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Topic6-3 Pump And PipingWater and refrigerant piping
Heat-conveying media in refrigeration systems:air water refrigerants
Water-distribution systems:Provide flow rate to all heat exchangers
Refrigerant pipe size: Flow through refrigerant system
Pipe size design: safe and low first and operating costinfluenced by - pressure drop, p as D- noise as D- operating cost as D- first cost as D (duct cost 10% system cost)
- Oil entrainment to compressor, v as D- heat transfer, optimum D
Pipe sizing (1)
(2)
(3)
Comparison of water and air as heat-conveying media
Final heat transfer: always from/to air in the conditioned space
Source or Sink(Heater or refrigeration unit)
Conditioned space
Heat source: electric or fuel-fired furnace
Heat sink: refrigeration unitHeat pump: source and sink of energy are the same equipment
Water
Air
Figure 7-1 Concept of heat-conveying process in an air-conditioning system
Located in or remotely located from conditioned space
Air heated/cooled → delivered to the space
Water heated/cooled → heat/cool the air in the space
Advantages of water-over-air distribution arrangement:1. Smaller size of heat source2. Less space by water pipes than by air ducts3. Water pipes are easier to insulate than air ducts
Example7-1 A heat-transfer rate of 250 kW is to be effected through a change in medium temperature of 15C. What must the cross-sectional area be to convey this energy flow if (a) a water pipe is used and the water velocity is 1 m/s and (b) an air duct is used and the air velocity is 10 m/s.
1.Data: A = ? for Q , t if (a) water, Vw (b) air, Va 2.Assumption:
Incompressible flow, constant
4.Methods: Q = mcpt = VAcpt
5.Properties: approximated for refrigerationwater, = 1000 kg/m3, cp = 4.19 kJ/kg.Kair , = 1.2 kg/m3, cp = 1.0 kJ/kg.K
6.Calculation(a) Aw = Q/(wVwcp,wt) = 250/(1000*1*4.19*15) = 0.00398 m2
(b) Aa = Q/(aVacp,at) = 250/(1.2*10*1.0*15) = 1.389 m2
7. Analysis & check: Aa/Aw > 349, Da/Dw = 1.33/0.0712 > 18
Air used in small plants, i.e. residential and small commercialWater used in large plants, i.e. chilled-water in large air conditioning systems
Water heaters
Heat source: electric resistance heating or fuel-fired furnace [combustion of fuel (natural gas, coal, oil)]
Fuel-fire water heater: steel with safety code, classified by operating pressure, lowest-pressure -- 100C
Fire-tube water heater: combustion gas in tubes, water in shell around tubes
Efficiency = qsupply/LHVLower heating value: water vapor in flue gasApprox. 80%
Heater size may be chosen larger than max. design for night/weekend setback, larger piping size also.
Heat distribution from hot-water systems
Heat exchanger: coil in warm air ducts, fan-coil units, natural convection convectors
A baseboard convector:Hot water flows through tube,air being heated flows by natural convection up over the tube and find and out of the louvers A fan-coil units (FCU)
Fig.7-4 heating capacity of a certain baseboard convector, based on room air temperature at 18C.
Example7-2 What mean water temperature is needed in the baseboard convector of Fig.7-4 to compensate the heat loss from a single-pane window glass wall when the indoor and outdoor design air temperature are 21 and -23C, respectively? The height of the glass is 2.4 m, and the convectors are placed along the entire length of wall.
Data: tmean = ? for Qloss through glass h, ti ,to
Assume: steady process, troom 18C
Method: q/w = Uh(ti - to) W/mGraph Fig.7-4: q/w & tmean
Prop.: single-glass U = 6.2 m2.K/W [Table 4-4]
Cal.: q/w = 6.2*2.4(21 – (-23)) = 655 W/mGraph Fig.7-4: 655 W/m → tmean = 77 C
Ana.: baseboard convectors can handle the extreme load from single-glass window.
