资源描述
表CG-10. 保温管道在大气21时输送饱和蒸汽时的凝结水量(假定保温效率为75%)% G% j) RR4 y4 M压力MPa0.1030.210.410.861.241.733.14.16.2公称通径DN管子外表面积m2/m每米管线每小时凝结水量Kg/m.h25320.1050.1320.070.090.090.100.100.130.150.180.180.210.210.250.2770.3440.3290.4060.4300.534400.1510.100.120.150.210.240.280.3880.4810.60450650.1900.2300.120.150.150.180.190.220.250.300.300.360.340.420.4760.5710.5640.6760.7410.887800.2780.180.210.270.380.420.490.6850.8131.083901000.3190.3590.190.220.240.270.300.330.400.450.480.540.570.640.7740.8600.9181.0211.2011.3351250.4440.270.330.400.550.650.781.0351.2251.6041502000.5290.6890.300.400.370.480.480.610.650.820.760.980.881.131.2041.5641.4271.8511.8652.4232500.8570.450.580.761.011.161.401.8362.2753.0053003501.0181.1190.570.680.680.760.860.971.191.291.371.531.651.802.2902.5122.7102.9753.5613.9054001.2800.700.851.101.471.772.052.8683.3854.4804505001.4361.6000.790.860.951.061.261.361.651.831.952.162.282.533.2013.5523.7954.2124.9875.5446001.9151.011.251.622.182.533.024.2165.0086.588c4 ?2 T4 e. n 表CG-11. 21时的预热负荷,管道规格为40。& m0 # Y. Y/ s( a9 q压力MPa0.0140.1030.2070.4130.8611.2401.723公称通径DNm2/mkg/mh25322.523.380.0450.0600.0550.0740.0640.0850.0760.1010.0940.1260.1060.1410.14180.158404.050.0710.0880.1080.1220.1500.1750.18950655.438.620.0970.1550.1190.1880.1370.2170.1640.2590.2020.3200.2280.3900.2540.4038011.270.1980.2480.2880.3380.4200.4700.5279010013.5616.060.2410.2830.2950.3480.3410.4030.4060.4810.5040.5950.5670.6710.6340.75212521.780.3480.5240.6040.6530.8100.9111.01815920028.2342.490.4990.7500.6150.9230.7081.0710.8471.2801.0491.5771.1831.7711.3131.99425060.241.0871.3101.5131.8012.2322.5152.81930035079.7793.761.4061.6521.7412.0392.0092.3512.3962.8132.9763.4823.3343.9293.7354.375400123.522.1732.6043.0953.7084.5345.1845.774450500156.26183.052.7533.2293.3933.9883.9144.5844.6885.4915.8046.8016.5487.6647.2928.557600254.484.4045.5366.3847.8349.45010.64111.908: L. k- W0 N6 x- N9 ?2 c B! H 6 p 表CG-12. 管道重量,kg/m$ 1 o4 n+ O5 f( H- g公称通径DN规格40#规格80#规格160#160#25322.523.533.234.464.245.605.457.75404.055.407.239.5450655.438.627.4711.4011.0914.9013.4420.378011.2715.2521.3127.659010013.5616.0618.6222.29-33.6334.0140.9812521.7630.9249.0557.3715020028.2342.492.5264.5767.41111.1779.11107.7725060.2481.46172.63-30035079.7793.76131.85159.24239.60282.76-400123.52203.88364.61-450500156.26183.05254.48311.03459.85546.02-600254.48441.99806.60-第五节 热量衡算与热交换计算 日期:2008-4-5 3:20:36 来源:来自网络 查看:大 中 小 作者:不详 热度: 2522一、热量衡算传热计算根据总传热方程进行:Q=KAtm对于一个热交换器,传热计算的内容有两种,一为设计计算,即根据给定的传热量,确定热交换器的几何尺寸和结构参数;二为校核计算,即对某些热交换器,根据它的尺寸和结构进行校核,看其能否满足传热量的要求。