汽轮机原理习题解答完整版

?汽轮机原理习题?参考答案

习题1(1)(2)(3)(4)(5)(6)(7)(8)损失已经计算，图略(9)图略(10)(11)(12)略。习题2(1)高压缸：级号内功率Pi级流量D0级前温度级前压力级前比熵级前焓全级压差级后压力级后焓级理想焓降级的相对内效率有效焓降级实际出口焓值调节级18555905.6537166.4393403.744.2511.753304.299.540.74101911273.76104243329.97895815365886.1483.811.66.45373311.270.88310.7173286.6824.590.88640247921.796636953289.47336325608886.1471.110.76.4583289.320.8599.8413263.74525.5750.89086547222.783884443266.53611635992886.1457.99.876.45953265.810.8579.0133238.5427.270.89270184224.343979233241.46602146325886.1443.89.016.4633241.2060.8388.1723212.43328.7730.89308983925.696873943215.50912656551886.1428.98.176.4673215.140.7977.3733185.48629.6540.89751988726.615054733188.52494566855886.1413.57.386.47063187.9820.7656.6153157.0330.9520.89978449827.850129783160.1318777194886.1397.66.616.4763160.30.735.883127.93232.3680.90297210129.227400973131.07259986345852.93815.896.483131.0560.6265.2643100.6830.3760.88166871126.781568773104.27443196653852.9365.25.266.4843103.280.5994.6613071.35531.9250.87961171328.081603943075.198396106930852.9348.84.666.4893074.4870.5634.0973041.3333.1570.8821905329.250791423045.236209117203852.9331.84.16.49433044.490.5253.5753010.1434.350.88509738930.403095323014.086905高压缸总的有效焓降为：366.592kj/kg，高压缸总的理想焓降为：428.53kj/kg由调节级前熵s＝6.439kj/(kg*k)和第11级后压力3.575Mpa，得焓h＝2978.087kj/kg高压缸重热系数：α=(428.53-(3403.74-2978.087))/(3403.74-2978.087)=0.00676调节级的相对内效率：73.761/99.54=0.741高压缸效率：366.592/(3403.74-2978.087)=0.86125(2)汽轮机高压缸各级效率逐渐增加，直到第一次回热抽汽后降低至最小，后又逐渐增加。原因：由于叶片变大，所以叶高损失变小，所以效率提高，直到有回热抽汽后，损失变大，所以效率降低。(3)高压进汽压力损失：16.7－16.00＝0.7Mpa再热器压力损失：3.575－3.12＝0.455Mpa中低连通管压力损失：0.791－0.769＝0.022Mpa中压缸级号内功率Pi级流量D0级前温度级前压力级前比熵级前焓全级压差级后压力级后焓级理想焓降级的相对内效率有效焓降级实际出口焓值17818747.6533.63.127.3123531.440.3362.7843490.37541.0650.91676293637.646869983493.7931328052758515.32.187.4353500.0230.3141.8663445.22954.7940.69791744838.241688653461.78131138520758496.42.477.3213454.9060.3022.1683410.12144.7850.90352528640.464379953414.4416248831758476.42.167.3283414.1680.2811.8793367.57446.5940.90014647441.94142483372.22657559158758455.51.887.3333371.690.2571.6233323.93447.7560.91076428343.49445913328.19554169062724.9433.71.637.33623327.480.2361.3943278.22449.2560.91366990145.003724653282.47627579651724.9411.11.397.3433282.070.221.173229.6152.460.91362595847.928817773234.141182810276724.9386.91.177.34933233.650.2030.9673177.76955.8810