High temperature water systems
High-temperature water (HTW): supply water 180 - 230CSaturation pressure 1000 – 2800 kPaTo transfer energy with lower flow rates (high p), Q = mcT = mh
Steam generator as heaterPressurization: no vaporization in heaterFlashing in to vapor – in distribution system
Method of pressurization: High-pressure nitrogen with level control of N2 to keep pressure in system
Saturated water
Pipe Standard
ANSI – American National Standard InstituteAPI – American Petroleum InstituteASHRAE – American Society of Heating, Refrigerating and Air-Conditioning EngineersASTM - American Society of Testing and MaterialsAWWA – American Water Works AssociationFM – Factory Mutual Research Corporation: electrical device safety
ISO – International Organization for StandardizationNFPA – National Fire Protection AssociationNFS – National Sanitation Foundation : Drinking water pipe
UL – Underwriters Laboratories : Safety
Ref: Engineering Piping System Design, http://www.me.engr.tu.ac.th/pipebook.pdf
Pipe Standard
ASME - American Society of Mechanical EngineersASTM B88 – Seamless Copper Water TubeASME B31 series – pressure piping & vessel (ANSI B31)B31.1 - 2012 - Power PipingB31.4 - 2012 - Pipeline Transportation Systems for Liquid HydrocarbonsB31.5 - 2013 - Refrigeration Piping and Heat Transfer Components
The Engineering Institute of Thailand Under H.M. The King’s Patronage (วิศวกรรมสถานแห่งประเทศไทยในพระบรมราชปูถมัภ,์ ว.ส.ท. (พ.ศ. 2486))สภาวิศวกร พระราชบญัญตัิวิชาชีพวิศวกรรม พ.ศ. 2505 → พระราชบญัญตัิวิศวกร พ.ศ. 2542ACAT – Air-Conditioning Engineering Association of Thailand (สมาคมวิศวกรรมปรบัอากาศแห่งประเทศไทย (พ.ศ. 2536))TSME - Thai Society of Mechanical Engineers (สมาคมวิศวกรเครือ่งกลไทย (พ.ศ. 2550))JSME - Japanese Society of Mechanical Engineers
Ref: The Engineering ToolBox, https://www.engineeringtoolbox.com/
Pipe Sizes
Iron Pipe Size (IPS) - an old pipe sizing system, early 1900; USA, UK- Pipe size: inside diameters of pipes (ID),i.e. IPS 6 = ID 6 in- Still used in some industries, and PVC manufacturers
In 1920s, Copper Tube Size (CTS) combined IPS standards- Pipe size: OD & wall thickness (STD, XS, XXS)STD - Standard, XS - Extra Strong, XXS - Double Extra Strong
In 1948, Nominal Pipe Size (NPS)[in], Diameter Nominal (DN)[mm]
→ NPS average between OD & ID for steel pipe
thicknesses of Steel pipe by Schedule (SCH), SCH40 = STD [ASTM}Schedule number 1000(Pressure)/(yield Stress)thicknesses of Steel pipe by Class A, B ,C (thick) [British Standard]thicknesses of Copper tube by Type K, L
NPS = 1” (DN = 25 mm)OD = 33.4 mm
SCH40 = STDID = 26.64 mm
SCH80 = XSID = 24.30 mm
SCH160ID = 20.70 mm
XXSID = 15.21 mm
3.4 4.5 6.7thick = 9.2
Available pipe and tubing Most common in refrigeration: Type L copper tubing – by ODSchedule 40 steel pipe – by NPS or DN
Sch40 NPS < 14” (356 mm) → NPS < ID < ODSch80 NPS < 14” → ID < NPS < ODNPS 14” → NPS (or DN) = OD
NPS2= DN50
DN = ODDN = NPS
Thickness of Type K > Type L
SCH5s, 10s, 40s, 80s for stainless steel
Ref: นายช่างมาแชร,์ https://naichangmashare.com/2020/03/04
NPS2 = DN50
Ref: Garth Denison, refrigerant-piping-handbook.pdf
Pipe and tube Materials’ properties
Ref: Engineering Piping System Design, http://www.me.engr.tu.ac.th/pipebook.pdf
Copper tubing with high k → Freon, CH3Cl, SO2 , CO2
steel pipe → NH3 for D > 4” (>100 mm)
Copper tubing
- Seamless Annealed (อ่อน), Drawn (แข็ง)- Dehydrated at mill (ก าจดัความชืน้ขณะรดี)- DN6 – DN150, seldom found for > D150- ผลิตแบบท่อปลายเรยีบ ปิดหวัทา้ยกนัความชืน้เขา้
Ref: Engineering Piping System Design, http://www.me.engr.tu.ac.th/pipebook.pdf
- Smooth surface, high k- Difficult to fouling- High density, high cost
- Joint: Flared type (บานปลายท่อ) D < ¾”, then Screwed (เกลียว)
- Joint: Blazing, solder or sweated capillary
Copper tubing
ASTM B280, type ACR (Air conditioning and refrigeration)1. ท่อทองแดงชนิดแข็ง (Hard Drawn Copper Tube)A/C > 60,000 Btu/h (>5 ton) แข็งแรง ทนแรงกระแทก ใชภ้ายนอกอาคาร2. ท่อทองแดงชนิดมว้น (Soft Drawn Copper Tube)A/C < 60,000 Btu/h (<5 ton) งานเดินท่อไม่ไกล ยบุตวัจากแรงกระแทกง่าย จดุยึดแขวนถ่ีขึน้ (hanger/support)
