Evaluation and Development of Air-conditioners using Low GWP Refrigerant

Transcription

1 th IEA Heat Pump Conference 07 Rotterdam Evaluation and Development of Air-conditioners using Low GWP Refrigerant Shigeharu TAIRA, Tomoyuki HAIKAWA *, Tomoatsu MINAMIDA DAIKIN INDUSTRIES, LTD. 6th, May, 07

2 Background Introduction of R3 reduced the GWP to one third of R40A. Important to reduce GWP as much as possible. Transition History of Refrigerant Montreal Protocol Kyoto Protocol CFC HCFC HFC40A HFC3 Next? ODP.0 GWP 0, ,80 Save Ozone Layer 0, Mitgate Global Warming IPCC AR4

3 Properties of refrigerants () Need another parameter to distinguish which refrigerant is better. Refrigerant R3 R40A =R3/R5 (50/50wt%) even -ze blend =R3 /R34ze(E) (70/30wt%) 3 worse -yf blend =R3/R5 /R34yf (67/7/6wt%) Global Warming Potential: GWP (AR4) < Temperature Glide: ΔT GL Discharge / Suction Pressure: P d / P s [MPa abs] /.07 /.087 /0.99 /.039 Discharge Temperature T d [ ] Refrigerating effect w r [kj/kg] 48 (00%) 64 (66%) 0 (85%) 9 (78%) Compressor Work: w s [kj/kg] 54.0 (00%) 36.5 (68%) 45.4 (84%) 4. (78%) Specific Volume in Suction v s [m 3 /kg] (00%) Coefficient of Performance: 4.59 COP = w r / w s (00%) (7%) 4.49 (97.8%) (0%) 4.63 (00.8%) (87%) 4.56 (99.%) Volume Capacity f = w r / v s [kj/ m 3 ] 78(00%) 665(9%) 609(83%) 648(90%) Calculation Conditions: T c = 45, T e =0, Suction line Temp.: T s =5, Condenser Outlet: T c.out =40, Compressor Adiabatic Efficiency: η comp =70%, in Cooling Operation.

4 4 Calculation of Actual Influence by Pressure Loss Pressure loss inside of pipe P loss = f L d Relation of Mass flow rate and Capacity of Rfrg. Cycle qmr = Φ 0 Relation of Specific volume and Density P P P DP loss vs = ρ v Adiabatic Compression on p-v diagram wr qmr/a ρ ΔP loss = f ʹ L d 5 Φ 0 Spec. of A/C A = p d /4: Flow cross-sectional area, Φ: Cooling Capacity, V s : Specific Volume, W r : Refrigeration Effect, f = 6 f / p Considering influence on COP, Convert ΔP loss [MPa] to ΔW P.loss [kw] ΔW P.loss : Work equivalent to Pressure Loss ΔP loss P P v k = C ΔW P.loss = qmr $ dh ΔW P P.loss f ʹ P = qmr $ v dp P Spec. of A/C Φ & v v wr v s P loss (When P )*++ = P, P. 0) L d 5 Φ 0 3 ΔW P.loss is Proportional to v s and Inversely Proportional to w r 3 vs wr Refrigerant s property vs wr 3 Refrigerant s property

5 Properties of refrigerants () 5 Easy calculation gives the information to find which refrigerants have the potential to raise the system performance. Refrigerant R3 R40A =R3/R5 (50/50wt%) Coefficient of Performance: 4.59 COP = w r / w s (00%) Refrigerating effect: 48 w r [kj/kg] (00%) Specific Volume in Suction: v s [m 3 /kg] (00%) 4.49 (97.8%) 64 (66%) (7%) better even -ze blend =R3 /R34ze(E) (70/30wt%) 4.63 (00.8%) 0 (85%) (0%) worse -yf blend =R3/R5 /R34yf (67/7/6wt%) 4.56 (99.%) 9 (78%) (87%) much worse R 4.8 (05%) 63 (66%) (04%) R90 (Propane) 4.78 (04%) 88 (6%) (7%) Pressure Loss at constant capacity : ΔP loss [% of kpa] ( v s / w r ) Work equiv. to Pressure Loss at constant capacity: ΔW P.loss [% of W] ( v s / w 3 r ) (00%) (65%) (4%) (44%) (4%) (6%) (00%) (80%) (69%) (60%) (38%) (30%)

