泵与压缩机课后题答案.pdf

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1、1-1解：1）基本计算π2π2−3吸入管过流断面面积：A=D=×0.1=7.854×1044Q34吸入管内平均流速：υ===1.203m/s，−3A3600×7.854×1022Lυ181.203吸入管的沿程损失：h=λ××=0.02××=0.266md2g0.12×9.82）求泵入口处的压强设吸水池液面为1-1断面，泵入口处为2-2断面。根据伯努里方程，有：2ppυ122=H+++hg1ρgρg2g2⎛υ⎞⎜2⎟p=p−ρgHg++h21⎜12g⎟⎝⎠25⎛1.203⎞=1.013×10−1000×9.8×⎜4.4++0.266⎟⎜⎟⎝2×9.8⎠55=1.013×10−

2、0.4645×105=0.5485×10Pa50.5485×10H=760−=760−411.54=348.46mmHg真133.285p0.5485×102H===5.597mHO=5597mmHO22980098001-2解：设泵入口处为1-1断面，泵出口处为2-2断面。根据伯努里方程，有：22pυpυ1122z+++H=z++12ρg2gρg2g入口和排出管径相同，有υ=υ。12pp21H=z−z+−21ρgρg5p=1.013×10−0.3×13600×9.8=61316Pa15p=1.013×10+147100=248400Pa2248400−61316H=0.5

3、+=25.95m750×9.8Q0.0251-3解：吸水管内平均流速：υ===3.183m/s1π2π2d×0.1144Q0.025排出管内平均流速：υ===5.659m/s2π2π2d×0.075244设泵入口处为1-1断面，泵出口处为2-2断面。根据伯努里方程，有：22pυpυ1122z+++H=z++12ρg2gρg2g泵的扬程为：22p−pυ−υ2121H=z−z++21ρg2g5(5)223.2373×10−−0.3924×105.659−3.183=0.8++1000×9.82×9.8=38.955mρgQH1000×9.8×0.025×38.955泵的有效功率

4、：N===9.544kWe10001000泵的轴功率：N=ηN=0.93×12.5=11.625kW轴电电Ne9.544泵的效率：η===0.82泵N11.625轴1-4解：设吸水罐液面为1-1断面，排水罐液面为2-2断面。根据伯努里方程，有：22pυpυ1122z+++H+H=H+H+++hw11m23ρg2gρg2g由于吸水罐液面和排水罐液面稳定，υ=υ=0。吸入和排出管路总水力损失为：12∆p1372930h=h+h+=1+25+=190.818mwsdρg850×9.8因此，泵的扬程为：p−p21H=H+H−H++hwm231ρg176479.7−196133=14

5、+4−8++190.8181000×9.8=198.813m31-5解：1）当泵流量Q=7.2m/h时TπDnπ×0.182×14502u===13.82m/s26060Q7.2Tc===0.555m/s2rπDBτπ×0.182×0.007×0.9×36002220c=u−cctgβ=13.82−0.555×ctg30=12.856m/s2u∞22r2A11H=uc=×13.82×12.856=18.13mT∞22u∞g9.832）当泵流量Q=28.8m/h时TQ28.8Tc′===2.221m/s2rπDBτπ×0.182×0.007×0.9×36002220c′=u−

6、cctgβ=13.82−2.22×ctg30=9.975m/s2u∞22r2A11H′=uc=×13.82×9.975=14.067mT∞22u∞g9.83）画出两种流量时叶轮出口的速度三角形（略）随流量Q的增大，c增加，c减小，H下降。T2r2uT∞πDnπ×0.22×290021-6解：u===33.406m/s26060QT0.025c===3.617m/s2r∞πDBπ×0.22×0.0122c3.6172r∞w===9.655m/s2∞0sinβsin222A0c=u−ccotβ=33.406−3.617×cot22=24.454m/s2u∞22r∞2Ac2r∞3

7、.6170α=arctan=arctan=8.4132∞c24.4542u∞()2()222c=c+c=24.454+3.617=24.72m/s22u2r∞11H=uc=×33.406×24.454=83.357mT∞22r∞g9.81-7解：1）求圆周速度和堵塞系数πD2nπ×0.32×1450u===24.295m/s26060zδ6×0.0052τ=1−=1−=0.93820πDsinβπ×0.32×sin2922AπD1nπ×0.162×1450u===12.3m/s16060zδ6×0.0051τ=1−