ASTM B819 – ท่อมว้น งานก๊าซทางการแพทย์ ท าความสะอาดจากโรงงานและมีจกุครอบหวั-ทา้ยASTM B88 – ท่อตรงยาว 6 m (20 ft) ท่อน า้ใช ้ระบบก๊าซ และระบบสญุญากาศ ท่ีไม่ตอ้งการความสะอาดเป็นพิเศษ งานเดินท่อ ระบบความเย็นและความรอ้นType M: บาง นิยมใชใ้นระบบท่อน า้รอ้น ระบบปรบัอากาศ การประปาType L: หนาปานกลาง ใชใ้นระบบท าความเย็น งานก๊าซทางการแพทย ์ระบบปรบัอากาศType K: หนา ใชใ้นอตุสาหกรรมท่ีทนแรงดนัสงู ระบบก๊าซอตุสาหกรรม ระบบน า้ยา
ASTM B75 – ท่อเลก็ DN6 ถึง DN50 (Capillary tubing) ต่อท่อจากอปุกรณต์่างๆ
Soft annealed copper tubing – ท่อเลก็มว้น ยาว 9-18 m (30-60 ft)
Steel Pipes ทอ่เหลก็เหนียวSeamless steel – by extrude
Welded steel – with seamBlack steel pipe ท่อเหลก็ด าGalvanized steel pipe ท่อเหลก็ชบุสงักะสีWrought iron +silicate, stainless, high cost
Stainless steel pipe +Cr Ni Mo, high cost
Cast iron pipe - C > 2%, water system
ASTM – A53Type F - Furnace-butt-welded, Grade A.Type E - Electric-resistance-welded, Grades A and BType S - Seamless, Grades A and B
NPS5 or DN125, SCH40 – ท่อยาว 6 m (20 ft) → Freon, CH3Cl, SO2 , CO2
Steel Pipe > 5” (>125 mm) → NH3
SCH40 – working fluid pressure < 250 psi (1500 kPa)For thickness < SCH40 → ท าเกลียวท่ีปลายไม่ได้
Steel pipe: > -50C (-60F)+ Ni or/and Cr : < -50C
Joints and Fitting, Hanger and support
Screwed: D < 3”, 250 psi
ASTM – B1.20.1 NPT National Pipe Thread Taper
NPT with Union
Flanged: D > 2” Butt-weld: D > 2”
Sleeve, Socket: D < 2”
Flared type: soft tube/pipe, D < 1”
Hanger and support
ท่อระบบความเยน็ – เดนิทอ่ใหต้รง ยดึใหแ้น่น ไมเ่กดิแรงเคน้ทีร่อยต่อทอ่– การยดืหดตวัทีเ่กดิจากอุณหภมู ิมน้ีอยกวา่ ทอ่ไอน ้า– ทอ่ทองแดงทนการโคง้งอหรอืสัน่สะเทอืนดกีวา่ทอ่เหลก็– มคีวามลาดลงไปตามทศิทางน ้ายาเขา้ compressor ประมาณ 1:100
จดุยึด – แบบแขวน รดั หนีบ – ผา่นโครงสรา้งตอ้งใสป่ลอก (sleeve)– หา่งกนัไมเ่กดิ 40*OD ตามกฎหมาย– นิยมยดึทุก 3 m ยดึทุกจุดทีเ่ปลีย่นทศิทาง ยดึทีว่าลว์และอุปกรณ์อื่นๆ
Noise and Vibration
Caused by1. Compressors and motors2. Opening/closing of suction and discharge valves3. Flow impact to valves and cylinders 4. Flow with high velocity and turbulence
Protect by1. Balancing of compressors and motors2. Installation spring and elastic anti-vibration
3. Short pipe – curved pipe to support vibrationLarge pipe – connected with soft/flexible pipe
4. Discharge muffler – ทีเ่กบ็เสยีงดา้นล่าง compressor5. Pipe, under vibration, hanged with spring/rubber hangers
Pressure drop of water flowing in pipe
Pressure drop in straight pipe: p = f(L/D)(V2/2) [Darcy-Weisbach equation]
in fitting pipe: p = K(V2/2) f(Leq/D)(V2/2), K f(Leq/D)Engineering calculation [D, p]: SI [mm, Pa/mm.wg], IP [in, psig/in.wg]
Example7-3 Compute the pressure drop when 3.0 L/s of 80C water flows through a steel pipe with a nominal diameter of 50 mm (ID = 52.5 mm) that is 40 m long.
Data: p = ? Water (Q, t) flows through pipe (ND, ID, L)
Assume: steady flow
Method: p = f(L/ID)(V2/2)f = f(Re,/D) Moody chart or formula:f = 0.25[log{/3.7ID+5.74/Re0.9}]-2
Re = VID/, V = Q/A, A = D2/4
Prop.: water@80C, = 971.63 kg/m3, = 0.358E-3 Pa.s, steel = 0.000046 m [Table 6.1]
Cal.: V = 3*10-3*4/(*52.52*10-6)
= 1.386 m/sRe = 197500/ID = 0.00088f = F(/D,Re) = 0.0208p = f(40/52.5)(V2/2)= 14.8 kPa
Ex7-4 Use Fig.7-6 and 7-7 to solve Ex7-3.
Assume: Schedule 40 steel pipe
Cal.: p = (p/L )*L*CFFig.7-6, t = 20 C : ND = 50 mm, Q = 3 L/s → p/L = 425 Pa/m
Fig.7-7 : t = 80 C , V = 1.386 1.4 m/s→ CF = 0.885
p = 425*40*0.885= 15.1 kPa
Head loss of water flowing in pipe at 20C
hf = (L/1000)(151Q/CD2.63)1.85 [Hazen-William equation]Type of tube roughness: (mm) : C
Ex7-3 p = ? Water (Q=3L/s, t=80C) flows through a steel pipe (ND=50, ID=52.5, L=40m)
Assume: steady flow
Method: p = ghf
hf = (L/1000)(151Q/CD2.63)1.85
Prop.: = 972 kg/m3, C = 120Fig.7-7 : t = 80 C, V 1.4 m/s→ CF = 0.885
Cal.: hf = (40*10-3)(151[3*10-3]/(120*0.05252.63)1.85
= 2.22 mp = 972*9.81*2.22(0.885) = 18.7 kPa
1. Darcy-Weisbach equation: p = 14.8 kPa2. Graph of steel pipe SCH40: p = 15.1 kPa (3%)3. Hazen-William equation: p = 18.7 kPa (27%)
Pressure drop in fittings
Fitting: change in area and directionEx: elbows, tee, open valves, etc.
