Cooling load and coe?cient of performance
optimizations for real air-refrigerators
Youming Tu a ,Lingen Chen
a,*,Fengrui Sun a ,Chih Wu b a
Postgraduate School,Naval University of Engineering,Wuhan 430033,PR China b Mechanical Engineering Department,US Naval Academy,Annapolis MD 21402,USA
Received 11November 2005;received in revised form 25February 2006;accepted 11March 2006
Available online 12June 2006
Abstract
Based on a simple irreversible variable-temperature heat reservoir air (Brayton)refrigeration cycle model,a performance analysis and optimization of a real air refrigerator is carried out using ?nite-time thermodynamics.To maximize the cooling load and the coe?cient of performance (COP)of the cycle,the allocation of a ?xed total heat-exchanger inventory and thermal-capacity rate matching between the working ?uid and heat reservoirs are optimized,respectively.The in?uences of pressure ratio,the total heat-exchanger inventory,the e?ciencies of the compressor and expander,the thermal capacity rate of the working ?uid and the ratio of the thermal-capacity rates of the heat reservoirs on the performance of the cycle are shown by numerical examples.The results obtained provide guidances for the design of practical air-refrigeration plants.
ó2006Elsevier Ltd.All rights reserved.
Keywords:Finite-time thermodynamics;Irreversible air-refrigeration cycle;Cooling load;COP;Generalized thermodynamic optimization
1.Introduction
In the literature of ?nite-time thermodynamic studies of refrigeration cycles,the per-formance optimizations for Carnot refrigeration cycles,vapor-compression refrigeration cycles,Stirling refrigeration cycles,absorption refrigeration cycles and thermoelectric 0306-2619/$-see front matter ó2006Elsevier Ltd.All rights reserved.doi:10.1016/j.apenergy.2006.03.003
*Corresponding author.Tel.:+862783615046;fax:+862783638709.
E-mail addresses:lingenchen@https://www.doczj.com/doc/f813119052.html, ,lgchenna@https://www.doczj.com/doc/f813119052.html, (L.
Chen).
Applied Energy 83(2006)
1289–1306APPLIED ENERGY
1290Y.Tu et al./Applied Energy83(2006)1289–1306
Nomenclature
C thermal-capacity rate,kW/K
c thermal-capacity rate matching,dimensionless
E1e?ectiveness of heat exchanger,dimensionless
k ratio of principal speci?c heats,dimensionless
m(kà1)/k,dimensionless
N1number of heat-transfer units,dimensionless
Q rate of heat transfer,kW
R cooling load,kW
T temperature,K
U heat conductance,kW/K
u distribution of heat conductances,dimensionless
x isentropic temperature ratio of working?uid,dimensionless
Greek symbols
e coe?cient o
f performance(COP),dimensionless
g e?ciency,dimensionless
p pressure ratio,dimensionless
Subscripts
c compressor
H hot-side
in inlet
L cold-side
max maximum
min minimum
opt optimum
out outlet
T total
t expander
wf working?uid
1,1s,2,3,3s,4state points
refrigeration cycles have been carried out.Many valuable results have been obtained,see the review articles and books[1–5].Recently,some progress in the research and devel-opment of air(Brayton)refrigeration cycles has been achieved[6–9].The method of ?nite time thermodynamic analysis has been also applied to the performance studies of simple endoreversible[10–12]and irreversible[13–17]Brayton refrigeration cycles, as well as regenerated endoreversible[18]and irreversible[19–21]Brayton refrigeration cycles.For a?xed total heat-exchanger inventory of heat exchangers(i.e.,the sum of heat conductances of the all heat exchangers)in the refrigeration cycle,the optimum per-formance of the cycle can be obtained by optimizing the distribution of the heat exchan-ger inventory between the hot-and cold-side heat-exchangers[10,20–25].Wu et al.[10] studied the optimum allocation of heat-exchanger inventory,the optimum thermal-
Y.Tu et al./Applied Energy83(2006)1289–13061291 capacity rate matching between the working?uid and heat reservoirs,and the corre-sponding relations between the cooling load and the coe?cient of performance(COP) of endoreversible constant-temperature and variable-temperature heat-reservoir Brayton refrigeration-cycles.Chen et al.[13]deduced the cooling load and COP of an irreversible simple Brayton refrigeration-cycle coupled to constant-temperature heat reservoirs. Later,Luo et al.[14]studied the optimum allocation of heat-exchanger inventory and its corresponding cooling load and COP performance of the same cycle on the basis of Ref.[13].Zhou et al.[11,12,15,16]analyzed and optimized the cooling load density of a simple endoreversible[11,12]and irreversible[15,16]constant-[11,15]and variable-[12,16]temperature heat-reservoir Brayton refrigeration-cycles.Chen and Su[17]opti-mized the exergetic e?ciency of an irreversible Brayton refrigeration-cycle.Saygin and Sisman[26]studied the performance of Brayton refrigeration-cycles working under quantum degeneracy conditions.
On the basis of Refs.[10,12,14,15,17],the optimum allocation of heat-exchanger inven-tory and thermal-capacity rate matching between the working?uid and heat reservoirs of the simple irreversible variable-temperature heat-reservoir air-(Brayton)refrigeration cycle is studied in this paper.The optimum performance characteristic is obtained.The in?u-ences of various parameters on the characteristic of the cycle are analyzed.The results may provide guidances for the design of practical refrigeration plants.
2.Analytical relation
A simple irreversible variable-temperature heat reservoir air(Brayton)refrigeration cycle and its surroundings to be considered in this paper are shown in Figs.1and2. The following assumptions are made for this model:
The working?uid?ows through the system in a steady-state fashion.The cycle is a Brayton refrigeration one,which consists of two isobaric processes(1!2,3!4) and two nonisentropic adiabatic processes(2!3,4!1).The two adiabatic processes (2!3s,4!1s)are the corresponding isentropic ones.Therefore,cycle1!
2!3!4!1is an irreversible one,and1s!2!3s!4!1s is an
endoreversible Array
Fig.1.The schematic of the real air-refrigerator.(A)The compressor;(B)The condenser;(C)The expander;(D) The heat source;(E)The heat exchanger.
one.The irreversible and endoreversible cycles are distinguished by using the e?ciencies of the non-isentropic compression and expansion processes.The compressor and expander e?ciencies are de?ned as:
g c ?eT 3s àT 2T=eT 3àT 2T;g t ?eT 4àT 1T=eT 4àT 1s Te1T The high-temperature (hot-side)heat-sink is considered as having a ?nite thermal-capacity rate C H .The inlet and outlet temperatures of the cooling ?uid are T Hin and T Hout ,respectively.The low-temperature (cold-side)heat-source is considered as having a ?nite thermal-capacity rate C L .The inlet and outlet temperatures of the cooled ?uid are T Lin and T Lout ,respectively.
The hot-and cold-side heat-exchangers are considered to be counter-?ow heat exchang-ers,and their heat conductances (heat-transfer coe?cient-area product)are U H and U L ,respectively.The working ?uid is an ideal gas having constant thermal-capacity rate (the product of mass ?ow rate and speci?c heat)C wf .