6 Experimantal Test System, Test Conditions 6 We used R3 Residential mini-split type Air-conditioner for drop in test. Test System Diagram P,T Accumrator Outdoor Unit V.F.D* P,T Compressor Heat Exchanger P,T 4-Way Valve Muffler Daikin Middle East Model FTXM4PVXM + RXM4PVXM Nominal Capacity Cooling 7.kW (4,000BTU/h) Heating 6.3kW (,600BTU/h) Fan Receiver Air (Counter flow) P,T Exp. Valve Muffler Stop Valve Ambient Temp. Conditions Indoor Unit Side Outdoor Unit Side Connection Pipe φ/inch, 7.5m Stop (Vapor side) Valve Indoor Unit P,T Heat Exchanger Fan P,T connection pipe φ/4inch, 7.5m (Liquid side) * V.F.D = Variable Frequency Drive P,T Pressure gauge & Thermo copule Ambient Temp. DB:7 / WB:9 DB:35 C Air (almost Parallel flow)

7 Outline of System Simulation 7 We used Energy Flow+M Core System in order to clarify the Influence on System Performance in each Refrigerant. EF+M Core System Ver Waseda UNIV. Saito lab. 09 EEV Expansion 4 6 fin tube heat ex DDPS Condenser 3 4 Q_CON fin tube heat ex DDPS Condenser A+B-C Suction 37 Tube DDPS θ Refrigerant Air Value 5 Reversing valve 3 4way valve 4 Energy Flow +M Core System made by Prof. Saito s lab, Waseda Univ., Japan. And JRAIA approved. *JRAIA: Japan Refrigeration and Air Conditioning Industry Association 7 Air vsptx inlet Connecting Indoor air 7 Air vsptx inlet 6 fin tube heat ex DDPS 3 Evaporator Exp. valve 900 PID controller Outdoor air 6 fin tube heat ex DDPS 3 Evaporator 4 37 Tube DDPS EVA_Q 9009 θ A-B-C Comp 900 PID controller + - COP 900 Compressor A*B/C 08 compressor Enables easy comparison by adding PID controller for SH and Capacity. *PID: Proportional, Integral, and Differential Controller

8 Certainty of Simulation 8 As for Simulation using R3, COP was well fitted from 00% capacity to 83% capacity. Relative COP vs R3 Base point Recognition of Certainty of Simulation 5% Speed down 0% 05% 00% 95% 90% 85% R3 experiment R3 simulation 80% 80% 85% 90% 95% 00% 05% Relative Capacity vs R3 Base point Capacity [W] Base Point in using R3 Capacity = 756W Compressor Speed = 78rps PID Control for Capcity and Suction SH Capacity 6.0 Suction SH 4K W Actual Time [s] *Unsteady calculation Suction SH [K]

9 Results of Drop-in test using Simulation 9 As for trends of COP in each refrigerant from R3, System Simulation gave good fitting to Experiment. Some errors of COP assumed to be caused by pressure drop. COP trend comparison of each refrigerant By Experiment Relative COP vs R3 measured 5% 0% 05% 00% 95% 90% Compressor speed = 78rps R3 Base 85% Capacity = Const. 80% 80% 85% 90% 95% 00% 05% Relative Capacity vs R3 measured 3% R3 measu 5% R40A mea Blend A me Blend B me Relative COP vs R3 measured By System Simulation 5% 0% 05% 00% 95% 90% Compressor speed = 78rps R3 Base 85% Capacity = Const. 80% 80% 85% 90% 95% 00% 05% Relative Capacity vs R3 measured R R B B

