设计一台往复式天然气压缩机计算题.docx

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1、设计一台往复式天然气压缩机，结构见图2-114。已知数据：吸气压 力0.3MPa(G),排气压力25MPa(G),行程95mm,转速力Or/min,排 气量并要求一级气缸直径175mm,活塞杆直径d=35mm。 一级进气温度为25,若用户要求排气压力为31MPa(G),则各级压 力如何变化？一、设计条件吸气压力：0.3MPa(G)=0.4MPa (绝压)排气压力：25Mpa(G)=25.1 MPa (绝压)要求排气压力：31MPa(G)=31.1 MPa (绝压)由题意知：Psi=0.4MPaPd4=25.1MPa行程 s=95mm=0.095m用=25=298. 15K转速 n=98Or/m

2、inDi=175mm=0.175m d=35mm=0.035m天然气：k=1.32次数级次%=2辱？/”匕J %PdJMP或%rslU%1I0.41.20803.010.771.49xl(y387.1%111.20803.05252.530.841.70x10-3III3.05258.55302.800.771.48x10-3IV8.553031.13.640.821.64x10-32I0.41.11282.780.801.54x1 O-398.7%II1.11283.06052.750.821.52x1 O-3III3.06058.05002.630.791.52x1 O-3IV8.0500

3、31.13.860.811.51x1(尸5、比较复算容积流量，得4 = 1.05 in/ min6、复算排气温度压力指示m=l.32,各级排气温度为T(n =382.1 A： T/2 =381.OXT&=376.9K&=413.7K二、确定各级直径总压比4总二总压比4总二殳3 匕04=62.75按等压比分配：1=嬉=#62.75=2.8146Ps1=0.4MPO Ps2Ps3Ps4LUHHHPd1Pd2 Pd3 Pd4=25.1 MPO级次IIImIVPs(MPa)0.41.12583.1685648.917924Pd(MPa)1.12583.1685648.91792425.1e2.8146

4、2.81462.81462.8146假设此压缩机冷却完善，所以/=Q=q=&=25=298.15K由图2-114可得:有已知条件带入（1）式得= -(0.1752-0.0352)x 0.095 4= 2.193x107 机 3又有公式(2-38)V =%x&(其中;I ： .=1、2=1、SInr、(M Clo, Psi看、i441.5980x2.193x10 31.5980x2.193x10 3= 0.698容积系数计算:容积系数计算:压力系数取乙=0.972/2 = 0.992p3 = 0.99% =1.0温度系数4X2, =0.954=0.96 2,3=0.974=0.98(3)泄露系数

5、(无油润滑)AZI =0.954=0.9643=6972/4=0.98(4)膨胀指数(由表21得)由公式级次InmIVPs(MPa)0.41.12583.1685648.917924Ps(bar)411.25831.6856489.17924m1.2251.3081.41.4% = 414Pl4410.698= 0.797必352少一1 2.8146画-1分别取% 二.30、二.35、% =0.145确定各级气缸直径:(H)级直径匕2=&25 = ?(。；-卜% =%x 用x2x! Ps2 几联立以上两式可得：带入数据可得:A 4 = -X 且 X 2 X!Ps211p2t2!2S1.50.4

6、,I-980 1.12580.99 x 0.96x 0.96 x 0.095= 6.274x10I九甘一%(物 T)Ap2=；-42)联立以上三式得:&2叭2=；(无-笛)1一%(-7)rBP 6.274x10-3 =?(D； -O.O352) 1-(). 130(2.8146而-1)解之得 D2 =0.103m=l 03 mm查表取气缸直径D2 = l()5mmo则实际行程容积为：V；2 = A/)2 S = ?(Q； c/2) S = 7(1052 - 0.0352) X 0.095 = 7.308 x 10-4加 活塞有效面积：Ap2= (- / ) = 5 (0.1052 - 0.03