According to the properties of the heat transfers between the heat reservoir and work-ing ?uid,the heat reservoirs,the working ?uid,and the theory of the heat exchangers,the rate of heat transfer (Q H )released to the heat sink and the rate of heat transfer (Q L )sup-plied by the heat source,i.e.,the cooling load (R),are,respectively:
Q H ?U H ?eT 3àT Hout TàeT 4àT Hin T =ln ?eT 3àT Hout T=eT 4àT Hin T
?C H eT Hout àT Hin T?C wf eT 3àT 4T?C H min E H1eT 3àT Hin T
e2TR ?Q L ?U L ?eT Lin àT 2TàeT Lout àT 1T =ln ?eT Lin àT 2T=eT Hout àT 1T
?C L eT Lin àT Lout T?C wf eT 2àT 1T?C L min E L1eT Lin àT 1T
e3Twhere E is the e?ectiveness of the heat exchangers:
E H1?
1àexp ?àN H1e1àC H min =C H max T 1àeC H min =C H max Texp ?àN H1e1àC H min =C H max T e4TE L1?1àexp ?àN L1e1àC L min =C L max T 1àeC L min =C L max Texp ?àN L1e1àC L min =C L max T e5Twhere C H min and C H max are the minimum and maximum of C H and C wf ,respectively,and C L min and C L max are the minimum and maximum of C L and C wf ,respectively,and N is the number of heat-transfer
units:
Fig. 2.The temperature-entropy diagram of a simple irreversible variable-temperature heat reservoir air-refrigeration cycle.
1292Y.Tu et al./Applied Energy 83(2006)1289–1306
N H1?U H=C H min;N L1?U L=C L min C H min?min f C H;C wf g;
C H max?max f C H;C wf g C L min?min f C L;C wf g;C L max?max f C L;C wf g
e6T
The property of endoreversible cycle requires T1s T3s=T2T4.The isentropic tempera-ture ratio x of the working?uid is de?ned as:
x?T3s=T2?T4=T1s?pekà1T=ke7Twhere k is the ratio of speci?c heat,and p is the cycle pressure-ratio.
The coe?cient of performance(COP)of the refrigerator is e=R/(Q HàQ L).The cool-ing load(R)and the COP(e)can be derived from above model.They are,respectively:
R?C wf E L1C Lmin f?g c C wfàeg t xà1àg tt1TeC wfàC Hmin E H1Textg cà1T T Linàg c C Hmin E H1T Hineg t xà1àg tt1Tg
C
wf
g càeC wfàC Hmin E H1TeC wfàC Lmin E L1Textg cà1Teg t xà1àg tt1T
e8T
1teà1?
C Hmin E H1fextg cà1TC Lmin E L1T Linà?g c C wfàeC wfàC Lmin E L1Textg cà1Teg t xà1àg tt1T T Hin g
C Lmin E L1f?g c C wfàeg t xà1àg tt1TeC wfàC Hmin E H1Textg cà1T T Linàg c C Hmin E H1T Hineg t xà1àg tt1Tg
e9T
3.Optimum allocation of heat-exchanger inventory
3.1.The existence of optimum distribution of heat conductance
For the?xed heat-exchanger inventory U T,that is,for the constraint of U H+U L=U T, de?ning a distribution of heat conductance u=U L/U T leads to:
U L?uU T;U H?e1àuTU Te10TThere exists an optimum distribution of heat conductance,which leads to the optimum cooling load or optimum COP.The performance optimization is performed using numer-ical calculations.In the calculations,k=1.4,U T=5.0kW/K,T Hin=303K,T Lin= 258K,g c=g t=0.8,C wf=0.8kW/K,C L=C H=1.0kW/K and p=0!15are set.
The corresponding three-dimensional diagrams among the cooling load(R),the distri-bution(u)of heat conductance and the cycle pressure ratio(p),as well as the comprehen-sive relationships among the COP(e),the distribution(u)of heat conductance and the cycle pressure ratio(p)are shown in Figs.3and4.They show the existence of an optimum distribution(u opt,R)of heat conductance corresponding to an optimum cooling
load Fig. 3.The comprehensive relationships among cooling load,pressure ratio and distribution of heat conductance.
Y.Tu et al./Applied Energy83(2006)1289–13061293
(R opt,u )and an optimum distribution (u opt,e )of heat conductance corresponding to an optimum COP (e opt,u )for a ?xed cycle pressure ratio (p ).From Figs.3and 4,one also can see that there does not exist any optimum pressure ratio corresponding to a maximum cooling-load,while there exists an optimum pressure-ratio (p opt,e )corresponding to an optimum COP (e opt,p )for a ?xed distribution (u )of heat conductance.Therefore,there exist a pair of optimum distribution (u opt,e )of heat conductance and optimum pressure ratio (p opt,e ),which lead to a maximum optimum COP (e max ).
It is easily known that u opt,R exactly equals u opt,e through calculations,that is,there exists an optimum distribution (u opt )of heat conductance that leads to both optimum cooling-load and optimum COP for a ?xed cycle pressure-ratio (p ).The optimum distri-bution (u opt )of heat conductance versus pressure-ratio (p )is shown in Fig.5.It can be seen that,with the increase of p ,u opt increases rapidly when p is small and decreases slowly when p is larger than a certain value.Furthermore,u opt is always less than 0.5.
3.2.The in?uences of the total heat-exchanger inventory
The in?uences of the total heat-exchanger inventory (U T )on the optimum cooling load (R opt,u )versus pressure ratio (p )and the optimum COP (e opt,u )versus pressure ratio (p )are shown in Figs.6and 7.The corresponding optimum distribution (u opt )of heat conduc-tance versus pressure ratio (p )is shown in Fig.8.In the calculations,k =
1.4,
Fig.4.The comprehensive relationships among COP,pressure ratio and distribution of heat
conductance.Fig.5.The optimum distribution of heat conductance versus pressure ratio.
1294Y.Tu et al./Applied Energy 83(2006)1289–1306
Y.Tu et al./Applied Energy83(2006)1289–13061295
inventory.
Fig.6.The optimum cooling load versus pressure ratio and heat-exchanger Array
inventory.
Fig.7.The optimum COP versus pressure ratio and heat-exchanger Array Fig.8.The optimum distribution of heat conductance versus pressure ratio and heat-exchanger inventory.
1296Y.Tu et al./Applied Energy83(2006)1289–1306
T Hin=303K,T Lin=258K,g c=g t=0.8,C wf=0.8kW/K,C L=C H=1.0kW/K and p=0–15are set.
The numerical calculations show that the optimum cooling-load(R opt,u),the optimum COP(e opt,u)and the optimum distribution(u opt)of heat conductance are all increasing functions of the total heat-exchanger inventory(U T),and they increase smoothly if U T is larger than a certain value.The optimum distribution(u opt)of heat conductance is always less than0.5.
3.3.The in?uences of the compressor and expander e?ciencies
The in?uences of the compressor and expander e?ciencies(g c=g t)on the optimum cooling load(R opt,u)versus pressure ratio(p)and the optimum COP(e opt,u)versus pres-sure ratio(p)are shown in Figs.9and10.The corresponding optimum distribution(u opt) of heat conductance versus pressure ratio(p)is shown in Fig.11.In the calculations, k=1.4,U T=5.0kW/K,T Hin=303K,T Lin=258K,C wf=0.8kW/K,C L=C H=
),
1.0kW/K and p=0!15are set.It is clearly seen that the optimum cooling load(R opt,u Array
e?ciencies.
Fig.9.The optimum cooling load versus pressure ratio,and compressor and expender Array Fig.10.The optimum COP versus pressure ratio,and compressor and expender e?ciencies.
the optimum COP (e opt,u )and the optimum distribution (u opt )of heat conductance are all increasing functions of the compressor and expander e?ciencies (g c =g t ).
