第十四章化工原理复习资料

解题思路：1.已知：t1=27℃，td1=22℃，t2=80℃，求：ϕ1，ϕ2解题思路：查水的饱和蒸汽压22℃2.668kPa27℃3.6kPa80℃47.38kPaϕ1=ϕ2=ϕPdPS1PdPS2ϕ1−ϕ2变化：ϕ12.已知：t1=65℃，tw1=40℃，t2=25℃，P=101.3kPa求：W水，Q解题思路：查水的饱和蒸汽压40℃7.375kParW=2401kJ/kg25℃3.168kPaHw=0.622pwP−pw1.09()H1=Hw−rwt1−tw1H2=HS2=0.622pSP−pSW水=H1−H2I1=（1.01+1.88H1)t1+2500H1I2=（1.01+1.88H2)t2+2500H2Q=I1−I23.已知：P=100kPa求：用焓——湿度图填充下表。解题思路：查焓——湿度图干球温度℃湿球温度℃湿度kg水/kg干空气相对湿度热焓kJ/kg干空气水汽分压kPa露点℃133−380400.031911.01654.832.560350.026201254.12940280.02043953.22557330.024211203.72850300.019625983.0234.已知：t1=80℃，H1=0.01kg水/kg干气，W水=0.1kg/S，θ=30℃，V干=10kg干气/S，忽略热损失求：（1）ΔI（2）t2（3）忽略ΔI，t2解题思路：（1）喷水后气体增加的焓即液体所带入的焓∆I=W水cpθ/V干=0.1×4.18×30/10=1.25kJ/kg干气（2）I1+ΔI=I2WH2=H1+V干=0.01+0.110=0.02kg水/kg干气∴（1.01+1.88H1)t1+2500H1+∆I=(1.01+1.88H2)t2+2500H2（∴t2=1.01+1.88H1)t1+2500(H1−H2)+∆I1.01+1.88H2（3）若忽略ΔI，则I1=I2（t2=1.01+1.88H1)t1+2500(H1−H2)1.01+1.88H25.已知：t1=30℃，td1=20℃，V湿=1000m3/h，W=2.5kg/h，t3=60℃，P=101.3kPa求：(1)t2，H2(2)ϕ3解题思路：（1）查水的饱和蒸汽压，20℃时Pd=2.338kPaH1=0.622PdP−PdυH1=(2.83×10+4.56×10−3H1)(t+273)134V湿∴V干=VHW水H2=H1−V干∵td2=t2∴H2=0.622得pS2pS2P−pS2查对应的饱和温度t2（2）查60℃，pS3=19.91kPa∴ϕ3=pS2pS36.已知：P=101.3kPa，t=25℃,ϕ=100%，X1*=0.02kg水/kg干料ϕ=40%，X2*=0.007kg水/kg干料,X=0.25，空气25℃，ϕ=40%求：自由含水量，结合水量，非结合水量解题思路：自由含水量=X-X2*结合水量=X1*非结合水量=X-X1*7.已知：N恒=1.1kg水/m2·h，GC=1000kg，A=55m2，X1=0.15kg水/kg干料，X2=0.005kg水/kg干料，X*=0，Xc=0.125kg水/kg干料，求：τ解题思路：∵X1XcX2∴干燥过程分恒速阶段与降速阶段两部分∴τ=τ恒+τ降=GCAN恒[(X1−XC)+XClnXC]X28.已知：浅盘n=50只，盘底面积70×70cm，厚度h=0.02m，ρ湿=1600kg/m3，X1=0.5kg水/kg干料，X2=0.005kg水/kg干料，X*=0，XC=0.3kg水/kg干料干燥条件：平行流过u=2m/s，t=77℃，ϕ=10%，N降∝(X-X*)求：τ解题思路：以一只盘为基准进行计算G=A·h·ρ湿=0.7×0.7×0.02×1600=15.68kgGC=G1+X1查焓—湿图，t=77℃，ϕ=10%时，H1351tw=38℃，rw=2411kJ/kg∴υH=(2.83×10−3+4.56×10−3H)(t+273)湿空气的密度ρ=1+HυH湿空气的质量流速G'=ρuα=0.0143(G')0.8N恒=α(t−tw)rw∴τ恒=GCA⋅X1−XCNAGCτ降=A⋅XCN恒lnXCX2∴τ=τ恒+τ降9.