Pressure drop in fitting equivalent length of straight ductin fitting pipe: p = K(V2/2) f(Leq/D)(V2/2), K f(Leq/D)
ตย1.
Pressure drop in fittings
Ref: Garth Denison, refrigerant-piping-handbook.pdf
ploss by enlargement (v) > by reduction/contraction (v) up to 2 times
ตย1.
Table Le Fitting in ft
Pressure drop in Accessories
- Filter-drier = 6 psi (41.4 kPa)- Solenoid valve = 4 psi (27.6 kPa)- Sight glass = 1 ft (0.3 m)
Refrigerant piping requirements
1. Return oil to compressor at adequate velocity at all steps of unloading
2. Ensure that only liquid refrigerant enter the expansion device3. Minimize system capacity and efficiency loss4. Minimize refrigerant charge5. Insulate line that are route through very warm spaces (Liquid line)6. Avoid excessive noise
General piping requirements1. Use clean type L copper tubing
- copper-to-copper joints: BCuP-6 without flux- copper-to-steel (or brass) joints: Bag-28, non-acid flux
2. Properly support piping to account for expansion, vibration, and weight
3. Avoid installing piping underground4. Test entire refrigerant circuit to leak
Sizing Rules
Suction line Discharge line Liquid line
Sizing Process: Suction line
1. Determine total length of piping2. Calculate refrigerant velocity at maximum and minimum capacities3. Select largest pipe diameter that results in acceptable velocity at both maximum and minimum capacities4. Calculate total equivalent length of straight pipe and fittings5. Determine pressure drop due to pipe and fittings6. Add pressure drop due to accessories
Oil trap : Suction line
U-trap
S-trap
Goose-neck
Evaporator below compressorRising suction line
Evaporator above compressor
Suction riser 1-5/8”For V > Vmin
Suction drop 2-1/8”
Oil trap every 5 m
Double Risers for part-load Reduce capacity→ velocity is too lowSolve by replacing a single Riser 1-5/8”by double risers 1-3/8” + 7/8”Same area, but 7/8” give 3.6*velocity of 1-5/8”
a single Riser 1-5/8”
double risers 1-3/8” + 7/8”
Suction line piping
Single suction line piping Multiple suction line piping
Discharge line
Discharge line oil traps every 3 m
Liquid line piping
Hot gas Bypass to Evaporator
Refrigerant piping
Discharge line: gas, p is a penalty on compressor power
Condenser
Evaporator
Expansion device
Compressor
Discharge gas
Suction gas
Liquid
Liquid line: liquid, smaller diameter, more p → liquid flashes in to vapor → expansion device will not work properly
Suction line: gas, p is a penalty on efficiency
Liquid line – oil miscible and back with liquid flow
Velocities in vertical suction line > 5 m/s to carry lubricating oil from evap. back to comp. (oil separated from vapor )
Ammonia is not miscible with oil – An oil separator at discharge
1
2
3
4
Liquid line: liquid, smaller diameter, more pdrop → more liquid flashes into vapor → expansion device will not work properly
Standard cycle
Pre
ssu
re,
kPa
1
23
4
Condensation
Exp
ansi
on
Evaporation
3a
4a
Entropy, kJ/kgK
Subcooling (3 – 3a) – 1. increases the refrigeration effect ( RE)
3. less vapor at the inlet to the evaporator → lower pressure drop in the evaporator
2. only liquid enters into the throttling device → efficient operation
Vapor flows with higher velocity than that of liquid
RE, h4-h4a
Standard cycle
Pre
ssu
re,
kPa
1
23
4
Condensation
Exp
ansi
on
Evaporation
3a
4a
Entropy, kJ/kgK
Superheating (1’ – 1a) – 1. increases both RE and Wc, COP = RE/Wc
2. higher discharge temperature, t2’
1a1’
Pressure drop
Suction line: gas, pdrop is a penalty on efficiency, 1. reduce ps to comp.