When g c =g t =1.0,the irreversible cycle becomes an endoreversible one,and Eqs.(8)and (9)become
R ?C H min C L min E H1E L1eT Lin àx à1T Hin T C H min E H1tC L min E L1àC H min C L min E H1E L1=C wf ;e ?ex à1Tà1e11T
Eq.(11)shows that the COP (e )is independent of u :it is only dependent on p .The COP is a monotonic decreasing function of the pressure ratio.This is di?erent from that for an irre-versible cycle.The optimum distribution (u opt )of heat conductance corresponding to the optimum cooling-load (R opt,u )is a constant and u opt =0.5.This can be obtained analytically
[10]:u opt =0.5is an upper bound of the optimum distribution (u opt )of heat conductance.
3.4.The in?uences of the thermal capacity rate of the working ?uid
The in?uences of the thermal capacity rate of the working ?uid (C wf )on the optimum cooling load (R opt,u )versus pressure ratio (p )and the optimum COP (e opt,u )versus pres-sure ratio (p )are shown in Figs.12and 13.The corresponding optimum distribution (u opt )of heat conductance versus pressure ratio (p )is shown in Fig.14.In the calculations,k =1.4,U T =5.0kW/K,T Hin =303K,T Lin =258K,g c =g t =0.8,C L =C H =1.0kW/K and p =0!15are set.It can be seen that with the increase of C wf ,the optimum cooling load (R opt,u )?rstly increases and then decreases while the optimum COP (e opt,u )and the optimum distribution (u opt )of heat conductance monotonically decrease.Also,the opti-mum distribution (u opt )of heat conductance is always less than 0.5.
4.Optimum thermal-capacity rate matching between the working ?uid and heat reservoirs
4.1.The existence of optimum matching
For ?xed thermal-capacity rates of heat reservoirs (C L and C H ),de?ning the thermal-capacity rate matching (c =C wf /C H )between the working ?uid and the heat
reservoirs.
Fig.11.The optimum distribution of heat conductance versus pressure ratio,and compressor and expender e?ciencies.
Y.Tu et al./Applied Energy 83(2006)1289–13061297
1298Y.Tu et al./Applied Energy83(2006)1289–1306
?uid.
Fig.12.The optimum cooling load versus pressure ratio and thermal-capacity rate of the working Array
?uid.
Fig.13.The optimum COP versus pressure ratio and thermal-capacity rate of the working Array Fig.14.The optimum distribution of heat conductance versus pressure ratio and thermal-capacity rate of the working?uid.
Y.Tu et al./Applied Energy83(2006)1289–13061299 From Fig.12,one can see that there exists an optimum thermal-capacity rate of the work-ing?uid(C wf)opt that leads to a maximum optimum cooling load(R max).That is,there exists an optimum thermal-capacity rate matching(c opt)between the working?uid and the heat reservoirs corresponding to the maximum optimum cooling-load(R max).From Fig.13,one can also see that the optimum COP(e opt,u)is a decreasing function of C wf for a?xed p.So,there is no optimum thermal-capacity rate matching between the working ?uid and the heat reservoirs that leads to a maximum optimum COP.The e?ects of the thermal-capacity rate matching between the working?uid and the heat reservoirs on cool-ing load are only analyzed by numerical examples.In the calculations,k=1.4, U T=5.0kW/K,T Hin=303K,T Lin=258K,g c=g t=0.8,C L=1.0kW/K,C L/C H=1 and p=5.0are set.
The corresponding three-dimensional diagram for the cooling load(R),the distribution (u)of heat conductance and the thermal-capacity rate matching(c)between the working ?uid and the heat reservoirs is shown in Fig.15.It can be seen that the curve of the cooling load(R)versus the thermal-capacity rate matching(c)between the working?uid and the heat reservoirs is a parabolic-like one.There exists an optimum thermal-capacity rate matching(c opt)between the working?uid and the heat reservoirs corresponding to opti-mum cooling load(R opt,c)for a?xed u.Meanwhile,there also exists an optimum distribu-tion(u opt)of heat conductance that leads to the optimum cooling load(R opt,u)for a?xed c. Therefore,there exist a pair of optimum distribution(u opt)of heat conductance and opti-mum thermal-capacity rate matching(c opt)between the working?uid and the heat reser-voirs,which leads to a maximum optimum cooling load(R max).
The optimum thermal-capacity rate matching(c opt)between the working?uid and the heat reservoirs corresponding to optimum cooling load(R opt,c)versus the distribution(u) of heat conductance is shown in Fig.16.It presents the maximum thermal-capacity rate matching(c max)between the working?uid and the heat reservoirs by adjusting the distri-bution(u)of heat conductance.
4.2.The in?uences of the total heat-exchanger inventory
The in?uence of the total heat-exchanger inventory(U T)on the cooling load(R)versus the thermal capacity rate matching(c)between the working?uid and the heat reservoirs is shown in Fig.17.In the calculations,k=1.4,T Hin=303K,T Lin=258K,g c=g t=0.8,
total
C L=1.0kW/K,C L/C H=1,u=0.5and p=5.0are set.The in?uence of the Array
Fig.15.The comprehensive relationships among cooling load,distribution of heat conductance and thermal-capacity rate matching between the working?uid and the heat reservoirs.
heat-exchanger inventory (U T )on the optimum cooling load (R opt,c )versus the distribu-tion (u )of heat conductance is shown in Fig.18.The corresponding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs versus the distribution (u )of heat conductance is shown in Fig.19.In the calculations,k =1.4,T Hin =303K,T Lin =258K,g c =g t =0.8,C L =1.0kW/K,C L /C H =1and p =5.0are set.
The numerical calculations show that the optimum cooling-load (R opt,c )and the cor-responding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs increase smoothly if U T is larger than a certain value,but their incremental changes become smaller.Furthermore,the optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs is always less than
unity.
Fig.16.The optimum thermal-capacity rate matching between the working ?uid and the heat reservoirs versus distribution of heat
conductance.
Fig.17.The cooling load versus thermal-capacity rate matching between the working ?uid and the heat reservoirs and heat exchanger inventory.
1300Y.Tu et al./Applied Energy 83(2006)1289–1306
4.3.The in?uences of the compressor’s and expander’s e?ciencies
The in?uence of the compressor and expander e?ciencies (g c =g t )on the cooling load (R )versus the thermal-capacity rate matching (c )between the working ?uid and the heat reservoirs is shown in Fig.20.In the calculations,k =1.4,U T =5.0kW/K,T Hin =303K,T Lin =258K,C L =1.0kW/K,C L /C H =1,u =0.5and p =5.0are set.The in?uence of the compressor and expander e?ciencies (g c =g t )on the optimum cooling-load (R opt,c )versus the distribution (u )of heat conductance is shown in Fig.21.The corresponding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat res-ervoirs versus the distribution (u )of heat conductance is shown in Fig.22.In the calcula-tions,k =1.4,U T =5.0kW/K,T Hin =303K,T Lin =258K,C L =1.0kW/K,C L /C H =1and p =5.0are
set.
Fig.18.The optimum cooling load versus distribution of heat conductance and heat-exchanger
inventory.Fig.19.The optimum thermal-capacity rate matching between the working ?uid and the heat reservoirs versus the distribution of heat conductance and heat exchanger inventory.
Y.Tu et al./Applied Energy 83(2006)1289–13061301
The numerical calculations show that the optimum cooling load (R opt,c )and the corre-sponding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs are decreasing functions of the compressor and expander e?ciencies (g c =g t ).When g c =g t =1.0,the irreversible cycle becomes an endoreversible one [10],the optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs corresponding to the optimum cooling load (R opt,c )is a constant,and c opt =1holds for the case of C L /C H =1except at the two extremities (u =0and u =1.0).
