(3)干燥器的热效率 Q1+Q2 100% Q1=m(2492+182-4.187m) ≈m0Cm2 Q=Q4+ Th,+2 tVh +moCmt+o +moMl+ c=10094185X @+o 9.899×(2492+1.8850-4187×25)+10×(1005185×0005)(50-25) 10[(1005185×005)×50-(1005+1.85×0005)×258 ⑶干燥器的热效率 1 2 100% + = Q Q Q t (2492 1.88 4.187 ) 1 w 2 m1 Q = m + t − t 2 0 2 2 1 ( ) Q m c t t  − m m m Vh0 + Qp =Vh1 Vh m c t Q Vh m c t Q 1 0 1 1 2 0 2 2 + + = + + m d m L Q Q Q = +d p c X = + 1.005 1.85 ( ) ( )( ) ( ) ( ) 1 2 9.899 2492 1.88 50 4.187 25 10 1.005 1.85 0.005 50 25 10 1.005 1.85 0.05 50 1.005 1.85 0.005 25 t d Q Q Q  + =  +  −  +  +  − =   +   − +    