECT

    UPRIGHT BOTTLE CABINET

    Student No. DRAWING

    267439818 REFRIGERATION PIPEWORK SCHEMATIC

    UNIT DRAWN BY DATE

    A4 NTS RS-001 A

    UEENEEJ032B SHAWN TAYLOR 8 OCT 2012

    COMMERCIAL DWG

    REFRIGERATION REFRIGERATION SCHEMATIC 001

    SYSTEM DESIGN SIZE SCALE DWG NO. REV