4.4.The in?uences of the ratio of the thermal-capacity rates of heat reservoirs
The in?uence of the ratio (C L /C H )of the thermal-capacity rates of heat reservoirs on the cooling load (R )versus the thermal-capacity rate matching (c )between the
working
Fig.20.The cooling load versus thermal-capacity rate matching between the working ?uid and the heat reservoirs and compressor’s and expender’s
e?ciencies.
Fig.21.The optimum cooling load versus distribution of heat conductance and compressor’s and expender’s e?ciencies.
1302Y.Tu et al./Applied Energy 83(2006)1289–1306
?uid and the heat reservoirs is shown in Fig.23.In the calculations,k =1.4,U T =5.0kW/K,T Hin =303K,T Lin =258K,g c =g t =0.8,C L =1.0kW/K,u =0.5and p =5.0are set.The in?uence of the ratio (C L /C H )of the thermal-capacity rates of the heat reservoirs on the optimum cooling-load (R opt,c )versus the distribution (u )of heat conductance is shown in Fig.24.The corresponding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs versus the distribution (u )of heat conductance is shown in Fig.25.In the calculations,k =1.4,U T =5.0kW/K,T Hin =303K,T Lin =258K,g c =g t =0.8,C L =1.0kW/K and p =5.0are set.
The numerical calculations show that the optimum cooling-load (R opt,c )is a decreasing function of the ratio C L /C H while the corresponding optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs is an increasing function of the ratio C L /C H .Also,the optimum thermal-capacity rate matching (c opt )between the working ?uid and the heat reservoirs is always less than
unity.
Fig.23.The cooling load versus thermal-capacity rate matching between the working ?uid and the heat reservoirs and ratio of the thermal capacity rates of the heat
reservoirs.
Fig.22.The optimum thermal-capacity rate matching between the working ?uid and the heat reservoirs versus distribution of heat conductance and compressor’s and expender’s e?ciencies.
Y.Tu et al./Applied Energy 83(2006)1289–13061303
In addition,all of the Figs.18,21and 24present a maximum optimum cooling-load (R max )for a ?xed p .
5.Conclusion
The distribution of the heat conductance and thermal-capacity rate matching between the working ?uid and heat reservoirs of the simple irreversible variable-temperature heat reservoir air (Brayton)refrigeration-cycle have been optimized in this paper.The relations between cooling load and distribution of heat conductance,between COP and distribution of heat conductance,between cooling load and thermal-capacity rate matching between the working ?uid and heat reservoirs are obtained.At the same time,the in?uences of the pressure ratio,the total heat-exchanger inventory,the e?ciencies of the
compressor
Fig.24.The optimum cooling load versus distribution of heat conductance and ratio of the thermal-capacity rates of heat
reservoirs.
Fig.25.The optimum thermal-capacity rate matching between the working ?uid and the heat reservoirs versus distribution of heat conductance and ratio of the thermal-capacity rates of heat reservoirs.
1304Y.Tu et al./Applied Energy 83(2006)1289–1306
Y.Tu et al./Applied Energy83(2006)1289–13061305 and expander,the thermal capacity rate of the working?uid and the ratio of the thermal capacity rates of heat reservoirs on the optimal performance of the cycle have been ana-lyzed.The results include those of endoreversible air(Brayton)refrigeration cycles obtained in the literature and can provide some valuable guidances for the performance analysis and optimization of practical Brayton refrigerator-systems.
Acknowledgement
This paper is supported by The Program for New Century Excellent Talents in the Uni-versities of P.R.China(Project No.NCET-04-1006)and the Foundation for the Authors of Nationally-Excellent Doctoral Dissertations of P.R.China(Project No.200136). References