这两种计算的关键都在于传热面积是否合适,计算的基本依据是总传热方程以及与之相关的热量衡算式,在第四节中,已对总传热方程进行了较为详细的讨论,下面介绍热交换中的热量衡算式。当热损失为零时,对热交换器作热量衡算可得到单位时间的传热量,此传热量又叫热负荷,即式3-20中的传热速率Q。热负荷分为两种,即工艺热负荷和设备热负荷,工艺热负荷是指工艺上要求的在单位时间内需要对物料加入或取出的热量,用QL表示,单位为W。设备热负荷是热交换器所具备的换热能力,所以设备热负荷也就是热交换器的传热速率Q。当热损失不可忽略时,为满足工艺要求,Q应大于QL。由热量衡算得到的是工艺热负荷QL。如果流体不发生相变化,比热取平均温度下的比热,则有:QL=whcph(T1-T2)=wccpc(t2-t1) (3-29)式中 w-流体的质量流量,kg/s;cp-流体的平均定压比热,kJ/(kgK);T-热流体温度,K;t-冷流体温度,K;(下标h和c分别表示热流体和冷流体,下标1和2表示热交换器的进口和出口)式3-29是热交换器的热量衡算式,也称为热平衡方程。若流体在换热过程中有相变,例如饱和蒸汽冷凝成同温度冷凝液时,则有:QL=whr=wccpc(t2-t1) (3-30) 式中 wh-饱和蒸汽的冷凝速率,kg/s;r-饱和蒸汽的冷凝潜热,kJ/kg;当饱和蒸汽在热交换器中冷凝后,冷凝液液温度继续下降到T2,两部分热量(即潜热和显热)要加起来计算,这时:QL=whr+cph(Ts-T2)=wccpc(t2-t1)式中 cph-冷凝液的比热,kJ/kgK;Ts-冷凝液饱和温度,K。为满足工艺要求,应该使热交换器的传热速率等于或略大于工艺热负荷,即:QQL (3-31)二、热交换计算将总传热方程与热量衡算式结合起来,可解决热交换计算中设计型和校核型的问题,分别举例如下:例3-7 在列管式热交换器中,用饱和温度为126的蒸汽将470m3/hr的某一溶液从40加热到45,采用38*3*2000mm的钢管(=46.5w/mK),试计算所需传热面积的管数n.已知:水蒸汽 1=11600w/m2K;某溶液2=3700w/m2K;=1320kg/m3;Cp=3.4kJ/kgK;解:以平均传热面积Am计算。(1)据式(3-29)某溶液470m3/hr 由 40升至45所吸收的热量为:QL=47013203.4(45-40)3106W(2)据式(3-28)求 tm126-126 由于一侧发生相变,故并流与逆流是相同的。40-45 t1=126-40=86t2=126-46=81 (3)据式(3-24)求得Km代入数据化简:解得:Km=2100w/m2K(4) 求A及n例3-8 某酒精蒸汽冷凝器的传热面积为5m2,酒精蒸汽冷凝温度为78,冷凝热为845KJ/kg。冷却水初温为20,终温为35,若K=800w/m2K,问此冷凝器能否冷凝350kg/hr的酒精蒸气?解:(1)求QL,据式(3-30)有:QL=Whr=(350/3600)84510382kw(2)求tm78-78 t1=78-20=5820-35 t2=78-35=43t1与t2没有相差一倍,可用算术平均值:tm=(58+43)/2=50.5(3)求QQ=KAtm=800550.5202 kwQQL 故能够冷凝。例3-9 水以2m/s的速度流过内径25mm,长5m的薄壁钢管,管外用某气体冷凝方式供热,其温度为350K,气很大。水的进出口温度分别为290K及295K。若水的流速增加一倍,则出口温度是多少?(tm可用算术平均值,管壁热阻可忽略)解:此题涉及到流速(即流量)变化前后的问题,必须找到变化前后K、tm之间的关系才能解答。(1)变化前后K的关系(设变化前为下标1,变化后为下标2)Re1=(du)/=(25*10-3*2*1000)/(1.005*10-3)=49751Re2=2Re1 均满足Dittus公式据式(3-17):1=A u10.8/d0.22=A (2u1)0.8/d0.2两式相比:1/2=(1/2)0.8=1/1.74再据:(3-22)因而 K1水=1 同理 K2=2亦即:K1/K2=1/1.74 (2) 变化前后tm的关系350-350 t1=350-290=60Ktm1=(60+55)/2=57.5K290-295 t2=350-295=55K350-350 t1=350-290=60Ktm2=(60+350-t2)/2=(410-t2)/2290-t2 t2=350-t2(3)求t2 流速增大1倍,u2=2u1则w2=2w1,因而:Q1=w1CP(295-290)=K1Atm1Q2=2w1CP(t2-290)=K2Atm2两式相比得:
展开阅读全文