0kJ/(kg•℃)Cp4=1.6×0.30+4.178×0.70=3.40kJ/(kg•℃)将以上数值带入焓平衡方程解得:W1=6220.92kg/hW2=5444.50kg/hW3=4704.99kg/hW4=4283.59kg/hD=7041.39kg/h求传热面积:(取K=2000W/(m2•℃))S1==140.55m2S2==152.67m2S3==151.27m2S4==156.88m2相对偏差εs=1->3%,故进行第四次核算重新分配有效温差S==149.39m2△t1'''==14.43℃,△t2'''==13.04℃△t3'''==11.59℃,△t4'''==10.15℃∑△t'''=14.43+13.04+11.59+10.15=49.21℃重新核算各效浓度及各沸点升高X1=0.5X2===0.40X3===0.34X4===0.30由经验公式:△i=1.78Xi+6.22Xi2得各效沸点升高△1,=1.78×0.5+6.22×0.52=2.45℃△2,=1.78×0.40+6.22×0.402=1.71℃△3,=1.78×0.34+6.22×0.342=1.32℃△4,=1.78×0.30+6.22×0.302=1.09℃重新核算各效加热蒸汽温度,二次蒸汽温度及各效沸点T1=120.2℃,t1=T1-△t1'''=120.2-14.43=105.77℃,T1'=t1-△1'=103.22℃T2=102.22℃,t2=T2-△t2'''=102.22-13.04=89.28℃,T2'=t2-△2'=87.57℃T3=86.57℃,t3=T3-△t3'''=86.57-11.59=74.98℃,T3'=t3-△3'=73.66℃T4=72.66℃,t4=T4-△t4'''=72.66-10.15=62.51℃,T4'=t4-△4'=61.42℃查表得: