参考答案

已知
q_m,1=1000kg/h，ω₁=0.4，ω₂=0.05，
则水分蒸发量W=q_m1×[(ω₁-ω₂)/(1-ω₂)]
=1000×[(0.4-0.05)/(1-0.05)]kg/h
=368.42kg/h$又知φ₀=60%，t₀=20℃，φ₂=80%，t₂=40℃
查饱和水蒸气表得，20℃时，p_s0=2.334kPa；40℃时，p_s2=7.375kPa
则H₀=0.622×[φp_s0/(p-φp_s0)]
=0.622×[(0.60×2.334)/(100-0.60×2.334)]kg水/kg绝干气
=0.009kg水/kg绝干气
H₂=0.622×[(φp_s2/(p-φp_s2)]
=0.622×[(0.80×7.375)/(100-0.80×7.375)]kg水/kg绝干气
=0.039kg水/kg绝干气
故q_m,L=W/(H₀-H₂)
=368.42/(0.039-0.009)kg绝干气/h
=12280.67kg绝干气/h
l=1/(H₀-H₂)
=1/(0.039-0.009)kg绝干气/kg水
=33.33kg绝干气/kg水$因风机装在预热器进口处，输送的是新鲜空气，其温度t₀=20℃，湿度H₀=0.009kg水/kg绝干气，则湿空气的体积流量
q_m,V=q_m,L(0.773+1.244H)×[(t+273)/273]
=12280.67×(0.773+1.244×0.009)×[(20十273)/273]m³/h
=10335.98m³/h$干燥产品量
q_m,2=q_m,1×[(1-ω₁)/(1-ω₂)]
=1000×[(1-0.40)/(1-0.05)]kg/h
=631.58kg/h