As ps → vsuc → mass flow (constant suction volume)
2’
2. prevents the entry of liquid droplets into the compressor.
or COP depends on nature of refrigerant
RE , h1a-h1
Wcomp , h2’-h2
T2’>T2
Standard cycle
Pre
ssu
re,
kPa
1
23
4
Condensation
Exp
ansi
on
Evaporation
3a
4a
Entropy, kJ/kgK
Discharge line: gas, pdrop is a penalty on 1. compressor power
1a1’
Pressure drop
Pressure drop
2a
Wcomp , h2a-h2
2. higher discharge temperature, t2a
T2a>T2
Syst
em p
ress
ure
System pressure: actual pressure change including the effects of pressure drop
Evaporatormanufacturer
TXV
liquid line
suction linecompressor
discharge linemanufacturercondenser
Liquid-Suction Heat exchanger (LSHX)
41
Subcool liquid from condenser with suction gas from evaporator h3 – h4 = h1 – h6
cl(t3 – t4) = cv(t1 – t6)
41
Pre
ssu
re,
kPa
Enthalpy, kJ/kg
1
2
5C = Te
50C = Tc4
5 8C Tsuperh
T=13C
3C Tsubc
T=47C
6
3
Since cl > cv
(tc – t4) < (t1 – te)tsubcool < tsuperheat
Actual VCRS systems
Irreversibilities1. Pressure drops (pdrop) in evaporator, condenser and LSHX 2. pdrop across suction and discharge valves of the compressor 3. Heat transfer in compressor 4. pdrop and heat transfer in connecting pipe lines
COPactual = cycleisentropicmotorCOPcarnot
Actual VCRS systems Process1. Pressure drop in evaporator 2. Superheat of vapour in evaporator 3. Useless superheat (heat gain from surrounding) in suction line 4. Suction line pressure drop
5. Pressure drop across suction valve6. Non-isentropic compression 7. Pressure drop across discharge valve8. Pressure drop in the delivery line 9. Desuperheating of vapour in delivery pipe 10. Pressure drop in the condenser 11. Subcooling of liquid refrigerant 12. Heat gain in liquid line
4 -1d1d -1c
1c -1b1b -1a
1a -11 -2
2 -2a2a -2b
2b -2c2c -33 -3a
3a -3b
44
Discharge line: vapor line
Liquid line: liquid
Suction line: vapor line
Standard cycle
4. Actual vapor-compression refrigeration cycle
2-2a → Pressure drop across discharge valve
2a-2b → Pressure drop in the delivery line
→ wc (compressor work)→ Tdischarge – compressor life
1a-1 → Pressure drop across suction valve
1-2 → Non-isentropic compression
suction pressure (ps) → specific volume → compressor volumetric efficiency (v)→ wc
→ wc
→ liquid flashes into vapor qe wc
→ expansion device will not work properly
4-1d → pressure drop in evaporator
1d-1c → Superheat of vapor in evaporator
1c-1b → Useless superheat in suction line 1b-1a → Suction line pressure drop
3-3a → Subcooling of liquid refrigerant
3a-3b → Heat gain in liquid line
→ efficient operation→ qe
2b-2c → Desuperheating of vapor in delivery pipe
2b-3 → Pressure drop in the condenser
→ wc but smaller than that in vapor line
need cooling
Pipe design for small pdrop
Reduce pdrop by velocity flow or pipe size→ heat transfer coefficient in evaporator*minimum velocity is required to carry lubricating oil back to compressor
Pressure drop & sat. temp. difference → Power, COP
Ex: Suction line Te = 4C (40F)Refrigerant T,C (F) p, kPa (psi)R12 1 (2) 11.2 (1.7)R22 1 (2) 18.4 (2.7)
R502 1 (2) 19.7 (2.9)R717 0.55 (1) 9.8 (1.5)
Ex: Liquid line Tc = 40C (104F)Refrigerant T,C (F) p, kPa (psi)R12 0.526 (1) 12.4 (1.8)R22 0.526 (1) 19.9 (2.9)
R502 0.526 (1) 21.4 (3.1)R717 0.526 (1) 22.7 (3.3)
Example: R12 Suction line Te = 4C (40F)Liquid line Tc = 40C (104F)Suction LineT,C RE, % Power/RE, %
0 100 1001 96.5 102.8
2 92.9 106.3
Discharge LineT,C RE, % 1/COP, %
0 100 1001 99.5 102.3
2 98.5 104.5
Suction line: gas, p is a penalty on efficiency→ps, vsuc, v, wc
Discharge line: gas, p is a penalty on compressor → wc, Tdischarge
Refrigerant Line Design: Oil entrainment
Because oil separated from vapor To carry lubricating oil from evap. back to comp- Velocities in vertical suction line > 5 m/s (> 1000 fpm)- Velocities in horizontal suction line > 2.5 m/s (> 500 fpm)
If Vel > 20 m/s (> 4000 fpm) → noise in pipe and friction/p