7、52 ) = 7.693 x 10-3m2(III)级直径A , = xx x!ps IJnr*n Ps3 Ts、Ap3A,l3Aj3s1.50.4,1-980 3.16860.99 x 0.97 x 0.97 x 0.095= 2.1835x1()-31又1-%(更-1)带入数据得2.1835xI0-3 =?(用jo 35(2.8146/ -1)解之得 3 = 0.057m=57mm查表圆整取= 60mm则实际行程容积为：V3= A 03s = 5 & 7 = 0.062 x 0.095 = 2.685 xl04w3活塞有效面积：A, = -D；.5 = -0,062 = 2.826x 10

8、-3 加2。3 4 34(W)级直径4 a =-x-x-x-P 亿 4 /444s1.50.411=xX 1 x-980 8.9179 =7.5246 xl0-4I又勾出二?团1-4(恭-1)带入数据得 7.5246 xlO-4 =-(/) 1-0.145(2.8146-1)解之得。4 = 0.034m=34mm查表得= 36mm则实际行程容积为：V；4= A 网$ = 5。：5 = 5 0.0362x 0.095 = 9.665 x 10-5in活塞有效面积：A =-Ds = -0.0362=1.017x10-3川川4 44由公式,=4会与计算出新的,与小，结果如下表:级次IIIIIIIV气

9、缸实际直径D (m)0.1750.1050.060.036有效实际面积Ap(m2)2.193x1()37.308 xlO-42.685 xlO-49.665 xlO-5气缸直径D (m)0.1750.1030.0570.034有效面积Ap(m2)2.193x1()36.998x1()72.423x1078.621 xlO-50.7972.3976.78218.487实际，;0.7972.2956.12116.490a0.1530.1300.1350.145实际优0.1530.1530.2080.238实际匕，2.193x10-37.308 xlCT42.685xl()T9.665x10，三、计

10、算活塞力由课本P22图2/4分别取:% =0.029 Jv2 =0.021 心=0.019*=0.0173 ia，c*为|=0.058(5；y2 = 0.043/3 = 0.027 g” =0.014分由公式各级进、排气压力和实际压力比级次公称压力压力损失实际压力/MPa实际压力比2P；I0.41.12580.0290.0580.38841.19113.067II1.12583.168560.0210.0431.10223.30482.998III3.168568.91790.0190.0273.10849.15872.946IV8.917925.10.0170.0148.766325.451

11、42.903玲=与(1+5/)压力比由彳二生计算得出，结果如上表。S止点气体力计算:列次内止点（ZN）外止点（ZN）I-IV16.27-19.23II-III16.06-17.99I -IV内止点气体力：% = p（n-a -居号（-4V2）一时,，咻=16.27kN（拉应力） 外止点气体力：Fcl = PAl-PryCD,- -Pdlv,4=-19.23kN （压应力） II -III内止点气体力：% =%-笳）-弋川咻= 16.06kN（拉应力）外止点气体力：Fc2 =舄-4q（Z）；-耳）-%J4i=-17.99kN（压应力）四、确定各级排气温度因假设冷却完善，故:级次IIIinIVt(

12、K)298.15298.15298.15298.15T(K)391.21389.09387.44386.06rt-l上表由Tdi= T (n= 1.32)计算得出各级排气温度。五、计算轴功率并选配电机各级指示功率为1n *百叫)-11OUHj 1计算结果如下表:总的指示功率为；=4*NiA14283W13498W13008W12447W53237WN产工57=1取机械效率偌 =。94 ,轴功率为：N:=- = 56635”)= 56.635伏 W)取电机功率余度10%, 56.635x 1.1 = 62.3(W),则选择电机功率为65%忆六、变工况热力计算1、确定与容积有关的系数2、设想容积的

13、计算级次压力系数（;温度系数4泄露系数为A)； = 4A4I0.970.950.950.8754II0.990.960.960.9124III0.990.970.970.9315IV1.000.980.980.9604级次TsX0Xk设想容积I298.150.875411.919xl03II298.150.912416.668x1()7m298.150.931512.501X10-4IV298.150.960419.282x10-53、压力比及容积系数4、各级压力比复算级次号二暇/MPaP或(j_ 劈T /I0.41.1512.8780.790II1.1513.0702.6660.829III3.0708.2712.6940.786IV8.27131.13.7600.625