问答题 按废物混合最适宜的C/N比计算:树叶的C/N比为50,与来自污水处理厂的活性污泥混合,活性污泥的C/N比为6.3。分别计算各组分的比例使混合C/N比例达到25。假定条件如下:污泥含水率=75%;树叶含水率=50%;污泥含氮率=5.6%;树叶含氮率=0.7%。
【正确答案】正确答案:对于1 kg的树叶: m =1×0.50 kg=0.50 kg m 干物质 =1 kg-0.50 kg=0.50 kg m N =0.50×0.007 kg=0.003 5 kg m C =50×0.003 5 kg=0.175 kg 对于1 kg的污泥: m =1×0.75 kg=0.75 kg m 干物质 =1 kg-0.75 kg=0.25 kg m N =0.25×0.056 kg=0.014 kg m C =6.3×0.014 kg=0.088 2 kg 计算加入到树叶中的污泥量使混合C/N比达到25。 C/N=25=[1 kg树叶中的C含量+x(1 kg污泥中的C含量)]/[1 kg树叶中的N含量+x(1 kg污泥中的N含量)] x为所需污泥的质量, 25=[0.175+x(0.088 2)]/[0.003 5+x(0.014)] x=0.33 kg 对于0.33 kg的污泥: m =0.33×0.75 kg=0.25 kg m 干物质 =0.33 kg-0.25 kg=0.08 kg m N =0.08×0.056 kg=0.004 kg m C =6.3×0.004 kg=0.03 kg 对于0.33 kg的污泥+1 kg的树叶: m =0.25 kg+0.5 kg=0.75 kg m 干物质 =0.08 kg+0.50 kg=0.58 kg m N =0.004 kg+0.003 5 kg=0.008 kg m C =0.03 kg+0.175 kg=0.205 kg 则C/N比为: C/N=0.205 kg(C)/0.008 kg(N)=25.6 则含水率为: 含水率=0.75 kg(水)/[0.75 kg(水)+0.58 kg(干物质)] =(0.75 kg/1.33 kg)×100% =56.39%
【答案解析】