RE = (h1-h4), kJ/kg m = mass flow rate. kg/sQ = refrigeration capacity, kWVol = Volume flow rate, m3/sVel = velocity, m/s
Suction line:m = Qe/(h1-h4) Vol = mv1
Vel = Vol/(D2/4)
1. Maximum D Q0.5
2. Maximum D v1 , 1/(h1-h4)
→ D2 =[4v1/((h1h4)*Vel)]Qe
→ Qe= [(h1-h4)*Vel/(4v1)]D2
1. Minimum Qe D2
2. Minimum Qe Vel3. Minimum Qe 1/v1 , (h1-h4)
Discharge line → Qe= [(h1-h4)*Vel/(4v2)]D2
Min. Qe for Oil flow up Suction, superheat 8C, Liquid Tc = 32C
Correction factor for Liquid Tc 32C
Qe= [(h1h4)*Vel/(4v1)]D
ตัวอย่าง1 Suction sizing R22, Qe = 35 kW, Te = 4C,
Tc = 49C, T3’ = 38C
ตย1
Min. Qe for Oil flow up Discharge, Te = -7C, SH 8C, subcool 8C
Correction factor for Te -7C
Qe= [(h1-h4)*Vel/(4v2)]D
Refrigerant Line Design: Friction/p
total length of friction loss of suction line from evaporator to compressor:
T = 1C (2F)T = 0.5C for NH3
p = 10-20 kPa (1-2 psi) Suction line:m = Qe/(h1-h4) Vol = mv1
Vel = Vol/(D2/4)pD/L = Const.Vel1.8
→ pD/L =Const.[4Qev1/(D2(h1-h4)*Vel)]1.8
for constant v1, h1, h4 in same pipe D = constant
pD/L Qe1.8 or Qe (pD/L)0.556
T = LQe1.8 or Qe (T/L)0.556
Discharge line → Qe (pD/L)0.556
pD/L =Const.[4Qev2/(D2(h1-h4)*Vel)]1.8
total length of friction loss of discharge line from compressor to condenser:
T = 0.5C (1F) p = 20-30 kPa (3-4 psi)
total length of friction loss of liquid line from condenser to expansion device:
T = 1C (2F)T = 0.5C for NH3
p = 10-20 kPa (1-2 psi)
Min. Qe for L = 30 m, T=1C , Liquid line Tc = 41C
Qe (pD/L)0.556 or Qe (T/L)0.556
Min. Qe for L = 30 m, T=1C , Liquid line Tc = 41C
Correction factor for Tc 41C
Qe (pD/L)0.556 or Qe (T/L)0.556
For other T and Le
T = Qe,table[(30/Le)(Tactual/Ttable)]0.556
For other Qe and Le
T = Ttable(Le/30)[Qe,actual/Qe,table]1.8
ตัวอย่าง1 Suction sizing R22, Qe = 35
kW, Te = 4C, Tc = 49C, T3’ = 38C
ตัวอย่าง1 จงหาขนาดทอ่ดดูขึน้ตามแนวดิ่ง (suction riser) โดยใหน้ า้มนักลบัไดดี้ และ ความเสียดทานในทอ่ดดูไมเ่กินมาตรฐาน ส าหรบัทอ่ยาว 15 เมตร ขอ้งอ 6 อนั, R22, Qe = 35 kW, Te = 4C, Tc = 49C, Tc, liquid = 38C
Data: suction D = ? For R22, Qe, Te, Tc, T3’
Assumption: suction riser, friction for T=1 K
Methods: Table 24 → 35.1 kW, Te = 4CChoose copper 1-3/8” (34mm) OD Type LTc = 49C → CF = 0.9 → Qe,table = 0.9*5.1 = 31.6Table Le Fitting in ft; elbow Le,el = 3.6 ft = 1.1 mTable 7-4; elbow Le,el = 1.2 m (chosen)→ Total Le = 15 + 6(1.2) = 22.2 m
Check T = Ttable(Le/30)[Qe,actual/Qe,table]1.8 < 1 KT = 1(22.2/30)[35/31.6]1.8 = 0.9 K < 1 K friction
Table 22 Oil entrainment up suction riser→ R22, Te = 4C, 1-3/8”OD → Qe,min = 9.496 kWTc, liquid = 38C → CF = 0.95 → Qe,min = 9.949*0.95 = 9 kW < 35 kW oil
Refrigerant Piping Nomograph’s
To get a proper pipe sizing of Dsuction line , Dliquid line
Known data to use a refrigerant nomograph:1. The system refrigerant type (Example R-22)2. System design capacity (Qe = 17 tons)3. Saturated Suction Temperature (SST) (Te = 0° F)4. Saturated Condensing Temperature (SCT) (Tc = 120° F)5a. Minimum allowable velocity (Vsuction line = 2000 fpm, Vliquid line = 200 fpm)5b. Maximum allowable pressure drop (psuction line = 2 psi, pliquid line = 4 psi)
Nomograph’s
Velocity
Example1. R-222. Qe = 17 tons3. Te = 0° F4. Tc = 120° F)5a. Vsuction line = 2000 fpm
Vliquid line = 200 fpm1
7 t
on
s
Vsu
c=
20
00
fp
m
Vliq
= 2
00
fp
m
Velocity in feet/minute
Dsuc = 7/8”Dliq = 1/2”
Nomograph’s
Pressure
Example1. R-222. Qe = 20 tons3. Te = -20° F4. Tc = 100° F)5b. psuction line < 1.5 psi/100 ft
pliquid line < 7.5 psi/100 ft
pliq
< 7
.5 p
si/1
00
ft
p
suc
< 1
.5 p
si/1
00
ft
Pressure drop in psi per 100 feet
Dsuc = 1-5/8”Dliq = 5/8”
Water piping
Figure 3.10 Friction Loss for Water in Commercial Steel Pipe (Schedule 40) [ASHRAE 2017F, CH 22, Fig 14]
Water flow for chiller, water cooling, etc.Velocity 200-600 fpm (1-3 m/s)Pressure loss in steel pipe 4 psi/100 ft (900 Pa/m)
Ref: ASHRAE Pocket Guide for Air Conditioning, Heating, Ventilation, Refrigeration, SI, 9th Ed.