已知：t0=20℃，H0=0.01kg/kg干气,t1=120℃，t2=70℃，H2=0.05kg/kg干气,θ1=30℃，w1=20%,θ2=50℃，w2=5%，cps=1.5kJ/(kg·℃)，G2=53.5kg/h，Q损=0求：（1）V空（2）QP（3）QD解题思路：（1）X=w1X=w221−w11−w2GC=G2(1−x2)W=GC(X1−X2)V空=WH2−H1=WH2−H01361干(2)I1=(1.01+1.88H1)t1+2500H1I0=(1.01+1.88H0)t0+2500H0QP=V空(I1−I0)(3)I2=(1.01+1.88H2)t2+2500H2i2=(cps+cpLΧ2)θ2i1=(cps+cpLΧ1)θ1QD=V空(I2−I1)+GC(i2−i1)10．已知：P=100kPa，w1=0.5，w2=0.01，G=20kg/s，t0=25℃，H0=0.005kg水/kg干气，t2=50℃，ϕ2=60%，理想干燥器求：（1）V（2）t1（3）η解题思路：（1）X=w1X=w221−w11−w2GC=G(1−x1)查焓—湿图，t2=50℃，ϕ2=60%时，H2V=GC(X1−X2)H2−H1V=V干（1+H1）（2）∵理想干燥器∴I1=I2（1.01+1.88H1)t1+2500H1=(1.01+1.88H2)t2+2500H2（∴t1=1.01+1.88H2)t2+2500(H2−H1)1.01+1.88H1（3）η=t1−t2t1−t011．已知：P=100kPa，w1=0.20，w2=0.01，G=1.75kg/s，t0=20℃，tw0=16℃，ϕ2=70%，1371求：（1）一次预热t1=120℃，V，η（2）先预热至t1=120℃达ϕ2=70%后，再加热至t3=100℃，再达ϕ4=70%后排出，求V，η解题思路：（1）X=w1X=w221−w11−w2GC=G(1−x1)W水=GC(X1−X2)∵理想干燥器∴由I—H图查得：H0，H2，t2W水=V干（H2−H0）∴V=W水H2−H0×（1+H0）η=t1−t2t1−t0（2）t0t1=120℃t2t3=120℃t4预热干燥加热干燥H0H1H2ϕ4=70%ϕ2=70%有中间加热，理想干燥器由I—H图得H4=0.0615kg水/kg干气，t4=51℃V=W水(H4−H0)×（1+H0）η=(I1−I2')+(I3−I4')(I1−I0)+(I3−I2)H0=H1H2=H3∴I1−I0=(1.01+1.88H0)(t1−t0)∴I3−I2=(1.01+1.88H2)(t3−t2)1381干I2=(1.01+1.88H0)t2+2500H0∴I1−I2=(1.01+1.88H0)(t1−t2)I4=(1.01+1.88H2)t4+2500H2∴I3−I4=(1.01+1.88H2)(t3−t4)∴η=(1.01+1.88H0)(t1−t2)+(1.01+1.88H2)(t3−t4)(1.01+1.88H0)(t1−t0)+(1.01+1.88H2)(t3−t2)12．已知：理想干燥器，V循=0.8V废，H0=0.0033kg水/kg干气，t0=16℃t2=67℃，H2=0.03kg水/kg干气,G1=1500kg湿料/h，w1=0.47，w2=0.05求：V，Q预解题思路：X=w1X=w221−w11−w2GC=G(1−x1)V=GC(X1−X2)H2−H1解法1：∵循环后，空气用量不变时，Q予不变，η不变∴I0=(1.01+1.88H0)t0+2500H0I1=I2=(1.01+1.88H2)t2+2500H2∴Q予=V干(I2−I0)V=V干（1+H0）解法2：由混点作物料恒算得5Hm=H0+4H2得HmI1=I2∴（1.01+1.88Hm)t1+2500×Hm得t1，算Q干=(1.01+1.88H2)×t2+2500×H2139