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一、选择题 1．I’d like to________the mall because it’s crowded and noisy. A．visit B．hang out C．walk D．go off 2．That path ________ directly to my house.You won't miss it. A．leads B．forms C．repairs D．controls 3．I don’t want to go. __________, I am too tired. A．However B．And C．Besides D．But 4．Some animals carry seeds from one place to another, ________ plants can spread to new places. A．so B．or C．but D．for 5．When I as well as my cousins __________ as a volunteer in Beijing, I saw the Water Cube twice. A．were treated B．treated C．was served D．served 6．He is wearing his sunglasses to himself from the strong sunlight. A．prevent B．stop C．keep D．protect 7．When you are________, you should listen to music to cheer you up. A．shy B．afraid C．strict D．down 8．Mr. Smith gave us some________on how to improve our speaking skills. A．advice B．news C．knowledge D．information 9．World Book Day takes place ________ April 23rd every year. A．at B．in C．on 10．More and more people have realized that clear waters and green mountains are as ________ as mountain of gold and silver. A．central B．harmful C．valuable D．careful 11．We loved the food so much, ________the fish dishes. A．special B．especial C．specially D．especially 12．—Oh, my God! I have ________ five pounds after the Spring Festival. —All of the girls want to lose weight, but easier said than done. A．given up B．put on C．got on D．grown up 13．—What do you think of the performance today? —Great! ________ but a musical genius could perform so successfully. A．All B．None C．Anybody D．Everybody 14．He ________ his homework________the morning of Sunday. A．doesn’t do; on B．doesn’t do; in C．doesn’t; on 15．Maria ________ speaks Chinese because she doesn’t know much Chinese. A．seldom B．always C．often D．usually 16．In 2018, trade between China and Hungary rose by 7.5 percent, and recently on Friday companies from China and Hungary________ several cooperation (合作) agreements under the
需要系数法确定计算负荷： 建筑电气图中：Pe，Pj，Ij，Kx，cosφ各代表什么 建筑电气图中：Pe设备容量，Pjs计算容量，Ijs计算电流，Kx需要系数，cosφ功率因数Pjs=Kx*Pe. Ijs=Pjs/(1.732*Ue*cosφ) , ue=380v -------------------------------------------- 关于：Pjs=Pe*Kx (单相)Ijs=Pjs/0.22CosΦ = (Pjs * 4.5454)/ CosΦ (三相)Ijs=Pjs/0.38*1.732*CosΦ(1.732为3开根号)=( Pjs*1.5193)/CosΦ 上面式中:Pe----负荷总功率；Kx----需用系数；CosΦ---功率因数。 另外也可以根据我提供的符合计算程序进行计算； --------------------------------------------- 负荷计算的目的是为了合理地选择导线截面，确保电气线路和设备经济、安全地运行。常用计算负荷的方法中有“需要系数”法，该法较简单、 精确度较高，且是实用的工程计算方法，因而得到广泛的应用。 一、电器负荷的计算 确定了各用电设备容量之后，将各用电设备分类，即将感性负荷与纯阻性负荷分类。现在民宅中的感性负荷主要有洗衣机、空调器、电冰箱、电风扇、荧光灯中的电感性镇流器；纯阻性负荷主要有电饭（火）锅、电热水器、电热取暖器、白炽灯、加热器等。要进行分类计算。 有功计算负荷等于同类用电设备的容量总和乘以一个需要系数，即 Ｐｊｓ＝Ｋｘ?∑Ｐｅ 式中Ｐｊｓ——有功计算负荷（ｋＷ） ∑Ｐｅ——同类设备的总容量（ｋＷ） Ｋｘ——设备的需要系数，它表示不同性质的民宅对电器负荷的需要和同时使用的一个系数，与用电设备的工作性质、使用效率、数量等因素有关。附表是推荐值，仅供参考。 二、工作电流的计算 由于各类用电设备的功率因素不完全相同，又存在感性负荷和纯阻性负荷，因此应分开计算。 １．计算电流 Ｉｊｓ＝Ｐｊｓ／Ｕｃｏｓ∮ ------------------------------- 电气负荷计算方法与公式 负荷计算的目的是为了合理地选择导线截面,确保电气线路和设备经济、安全地运行。常用计算负荷的方法中有“需要系数”法,该法较简单、精确度较高,且是实用的工程计算方法,因而得到广泛的应用。 一、电器负荷的计算 确定了各用电设备容量之后,将各用电设备分类,即将感性负荷与纯阻性负荷分类。现在民宅中的感性负荷主要有洗衣机、空调器、电冰箱、电风扇、荧光灯中的电感性镇流器;纯阻性负荷主要有电饭(火)锅、电热水器、电热取暖器、白炽灯、加热器等。要进行分类计算。 有功计算负荷等于同类用电设备的容量总和乘以一个需要系数,即 Pjs=Kx?∑Pe 式中Pjs——有功计算负荷(kW)
现在进行时 撰稿：王红艳审稿：白雪雁 【概念引入】 I. 什么是现在进行时？ 1）现在进行时表示说话时正在进行或发生的动作。 例如：I am reading a book. 我正在看书。 2）表示现阶段正在进行而说话时不一定在进行的动作。 例如：I am learning English hard these days. 这些日子我正在努力学习英语。 II. 现在进行时的标志词。 现在进行时常和now、at the moment、look、listen等连用。 【用法讲解】 I.现在进行时的结构。 现在进行时的结构是：助动词be（am，is，are）＋现在分词v-ing 现在分词的构成： 1）动词的后面直接加－ing。例如：work－working，study－studying 2）以不发音的字母e结尾的动词，先去掉字母e，再加－ing。例如：live－living 3）以重读闭音节结尾并且只有一个辅音字母的动词，先双写这个辅音字母，再加－ing。 例如：stop－stopping，swim－swimming，run－running II. 现在进行时的用法。 1）现在进行时表示说话的时候正在进行的动作，经常和now，right now，at the moment 等时间状语或者动词look，listen等连用。 例如：My father is watching TV now.我爸爸现在在看电视。 Look! My brother is playing basketball there. 看！我弟弟正在那里打篮球。 2）现在进行时可以表示目前一段时间内一直进行的动作，经常和these days，this week，at present等时间状语连用。 例如：My parents are working on a farm these days. 这些天我的父母在农场干活。 3）现在进行时还可以表示现在不断发展变化的事情，表示不断发展变化的动词有get，grow，turn，become等。 例如：The leaves are turning yellow. 树叶在变黄。 4）现在进行时还可以表示将要发生的动作，只限于动词arrive，begin，go，come，leave，fly等动词。 例如：I am coming soon. 我马上来。 Ⅲ. 现在进行时的句式变化。 肯定句式：主语+be( am, is, are)+现在分词+其它. 否定句式：主语+be(am, is, are) +not +现在分词+其它. 一般疑问句：Be(am, is, are) +主语+现在分词+其它？ 特殊疑问句：疑问词+be(am, is, are)+主语+现在分词+其它？ 对现在进行时的特殊疑问句的回答，它不可以用Yes或No直接作答，要根据实际情况回答。 Ⅳ. 现在进行时的特殊用法。 表示位置移动的动词，如：leave/ come/go/begin等用于现在进行时，表示按计划或安排近期将要进行的动作，常与表示将来的时间状语连用。 —Can you help me? 你能帮我吗？
