化工原理 固体干燥 计算题-答案（修订版）.docVIP

11.固体干燥-计算题（答案） 层次：A [2] j11a10082 (题分：10) (1)水分蒸发量W x1=w1/(1-w1)=0.40/(1-0.40) =0.667 kg水/kg x2==w2/(1-w2)=0.05/(1-0.05) =0.053 kg水/kg Gc=G1(1-w1)=1000(1-0.40)=600kg绝干料/h ∴W=Gc(x1-x2)=600×(0.667-0.053) =368.4kg水/h (2)L=W/(H2-H1)=368.4/(0.039-0.009) =12280kg绝干气/h L’=L(1+H1)=12280(1+0.009) =12390.5kg原空气/h (3)G2=G1-W=1000-368.4=631.6kg/h [3] j11a10084 (题分：10) 1.求W W=Gc(x1-x2) G1=1000kg/h=0.278kg/s ∴Gc=0.278(1-0.4)=0.167kg/s x1=w1/(1-w1)=0.40/(1-0.40) =0.667kg/kg绝干料 x2=w2/(1-w2)=0.05/(1-0.05) =0.053kg/kg绝干料 ∴W=Gc(x1-x2)=0.167(0.667-0.053) =0.1025kg/s 2.求H1 H1=0.622p/(P-p) =0.622×1.0/(101.3-1.0) =0.0062kg/kg干空气 3.求H2 H2=0.622ψp/(P-ψp) =0.622×0.7×7.4/(101.3-0.7×7.4) =0.0335kg/kg干空气 4.求L L=W/(H2-H1)=0.1025/(0.0335-0.0062) =3.76kg干空气/s 进口的湿空气量=L(1+H1)=3.76(1+0.0062) =3.783kg/s [5] j11a10101 (题分：10) 由露点温度 td = 15℃ lg pd = 7.12120－1204.67 / (15＋223.5) pd = 117.5mmHg 该体系温度下丙酮的饱和蒸汽压 ps = 117.5 / 0.4 = 293.75 mmHg 由lgps = A－B/ (t＋C)得：lg293.75 = 7.12120－1204.67 / (t＋223.5) 所以体系温度：t = 35.4 [6] j11a10102 (题分：10) 混合气分子量 饱和湿度 ?s = 50% p = 0.5×15=7.5 kN ? m2 故湿度为H = 7.5 ? (101.3－7.5)×18 ? 8.5 = 0.169 kg ? kg干气。 [7] j11a15081 (题分：15) (1)如答图所示 (2) Gc=G2(1-w2) =0.126(1-0.04)=0.121kg/s X1=42/58=0.724kg X2=4/96=0.0417kg W=Gc(X1-X2)=0.121(0.724-0.0417) =0.0826kg L=W/(H2-H0)=0.0826/(0.03-0.008) =3.76 kg (3)Qp=L(I1-I0) I1-I0=(1.01+1.88H)(t1-t0) =(1.01+1.88×0.008)×(93-21) =73 kJ/kg绝干气 ∴预热器供应热量 Qp=L(I1-I0)=3.76×73=274 kw 层次：B [8] j11b10003 (题分：10) （1） 饱和湿度为： kg水/kg干气 （2）60℃ 时饱和湿度 kg水/kg干气 （3）60℃ 时饱和湿度 kg水/kg干气 [9] j11b10006 (题分：10) 由