.91323874251.032694163182.617306910666724.9361.10.977.3553182.2960.1790.7913124.87757.4190.92250845652.969513043129.326487中压缸总的有效焓降为：398.7236kj/kg由第1级前熵7.312和第9级后压力0.791Mpa得出h＝3099.103kj/kg中压缸的效率：398.7236/(3531.44-3099.103)=0.92225低压缸级号内功率Pi级流量D0级前温度级前压力级前比熵级前焓全级压差级后压力级后焓级理想焓降级的相对内效率有效焓降级实际出口焓值18610329.3331.40.7697.3663123.7080.2420.5273024.2599.4580.86758229686.2883037.4229197329.3287.20.5417.3773037.040.1910.352931.46105.580.79243701583.66552953.374538585294.3243.70.3727.39052952.60.140.2322847.434105.1660.8986459594.5072858.09349262275.4194.50.2327.413432858.0930.10.1322745.626112.4670.910542648102.4062755.68758297258.5140.80.1337.434242755.60.06080.07222624.28131.320.843055133110.712644.89613222246.591.80.07367.4652662.6830.05270.02092522.65140.0330.845514986118.42544.283711308246.563.40.02287.8652614.860.01680.0062463.047151.8130.830956506126.152488.711个低压缸总的有效焓降为：722.1265kj/kg由第1级前的熵7.366和第7级后压力0.006Mpa得出h＝2268.757kj/kg低压缸的效率：722.1265/(3123.708-2268.757)=0.84464(4)dmu级的理想焓降ca速比相对内效率调节级0.741高压缸第一级0.85419134.175924.59221.76560.6050340.8864高压缸末级1.07704169.18134.35262.10680.6454660.8851中压缸第一级1.08966171.163441.065286.58330.5972550.916763中压缸末级1.32736208.501257.419338.87760.615270.92251低压缸第一级1.91978301.558399.4584460.676140.9464低压缸末级2.64287415.141191.813619.37550.6702570.861沿蒸汽膨胀流程，级焓降逐渐增大。原因：沿蒸汽膨胀流程，轮周速度增大，为了保证较高的效率，各级均应在最正确速比附近工作，那么相应的理想焓降要增大。(5)(6)Pe发电机功率Pel发电机效率汽耗率给水焓热耗率汽轮机总理想焓降凝结水的焓总的吸收热量循环热效率汽轮机总实际焓降内效率热效率3085723000000.9722204223.034011146.9648122.0191712.941137.773779.3130.453241487.440.868360.38264习题3〔1〕设平衡活塞压力为P1，平衡活塞汽封漏气量为DG1假设调节级和平衡活塞汽封均未达临界查得，\假设成立C时，查得〔2〕\D=495mm的轴封最后一片达临界查得\D=432mm的轴封未达临界查得〔3〕大气环境：\最后一个轴封未达临界〔4〕方法如〔1〕解得〔5〕〔6〕要外泄，那么作业4〔1〕〔2〕假设调节级为临界工况时，算开阀点4后的流量：每个汽门流量为：阀点5：每个汽门流量为：〔3〕由0’点等熵交P2于点2t，得h2t=3288.99kJ/kg阀点1：由0’点等熵交P21于点2t1，得h2t1=2888.09kJ/kg阀点2：阀点3：阀点4：阀点5：〔4〕查得：5个调门全开：斥汽损失忽略不记3个调门全开：速度三角形：5个阀门全开：3个阀门全开：图略。〔5〕\由第〔2〕问知，前三个阀门全开，第四个阀门局部开启

习题5〔1〕由第4题知，调节级效率为高压缸非调节级：〔2〕〔3〕〔4〕〔5〕〔6〕由〔2〕知，〔7〕

习题6〔1〕〔2〕〔3〕〔4〕〔5〕

习题7〔1〕其中由凝汽器压力不高于4.9kPa，对应的蒸汽饱和温度为32.54C，由此求得凝汽器的传热端差不大于3.09C。由循环倍率计算出循环冷却水流量，并计算出冷却管数〔2〕改为双压凝汽器后，高、低压凝汽器的传热面积与凝汽器相同。由于原凝汽器为双流程，改双压运行后，变为单流程运行。这样，高、低压凝汽器的总传热面积和冷却水的流速均与原方案一致。对于高、低压凝汽器，由于现为单流程凝汽器，故增加流程修正项对高压凝汽器，；对低压凝汽器其中：Why?Why?查得：凝汽器平均为4.669kPa，较单压凝汽器降低0.2315kPa。凝结水的混合温度31.565C，习题8解：(1)基于汽轮机调节系统静态特性，一次调频的功率增量，当电网频率由50.0Hz升高至50.1Hz时，对1000MW机组，功率增量对600MW机组，功率增量/