boring 令人厌烦的，乏味的，无聊的 tedious 乏味的，单调的，冗长的 flat 单调的，沉闷的 dull 乏味的，单调的 troublesome 令人烦恼的，讨厌的，麻烦的 tired 疲劳的，累的 bored 无聊的，无趣的，烦人的 exhausted 极其疲倦的 weary 疲劳的 bright 聪敏的，机灵的 apt 聪明的，反应敏捷的 intelligent 聪明的，有才智的 shrewd 机灵的，敏锐的，精明的（表示生意上的精明） ingenious （人，头脑）灵巧的 alert 警觉的，留神的 cute 聪明伶俐的，精明的 acute/cute acute 指的是视力，感觉的敏锐 dull 愚钝的，笨的 awkward 笨拙的，不灵巧的 absurd 荒谬的 ridiculous 可笑的，荒谬的 idiotic 白痴般的 blunt 率直的，直言不讳的 clumsy 笨拙的，粗陋的 happy 快乐的，幸福的 cheerful 欢乐的，高兴的 content 满意的，满足的 merry 欢乐的，愉快的，快乐的 pleasure 高兴，愉快，满足 enjoyment 享乐，快乐，乐趣 cheer 喝彩 applause 鼓掌，掌声 optimism 乐观，乐观主义 delight 快乐，高兴 kick 极大的乐趣 paradise 天堂，乐园 instant 立即的，即刻的 instantaneous 瞬间的，即刻的 immediate 立即的，即刻的 simultaneous 同时发生的，同时存在的，同步的punctual 严守时刻的，准时的，正点的 pick 挑选，选择 select 选择，挑选 single 选出，挑出 elect 选举，推举 vote 投票，选举 appoint 任命，委派 nominate 提名，任命 propose 提名，推荐 recommend 推荐，举荐 designate 指派，委任 delegate 委派（或选举）…为代表 install(l) 使就职，任命 ballot 使投票表决 dub 把…称为 choice 选择（权） option 选择 selection 选择，挑选 alternative 取舍，供选择的东西 favorite 特别喜爱的人（或物） inclination 爱好 preference 喜爱，偏爱，优先 observe 注意到，察觉到 perceive 认识到，意识到，理解 detect 察觉，发现 appreciate （充分）意识到，领会，体会 alert 使认识到，使意识到 awake 意识到，醒，觉醒 scent 察觉 ancient 古代的，古老的 primitive 原始的 preliminary 预备的，初步的 preliminary trial初审 primary 最初的，初级的 initial 开始的，最初的 original 起初的 former 在前的，以前的 previous 先，前 prior 在前的，优先的 beforehand 预先，事先 medieval 中世纪的，中古（时代）的preceding 在先的，在前的，前面的 senior 资格较老的，地位较高的 following 接着的，下述的 attendant 伴随的 subsequent 随后的，后来的 succeeding 以后的，随后的 consequent 作为结果（或后果）的，随之发生的 resultant 作为结果的，因而发生的therefore 因此，所以 consequently 所以，因此 then 那么，因而 thus 因此，从而 hence 因此，所以 accordingly 因此，所以，于是 thereby 因此，从而
利用需要系数法来确定负荷计算 许振西 （厦门华电开关有限公司，福建厦门361006） 1负荷计算的意义和目的 工厂进行电力设计的基本原始资料是各工艺部门提供的用电设备安装容量，这些用电设备种类多，数量大，工作情况复杂。如果只是简单的把各用电设备的容量加起来作为选择导线和电气设备容量的依据，那么必然会造成投资和资源的浪费。因为，各用电设备通常并不同时运行，使用时也并不是都能达到额定容量，工作制也不一样。若估算过小，将造成导线及电气设备因过流而发热，加速绝缘老化，降低使用寿命，严重时引起火灾事故，影响供电系统的正常可靠运行。 为避免这种情况，设计时用的总负荷是一个假想负荷，即计算负荷。计算负荷也称为需要负荷或最大负荷，是根据已知的用电设备安装容量确定的、用于按发热条件选择导体和电气设备时所使用的一个假想的持续负荷。计算负荷产生的热效应与某一段时间内实际变动负荷产生的最大热效应相等。 求计算负荷的这项工作称为负荷计算。负荷计算的目的是为了合理的选择供电系统的导线、开关电器、变压器等设备，使电气设备和材料既能充分得到利用，又能满足电网的安全运行。同时，也是选择仪表量程，整定继电保护的重要依据。 2负荷计算的方法 目前普遍采用的确定计算负荷的方法有需要系数法和二项式法。需要系数法的优点是简便实用，适用于全厂和车间变电所负荷的计算；二项式法适用于机加工车间，有较大容量设备影响的干线和分支干线的负荷计算。《民用建筑电气设计规范》（JGJ/T 16-2008）将需要系数法作为民用建筑电气负荷计算的主要方法，下面将加以详细介绍。 1、将用电设备按类型分组，同一类型的用电设备归为一组，并算出该组用电设备 的设备容量Pe。 对于长期工作制的用电负荷（如空调机组等），其设备容量就是设备铭牌上所标注的额定功率。 对于断续或短时工作制的用电设备，是将额定功率统一换算到负载持续率为25%时的有功功率。电焊机的设备功率是指将额定功率换算到负载持续率为100%时的有功功率。 对于照明设备：白炽灯的设备容量按灯泡上标注的额定功率取值；带自感式镇流器的荧光灯和高压汞灯等照明装置，由于自感式镇流器的影响，不仅功率因数很低，在计算设备容量时，还应考虑镇流器上的功率消耗。因此，对采用自感式镇流器的荧光灯装置，其设备容量取灯管额定功率的倍，高压汞灯装置的设备容量取灯泡额定功率的倍。 另外，成组用电设备的设备功率不应包括备用设备。 2、根据用电设备组的设备容量Pe，即可算出设备的计算负荷： 计算有功功率P c=K x P e 计算无功功率Q c=P c tanφ 计算视在功率S c= 计算电流I c=S c/U n 式中：K x为需要系数，φ为功率因数角，U n为额定线电压。
现在进行时特殊用法展现 现在进行时表示说话时正在进行的动作，这是我们平时接触最多的。然而除此以外，现在进行时还有以下几种用法。 1．表示“在做某事的过程中”，此时动作不一定正在发生。例如： Next I'll give you a few minutes to read the article．When you are reading，make a mark where there is a new word．现在我给你们几分钟时间读一下这篇文章。在读的过程中，在有生词的地方作以标记。 2．表示按计划、方案或安排而进行的将来的动作。在这种情况下谓语动词多为非延续性动词，如come，go， leave，move，die，start，stop，arrive等，及少数延续性动词，如spend，stay等。例如： She is leaving for Guangzhou next week．她下星期就要去广州了。 We are spending the whole summer holiday inBeijing soon．不久我们要在北京度过整个暑假。 Where are you staying in Guangzhou？在广州你打算住什么地方？ 3．与副词forever，always，constantly等连用，表示赞成、厌烦、生气等情绪。例如： She is always talking loudly in our class．她总是在我们班上吵吵嚷嚷的。（表示厌烦） She's constantly changing her mind．她老是改变主意。（表示不以为然） He is forever complaining about his job．他总是对他的工作提出抱怨。（表示厌烦） 4．teach，work，live，study等表示状态的动词使用现在进行时可表状况，与一般现在时区别不大。例如： I'm studying in No．1Middle School．我在一中学习。（相当于：I study in No．1Middle School．） My brother is working in a big factory．我哥哥在一家大工厂工作。（相当于：My brother works in a big factory．） 5．表示目前经常发生的动作，然而此时动作不一定正在进行之中。例如：
电气设计的负荷计算方法及其应用范围 电气负荷计算方法有:需要系数法,利用系数法,二项式系数法,单位面积功率计算法,单位产品功率计算法等. (1),需要系数法:用设备功率乘以需要系数和同时系数,直接求出计算负荷; (2),利用系数法:采用利用系数求出最大负荷班的平均负荷,再考虑设备台娄和功率差异的影响,乘以与有效台数有关的最大系数求得计算负荷; (3),二项式系数法:将负荷分为基本负荷和附加负荷,后者考虑一定数量大容量设备影响; (4),单位面积功率法,单位指标法,单位产品耗电量法等,可用于初步设计用电量指标的估算,对于住宅建筑,在设计各阶段均可使用单位面积功率法. 它们的应用范围各不一样,按《民用建筑电气设计规范》3.4.2.1."在方案设计阶段可采用单位指标法;在初步设计阶段,宜采用需要系数法."可见:民用建筑电气计算负荷推荐采用需要系数法;这是因为民用建筑中电气设备很少有特别突出的大功率设备,而按照需要系数法简单易行;而在工业建筑中,由于各设备的用电量存的很大差异,用需要系数法进行计算与实际就存在很大出入. 例如:某车间用电设备如下: 电焊机25台,功率分别 为:3.0KVA*8;8KVA*6;16KVA*5;30KVA*2;180KVA*2;200KVA*2;ε=50% 风机:50台,功率均为:2.2KW 机床:66台,功率分别为:7.5Kw*30;15KW*30;30KW*2;45KW*2;90KW*2 吊车:2台,分别为15KW,22KW. 本车间的总配电计算负荷用上述(1),(2),(3)分别如下: (一),采用需要系数法: 电焊机,Kx=0.35, Pjs=Kx*Pe =0.35*972**cosΦ =0.35*972**0.7=168.39Kw Qjs=Pjs*tgΦ=1.02*168.39=171.76Kvar 风机:Kx=0.75 Pjs=Kx*Pe=0.75*50*2.2=82.5KW Qjs=Pjs*tgΦ=0.75*82.5=61.9Kvar 机床:Kx=0.12 Pjs=Kx*Pe=0.12*1005=120.6KW Qjs=Pjs*tgΦ=1.73*120.65=208.6Kvar 吊车:Kx=0.1 Pjs=Kx*Pe=0.1*37=3.7KW Qjs=Pjs*tgΦ=1.73*3.7=6.4Kvar P∑=K∑p*∑Pjs=0.9*374.8=375.19KW Q∑=K∑q*∑Qjs=0.95*374.8=448.66KW S∑==584.86KVA cosΦ∑=0.505