Example: Steel pipeAt Q = 6 L/s p/L = 900 Pa/mD = 67 mm (2-5/8”)V = 2.2 m/s (433 fpm)
Figure 3.8 Friction Loss for Water in Copper Tubing (Types K, L, M)
Figure 3.9 Friction Loss for Water in Plastic Pipe (Schedule 80)
Example: Copper tubeAt Q = 6 L/s p/L = 1000 Pa/mD = 68 mm (2-5/8”)V = 2.5 m/s (492 fpm)
Example: Plastic pipeAt Q = 6 L/s p/L = 150 Pa/mD = 87 mm (3-1/2”)V = 1.2 m/s (236 fpm)
Pump characteristics and selection
Pump data: pressure differences at various flow ratesPressure rise (p) & Flow rate (Q) of pump curve on isoefficiency line
Pideal = wvdp = wv(p2 – p1) = Q(p2 – p1)Pactual = Pideal / = Q(p2 – p1)/ Ex7-5 Using the efficiency curve shown
for the pump in Fig.7-9, compute the power required by the pump when the water flow rate is 6 L/s.
Method: Pa = Q(p2 – p1)/ = Qp/
Pump curve: at Q → p,
Prop.:Pump curve: at Q = 6 L/s →p = 240 kPa, = 0.78
Cal.: Pa = 0.006*240/0.78 = 1846 W
Not in catalog, for understanding only
Pump curve: Pdata 1900 W
Ref: ASHRAE Pocket Guide for Air Conditioning, Heating, Ventilation, Refrigeration, SI, 9th Ed.
Pump curve 180 mm: At Q = 6 L/s p = 135 kPa = 0.675P = 1.6 kW
Pa = Qp/
= 1.2 kW
Pump characteristics & System curve
Pump curves & System curve (pipe characteristics)
pipe: p = f(L/D)(V2/2) → p CpfQ2
Cpf : friction on pipes in the systems, depends on pipe design
Pump curves & System curve @balance point
Valve control:
Pipe, not throttled = fully open valve
Design of water distribution systems
Design process: determination of sizes and arrangement of water system components (straight pipe, valves, fittings, pumps, heat exchanger, expansion tank)
Direct return
Reversed return
Design process:
1. decide on component’s location
2. Select pipe size
3. Select pump
4. Choose expansion tank
Direct return: nonuniform p in HX
pA > pD;
1.control valveA –nearly closed
2.pD– insufficient to provide flow rate
Reversed return: uniform p in HX
But additional pipe
Sequence of water distribution systems
Heater
C.Expansion
tankConst. p
Pump
Relative location of heater, expansion tank and pump Low p at pump inlet
High p at pump outlet
Constant p at expansion tank
A. pump - Exp. Tank
→ Inlet p drop → cavitation
A.Expansion
tankConst. p
“Pump away from Exp. tank”
B. If tank – pump - heater
→ high p enter heater → open relief valve
B. Heater
Usual sequence:
C. heater - tank – pump
Cavitation : water flashing into vapor at low p → poor performance and accelerated wear
Pump pinlet > psat, NPSH
Net pressure suction head
Low pHigh p
NPSH: Net Positive-Suction Head
63
Head required at the pump inlet to keep from cavitating.
g
p
g
V
g
pNPSH vii
R
−+=2
2
If the pump inlet is placed at a height zi above a reservoir where free surface is at pressure pa.
g
phz
g
pNPSH v
iLossia
A
−−−= ,
-zi
+zig
phz
g
pNPSH v
iLossia
A
−−+= ,
System characteristics & Pump selection
64
g
VK
D
fLzzhzzH Losss
2)(
2
1212 ++−=+−=
bQaHs +=
2bQaH s +=
laminar
turbulent
Pump Selection Regions [ASHRAE 2016S, CH 44, Fig 35]
Sizing and expansion tank
Expansion tank: cushion of air in water as volume change by t Expansion tank size: Vtank = {v/vc}Vsystem/[(pi/pc)-(pi/phot)]
VB - VC = Vsv/vc
Const. T, air as ideal gas:
1 /[(pi/pc)-(pi/ph)]
= 1/[(VB/Vt)-(VC/Vt)]
= Vt/[VB-VC]
Ex7-6 What is the size of an expansion tank for a hot-water system with a volume of 7.6 m3 if the highest point in the system is 12 m above the expansion tank, the system is filled with 20C water, the operating temperature is to be 90C, and the maximum pressure in the system is to be 250 kPa gauge?
Data: Vt = ? Hot-water system Vs, static headc, tc, th, ph
Method: Vt = {v/vc}Vs/[(pi/pc)-(pi/ph)]
v = vh – vc = vliq@90C – vliq@20C
pi,abs = ghc +patm , ph,abs = ph +patm
Prop: water properties vliq@90C , vliq@20C
Assume: patm = 101 kPa
Problem7-3 In the piping system shown below the common pipe has a nominal 75 mm diameter, the lower branch has 35 mm, and the upper branch 50 mm. The pressure of water at the entrance is 50 kPa above atmospheric pressure, and both branches discharge to atmospheric pressure. The water temperature is 20C. What is the water flow rate in liters per second in each branch?