10 Comparison of System Loss by Loss Analysis 0 The ratio of Pressure Loss at Suction pipe almost followed theatprediction by the properties comparison. cooling operation & at the nominal rated capacity (@756W)at cooling operation & at the nominal rated capacity (@756W) Loss, Input Wloss [%] (00%@R3's SUM) 0% 00% 90% 80% 70% By Experiment SUM: 00.0% 00.0%.9% 3.7%.6% 30.9% 30.9% 40% 30% 0% 0% 0% 3.7%.6% 36.8% 36.8% 3.7%.6% 37.% 37.% 0.9%.9% 9.3%.5%.3% 8.% 8.%.% 0% 3.7%.6% 00% 90% 34.0% 34.0% η=66.% h=66.% η=65.6% h=65.6% h=67.7%80% η=67.7% η=67.0% h=67.0% 60% 50% 4.9% 4.3% 4.3% 4.9%.5%.0%.8%.5% 0% Loss, Input Wloss [%] (00%@R3's SUM) 0% 3.5% 3.5%.% 3.7% 3.7% 70% By System Simulation SUM: 00.0% 00.0% 07.5% 07.5%.4% 40% 30% 33.0% 33.0% 0% 0%.%.% 4.9% 4.9% 4.% 4.% 4.0% 4.0% R3 R3 R40A R40A A -zeblend blend B -yfblend blend 0% 04.7% 04.7%.9%.0%.0% 3.7% 3.7% 3.7%.6%.3%.6% Others.6% 3.7%.6% Others, Others Fan Inputs Outdoor fan input Outdoor fan input 9.6% 9.6% Indoor fan input h=68.4% η=68.4% 3.0% 3.0% η=68.4% h=68.4% Compressor loss Relatively Smaller 0.6% 60% 0.7% 50% 08.3% 08.3% 9.9% loss Condenser.5% Theoretical.% comp. work in ideal condition Evaporator loss 3.% 3.% 8.3% 8.3% Suction pipe pressure loss.%.% 3.9% 3.9% R3 R3 R40A R40A 3.% 3.% η=68.4% h=68.4% 3.% 3.% η=68.4% h=68.4% Compressor loss.5% 0.%.%.3% 3.4% 3.4% 3.9% 3.9% A -zeblend blend Compressor Loss Indoor fan input Condenser Loss loss Condenser Ideal condition Theoretical comp. work inwork ideal condition Evaporator loss 30.4% 30.4% 3.5% 3.5% B -yfblend blend Evaporator Suction Loss pipe pressure loss Suction Pressure Loss

11 Refrigerant Mass Increases by Pressure Loss Pressure Loss increases mass of refrigerant use. Pressure Loss has to be considered for Global Warming and Safety. Thinking of making pressure loss constant, by adjusting inside diameter of pipe, with using the equation of Work equivalent to pressure loss ΔW P.loss. Refrigerant R3 R40A =R3/R5 (50/50wt%) Work equiv. to Pressure Loss at constant capacity: ΔW P.loss [% of W] ( v s / w r 3 ) Inside Dimeter of pipe for constant Pressure Loss: d [% of m] ( / ΔW P.loss5 ) Inside Volume of pipe for constant Pressure Loss: V [% of m 3 ] ( d ) even worse much worse -ze blend =R3 /R34ze(E) (70/30wt%) -yf blend =R3/R5 /R34yf (67/7/6wt%) Spec. of A/C R Refrigerant s property R90 (Propane) 00% 80% 69% 60% 38% 30% 00% % % 0% 9% 5% 00% 6% 3% % 65% 56%

12 Conclusion Conclusion The following results were revealed through our examination and analysis. In case of judging the performance of refrigerant in system from refrigerant property, it is effective to consider Work equivalent to Pressure Loss. it is Proportional to v s (Specific Volume), and Inversely Proportional to w r3 (Refrigerating Effect) at a Constant Capacity. The more Accurate calculation of Pressure Loss is necessary to be considered for the more accurate simulation. As for GWP, HFO-blend refrigerant has similar value to R3, however the Actual Climate Impact Increases. Because Refrigerant Mass has to be Increased in order to Reduce Pressure Loss.

13 RAC Manufacturers in Japan 3 Daikin is the st to launch R3 Air Conditioners. Mitsubishi Electric, Hitachi, Panasonic, etc., followed us Launch date by Company Press Release Launch 0/ 03/04 All RAC RAC.-8.0kW Wall-mounted type MITSUBISHI ELECTRIC HITACHI JC Panasonic 03/08 03/09 03/ RA.-8.0kW 03/ RA.-9.0kW 03/09 03/0 RA.-7.kW FUJITSU GENERAL 03/ SHARP 04/ RAC Multi Split type, Floor-standing type, etc. 04/0 RA.-8.0kW 03/ 04/0 RA.-7.kW

14 Promoting Lower GWP Refrigerant R3 4 Daikin have sold more than 0 million R3 units in 5 countries so far About 3 million R3 units were penetrated over the world under Daikin estimation

15 5 Thank you for your kind attention

16 Maldistribution problem of HEXs 6 Expansion of paths of HEX. is just the means for every refrigerant. It causes maldistribution problem and reduction of HEX s performance. Thinner tubes are better to reduce refrigerant mass. Moreover raise HEX s performance. Keeping pressure loss ΔW P.loss constant by adjusting number of paths *Face area = const. R40A, -ze blend R3 R -yf blend R90 4 paths 5 paths 6 paths Heat Exchanger Easy Even distribution Difficult