不能片面说人做主语用ed，物做主语ing ing形式是修饰引起这种感觉的人或物；ed形式是描写人或物的感受。（当然物一般是动物） 翻译的话 ing形式的词译为“令人……的”；ed形式译为“……的” boring是令人感到厌烦的；bored是厌烦的。 a boring person 能够指一个了无情趣的人，让人觉得无趣的人 a bored person 则是说这个人自己感到很无趣 1.bore 1）vt.使厌烦；挖 e.g. I'm bored with this job. 这件工作厌烦了。 The oldier bore the sharp pain in the wound with great courage. 这士兵以巨大的勇气忍受着伤口的剧烈疼痛。 2）n.令人讨厌的人（或事） e.g. It's a bore having to go out again. 外出真是讨厌。 boredom n.厌倦，无趣 e.g. in infinite boredom 极其无趣 boring n. 钻(孔) adj. 令人厌烦的(事或物） e.g. The play was boring. 这部短剧很一点意思都没有。 bored adj. 无聊的, 无趣的, 烦人的 e.g. Jack is so bored. 杰克是个没有趣的人。 2.surprising 是针对事或物感到惊奇。 surprised 则是针对人。 3.pleasant adj. 愉快的, 快乐的, 舒适的, 合意的可爱的, 举止文雅的, 活泼的滑稽的, 有趣的 (天气)晴朗的, 美好的容易相处的, 友爱的 e.g. a pleasant voice 悦耳的声音 a pleasant companion 可爱的伴侣 a pleasant time 愉快地度过时光 pleasing adj. 舒适的, 使人愉快的; 满意的; 惹人喜欢的, 可爱的 e.g. a pleasing look 使人愉快的神情 a very well mannered and pleasing young man 彬彬有礼而令人喜爱的年轻人
需要系数法计算负荷 1、车间变电所的计算负荷 有功、无功功率的同时系数分别取0.8-1、0.93-1 2、配电所或总降压变电所的计算负荷为各车间变电所计算负荷之和再乘以同时系数。配电所的同时系数分别为0.85-1、0.95-1，总降压变电所的同时系数分别为0.8-0.9、0.93-0.97. 单位面积功率法确定计算负荷 Pe=Pe'.S/1000 kW Pe----计算有功功率 Pe'----单位面积功率（负荷密度）w/m2（见下表1） S------建筑面积m2 用以上方法计算负荷时，还应结合工程具体情况，乘以不同的系数，系数见下表2 电能计算 W=η*Pe*365*24 η-----平均负荷系数，缺少经验数据时取0.25-0.35 民用建筑负荷密度指标（表1） 建筑类别负荷密度（w/m2) 住宅建筑 基本型50 提高型75 先进型100 公寓建筑30-50
旅馆建筑40-70 办公建筑30-70 商业建筑 一般40-80 大中型60-120 体育建筑40-70 剧场建筑50-80 医疗建筑40-70 教育建筑 大专院校20-40 中小学校12-20 展览建筑50-80 演播室250-500 汽车库8-15 住宅建筑用电负荷需要系数（表2） 户数系数户数系数 3 1 18 0.50 4 0.9 5 21 0.50 6 0.80 24 0.45 8 0.70 25-100 0.45 10 0.65 125-200 0.35 12 0.60 260-300 0.30 14 0.55 16 0.55
注：1、表中户数是指单相配电时接于同一相上的户数，按三相配电 时连接的户数应乘以3。 住宅的公用照明和电力负荷需要系数可按0.8选取。 国家发改委2010年10月出台的居民阶梯电价方案中： 居民生活阶梯电价全国平均电量分档标准表 第一档第二档第三档 全国平均分档标准 全国 平均 分档 标准 用户 覆盖 率 全国 平均 分档 标准 合计城市合计城市农村 % % % 度/月% % % 度/月% 70% 51% 79% 110 90% 82% 95 % 210 100 % 80% 65% 88% 140 95% 90% 98 % 270 100 % 天然气每户每月9-12立方米 管道煤气每户每月30立方米左右
现在进行时动词加ing的变化规律 1)一般情况下，直接加 -ing: 如：go—going answer—answering study—studying be—being see—seeing [注一] 和名词复数、一般现在时动词第三人称单数加-s(-es)不同， 动词末尾如为“辅音字母 + y”时，y不变，其后直接加ing。 如: study—studying fly—flying carry—carrying [注二] 动词结尾为辅音字母r时，加-ing，r在此必须发音。 如: water—watering answer—answering wear—wearing 2)以不发音的e结尾的动词，去掉e，再加ing 如：come—coming write—writing take—taking become—becoming 3)动词是闭音节的单音节词，或是以重读闭音节结尾的多音节词， 而末尾只有一个辅音字母时，这个辅音字母须双写，然后再加ing。 如：sit—sitting run—running stop—stopping begin—beginning admit—admitting forget—forgetting [注一] send，think，accept等动词虽是闭音节或以重读闭音节结尾， 但末尾有一个以上的辅音字母，因此，这个辅音字母不双写，应直接加ing。 如：sending thinking accepting 4)少数几个以-ie结尾的动词，须将ie变作y，再加ing。 如：die—dying tie—tying lie—lying躺，说谎 5)少数以-c结尾的动词变为现在分词时和过去式，须先将-c变为ck,然后再加-ing 或-ed 。 如：picnic—picnicking (picnicked) traffic—trafficking (trafficked)
负荷计算方法 供电设计常采用的电力负荷计算方法有：需用系数法、二项系数法、利用系数法和单位产品电耗法等。需用系数法计算简便，对于任何性质的企业负荷均适用，且计算结果基本上符合实际，尤其对各用电设备容量相差较小，且用电设备数量较多的用电设备组，因此，这种计算方法采用最广泛。二项系数法主要适用于各用电设备容量相差大的场合，如机械加工企业、煤矿井下综合机械化采煤工作面等。利用系数法以平均负荷作为计算的依据，利用概率论分析出最大负荷与平均负荷的关系，这种计算方法目前积累的实用数据不多，且计算步骤较繁琐，故工程应用较少。单位产品电耗法常用于方案设计。 一、设备容量的确定 用电设备铭牌上标出的功率（或称容量）称为用电设备的额定功率P N ，该功率是指用电设备（如电动机）额定的输出功率。 各用电设备，按其工作制分，有长期连续工作制、短时工作制和断续周期工作制三类。因而，在计算负荷时，不能将其额定功率简单地直接相加，而需将不同工作制的用电设备额定功率换算成统一规定的工作制条件下的功率，称之为用电设备功率P Nμ。 （一）长期连续工作制 这类工作制的用电设备长期连续运行，负荷比较稳定，如通风机、空气压缩机、水泵、电动发电机等。机床电动机，虽一般变动较大，但多数也是长期连续运行的。 对长期工作制的用电设备有 P Nμ=P N （2－9） （二）短时工作制 这类工作制的用电设备工作时间很短，而停歇时间相当长。如煤矿井下的排水泵等。 对这类用电设备也同样有 P Nμ=P N （2－10） （三）短时连续工作制用电设备 这类工作制的用电设备周期性地时而工作，时而停歇。如此反复运行，而工作周期一般不超过10分钟。如电焊机、吊车电动机等。断续周期工作制设备，可用“负荷持续率”来表征其工作性质。 负荷持续率为一个工作周期内工作时间与工作周期的百分比值，用ε表示 100%100%t t T t t ε=?=?+ （2－11） 式中 T ——工作周期，s ； t ——工作周期内的工作时间，s ； t 0——工作周期内的停歇时间，s 。 断续周期工作制设备的设备容量，一般是对应于某一标准负荷持续率的。 应该注意：同一用电设备，在不同的负荷持续率工作时，其输出功率是不同的。因此，不同负荷持续率的设备容量（铭牌容量）必须换算为同一负荷持续率下的容量才能进行相加运算。并且，这种换算应该是等效换算，即按同一周期内相同发热条件来进行换算。由于电流I 通过设备在t 时间内产生的热量为I 2Rt ，因此，在设备电阻不变而产生热量又相同的条 件下，I ∝P ∝I 。由式（2－11）可知，同一周期的负荷持 续率ε∝t 。因此，P ∝ε
现在进行时 一、定义及用法： 1定义：（1）表示说话时正在进行的动作及行为。（2）表示现阶段正在进行的动作。 2基本用法： (1)现在进行时主要表示说话人的说话时刻正在进行的动作、不断重复的动作或目前这个阶段（不一定是说话时刻）正在进行的动作，如： We’re having a meeting. 我们在开会。（说话时正在进行的动作） Be quiet！The baby is sleeping.安静，孩子在睡觉。 He is teaching in a middle school. 他在一所中学教书。（目前阶段在进行的动作） (2)现在进行时表将来: 现在进行时表将来，主要表示按计划或安排要发生的动作： I’m leaving tomorrow. 我明天走。 They’re getting married next month. 他们下个月结婚。 注意：现在进行时与一般现在时均可表示将来，区别是：用现在进行时表示将来，其计划性较强，并往往暗示一种意图；而一般现在时表示将来，则其客观性较强，即通常被视为客观事实，多指按时刻表或规定要发生的情况： I’m not going out this evening. 今晚我不准备出去。 What time does the train leave？火车什么时候开？ (3)现在进行时表示感情色彩: 现在进行时有时可表示满意、称赞、惊讶、厌恶等感情色彩，通常与always，forever，constantly，continually等副词连用。比较： She’s always helping people. 她老是帮助别人。（表赞扬） She always helps others. 他总是帮助别人。（陈述一个事实） The boy is constantly lying. 这孩子老是撒谎。（表示厌恶） The boy often lies. 这孩子常撒谎。（指出缺点） 二、结构: 现在进行时常有三种句型： （1）肯定式：主语＋be＋v-ing＋其它。 如：He is mending his bike．他正在修自行车。 （2）否定式：主语＋be＋not＋v-ing＋其它。 如：He is not（isn't）mending his bike．他没在修自行车。 （3）疑问式：主要分一般疑问句和特殊疑问句两种句式。 一般疑问句：Be＋主语＋v-ing＋其它？ 如：—Is he mending his bike？他正在修自行车吗？—Yes，he is．（No，he isn't．）特殊疑问句：疑问词＋be＋主语＋v-ing＋其它？ 如：—What is he doing？他正在干什么？ 三、何时用现在进行时？ （1）以Look！或Listen！开头的句子提示我们说话时动作正在进行，应用现在进行时。 如：Look！The children are playing games over there． Listen！Who's singing in the classroom？ （2）当句子中有now（现在）时，常表示说话时动作正在进行，这时也常用现在进行时。 如：We are reading English now． （3）描述图片中的人物的动作时常用现在进行时，以示生动。 如：Look at the picture．The girl is swimming． （4）表示当前一段时间内的活动或现阶段正在进行的动作时常用现在进行时。这时常与时间状语these days，this week等连用。