Data: pipe diagram, pipe size , pentry, pdis, tw , Qm1, Qb2, Qb3 = ?
Assume: steady flow, no heat, no work done
Met: energy balance, p1+V21/2+gz1= p2+V2
2/2+gz2+ploss,1-2
Same level, z1 = z2 = z3, p1+V2
1/2 = p2+V22/2+ploss,1+ploss,2 --Eq(1)
p1+V21/2 = p3+V2
3/2+ploss,1+ploss,3 --Eq(2)mass balance, Q1 = Q2 + Q3 --Eq(3)
Main, L1 = 8+4+5+7+15=39 mNominal D = 75 mmFour 90 elbow
Straight branch, L2= 30 m, Nominal D = 50 mm
Side branch, L3 = 6+18 = 24 m, Nominal D = 35 mmOne 90 elbow
Problem7-3: pipe diagram, pipe size , pentry, pdis, tw , Qm1, Qb2, Qb3 = ?
Calculation; L1t =L1+4*Lelbow1=39+4*3=51 mL2t =L2+Lstraight branch=30+0.9= 30.9 mL3t =L3+Lelbow3 +Lside branch =24+1.2+4.6= 29.8 m
Eq(1); p1 - p2 = (V22-V2
1)/2 + ploss,1+ploss,2
Eq(2); p1 – p3 = (V23-V2
1)/2 + ploss,1+ploss,3
Bcaz; p1 - p2 = p1 – p3 = p = 50 kPaEq(3); Q = AV, A=D2/4, D2
1V1 = D22V2 + D2
3V3
ploss = f (L/D)(V2/2) = F(D, V)f = 0.25[log{/3.7D+5.74/Re0.9}]-2
Re = VD/
f = 0.25[log{/3.7D+5.74(/VD)0.9}]-2
F1(V1,V2,V3) = p1 – p2 – (V22-V2
1)/2000 – f1(L1t/D1)(V2
1/2000) – f2(L2t/D2)(V22/2000)
F2(V1,V2,V3) = p1 – p3 – (V23-V2
1)/2000 – f1(L1t/D1)(V2
1/2000) – f3(L3t/D3)(V23/2000)
F3(V1,V2,V3) = D21V1 – D2
2V2 – D23V3
F3 = V1 – (D2/D1)2V2 – (D3/D1)2V3
Prop: [Table 7-2] Schedule 4075 mm D, D1 = 77.92 mm50 mm D , D2 = 52.51 mm35 mm D, D1 = 35.04 mm[Table 7-3] water 20C, = 998.2 kg/m3, = 1.008 mPa.s[Table 7-4] Leq of straight pipe and fittingsLeq,elbow,1 = 3.0 m [75 mm D]Leq,elbow,3 = 1.2 m [35 mm D]Leq,straight branch,2 = 0.9 m [75 mm D]Leq,side branch,3 = 4.6 m [75 mm D]Steel, = 0.000046 m [Table 6.1]
Properties: [Table 7-2] Schedule 4075 mm D, D1 = 77.92 mm50 mm D , D2 = 52.51 mm35 mm D, D1 = 35.04 mm[Table 7-3] water 20C, = 998.2 kg/m3, = 1.008 mPa.s[Table 7-4] Leq of straight pipe and fittingsLeq,elbow,1 = 3.0 m [75 mm D]Leq,elbow,3 = 1.2 m [35 mm D]Leq,straight branch,2 = 0.9 m [75 mm D]Leq,side branch,3 = 4.6 m [75 mm D]
Problem7-3: pipe diagram, pipe size , pentry, pdis, tw , Qm1, Qb2, Qb3 = ?
F1 = 50 – 0.4991(V22-V2
1) – 81.67[log(0.00016+0.0002292V0.91)]-2V2
1
–73.43[log(0.00024+0.000327V0.92)]-2V2
2
F1 = 50 – 0.4991(V22-V2
1) – 81.67[log(.00016+0.0002292V0.91)]-2V2
1
–106.1[log(0.00035+0.0004706V0.93)]-2V2
3
F3 = V1 – 0.4541V2 – 0.2022V3
Excel Solver (Iteration process); f1, f2, f3 = 0.02055, 0.0215, 0.02435Re1, Re2, Re3 = 110808, 121478, 64283 > 10000 → turbulent flowV1, V2, V3 = 1.436, 2.336, 1.852 m/s, Q = 1000VD2/4 L/s; Q1, Q2, Q3 = 6.848, 5.059, 1.786 L/s
For easier calculation, by hand calculator1. Assume (V2
2-V21)/2 << ploss,1+ploss,2 and (V2
3-V21)/2 << ploss,1+ploss,3
Then Eq(1) = Eq(1); ploss,2 = ploss,3
2. Assume f1 = f2 = f3 0.02 roughly, ploss = 0.02 (L/D)(V2/2)Q1, Q2, Q3 = 7.25, 5.29, 1.96 L/sError: Q1, Q2, Q3 = 5.8%, 4.6%, 9.7%