一、选择题 1．Is this a photo of your son? He looks________ in the blue T-shirt. A．lovely B．quietly C．beautiful D．happily 2．—Jerry looks so tired. He works too hard. —He has to ________ a family of four on his own. A．offer B．support C．provide D．remain 3．— Mr. Wilson, can I ask you some questions about your speech? — Certainly, feel __________ to ask me. A．good B．patient C．free D．happy 4．Some animals carry seeds from one place to another, ________ plants can spread to new places. A．so B．or C．but D．for 5．— Can you tell us about our new teacher? —Oh, I’m sorry. I know________ about him because I haven’t seen him before. A．something B．anything C．nothing D．everything 6．—Help yourselves! The drinks are ________ me. —Thank you. You’re always so generous. A．above B．in C．on D．over 7．Gina didn’t study medicine. ________, she decided to become an actor. A．Instead B．Again C．Anyway D．Also 8．—Have you got Kathy’s________ for her concert? —Yes, I’d like to go and enjoy it. A．interview B．information C．invitation D．introduction 9．More and more people have realized that clear waters and green mountains are as ________ as mountain of gold and silver. A．central B．harmful C．valuable D．careful 10．Kangkang usually does her homework ________ it is very late at night. A．until B．when C．before D．after 11．He ________all the “No Smoking” signs and lit up a cigarette. A．requested B．attacked C．protected D．ignored 12．一Where is Mr. Brown? 一I think he's _____________ the music hall. A．on B．in C．over D．from 13．— Is your home close to the school, Tom? — No, it's a long way, but I am________ late for school because I get up early daily. A．always B．usually C．never D．sometimes 14．—Mum, I don’t want the trousers. They’re too long.
英语现在进行时用法 初中英语――现在进行时 1】现在进行时的构成 现在进行时由"be+v-ing"构成。be应为助动词，初学者最容易漏掉，它应与主语的人称和数保持一致。 2】现在进行时的应用 在实际运用时，现在进行时常用以下几种情况： (1)当句子中有now时，常表示动作正在进行，这时要用现在进行时。如： They are playing basketball now.现在他们正在打篮球。 (2)以look, listen开头的句子，提示我们动作正进行，这时要用现在进行时。如： Listen！She is singing an English song.听，她正在唱英语歌。 (3)表示当前一段时间或现阶段正在进行的动作，且此时有this week, these days等时间状语，这时常用现在进行时。如： We are making model planes these days.这些天我们在做飞机模型。 (4)描述图片中的人物的动作，也为了表达更生动。此时也常用现在进行时。如： Look at the picture. The children are flying kites in the park.看这幅图，那些孩子正在公园放风筝。 3】现在进行时的变化 肯定句式：主语+be( am, is, are)+现在分词+其它. 否定句式：主语+be(am, is, are) +not +现在分词+其它. 一般疑问句：Be(am, is, are) +主语+现在分词+其它？ 特殊疑问句：疑问词+be(am, is, are)+主语+现在分词+其它？ 对现在进行时的特殊疑问句的回答，它不可以用Yes或No直接作答，要根据实际情况回答。 注意事项 1.在英语中，并不是所有的动词都要使用正在进行时。例如一些表示状态和感觉的动词，一般不用进行时态，而是用现在一般时表示。例如： I hear someone singing. 我正听见有人唱歌。
【英语】英语形容词常见题型及答题技巧及练习题(含答案)及解析 一、初中英语形容词 1．My deskmate is really _____．She likes to attend different activities after school． A. active B. quiet C. lazy D. honest 【答案】 A 【解析】【分析】我的同桌同学非常活跃，放学后喜欢参加很多不同的活动． 句中提到"She likes to attend different activities after school"放学后喜欢参加很多不同的活动，由此推测此人非常活跃，A积极的，活跃的；B安静的；C懒惰的，D诚实的，根据句意可知选择A． 2．Wang Wei speaks English as ________ as Yang Lan. They both study English hard. A. good B. well C. better D. best 【答案】 B 【解析】【分析】句意：王伟的英语讲的和杨澜的一样好。他们学习英语都努力。可知as…as中间用形容词或副词原级；此处是副词修饰动词speak。good好的，形容词原形；well好地，副词原形，better比较级；best最高级，故选B。 【点评】此题考查形容词原级。注意as...as中间跟形容词或副词原级。 3．—If there are ________ people driving, there will be ________ air pollution. —Yes, and the air will be fresher. A. less; less B. less; fewer C. fewer; fewer D. fewer; less 【答案】 D 【解析】【分析】句意：——如果开车的人越少，空气污染越少。——是的，空气将会更新鲜。little少的，形容词，其比较级是less，修饰不可数名词，few几乎没有，形容词，其比较级是fewer，更少，修饰可数名词，people，可数名词，用fewer修饰，air pollution，空气污染，不可数名词，用less修饰，故选D。 【点评】考查形容词的辨析。注意less和fewer意思和用法。 4．—Guess what? The university has accepted my application! —Wow! That's ________ new I've heard this year, Boris! Let's celebrate. A. a worse B. the worst C. a better D. the best 【答案】 D 【解析】【分析】句意：——猜猜什么？那所大学已经接受我的申请了。——哇喔，那是今年我听到的最好的消息，Boris，让我们庆祝一下。A.一个更糟的，比较级；B.最糟的，最高级；C.一个更好的，比较级；D.最好的，最高级。因为大学接受申请了，所以是好消息，排除A、B。根据 I've heard this year，今年我听到的，可知是最高级，故选D。 【点评】考查形容词辨析，注意平时识记最高级结构，理解句意。