水力学第六章答案（吕宏兴裴国霞等）.doc

'第六章明渠恒定均匀流6.2已知h=1.2m,b=2.4m,m=1.5,n=0.025,i=0.0016求v和Q解：A=（b+mh)h=(2.4+1.5×1.2)×1.2=5.04m2=b+2h=2.4+2×1.2×=6.73mR==0.75mV=C==×=1.32(m3/s)Q=vA=1.32×5.04=6.65(m3/s)6.4已知：n=0.028,b=8m,i=,求Q（断面为矩形m=0)=2(－m)=2==4mAm=bmhm=8×4=32=＋2=16m==2mC==×=40.09Q===20.28(m3/s)6.7已知：矩形渡槽，b=1.5m,=116.5m,Q=7.65m3/s,D进口=52.06mh=1.7m,求i及D出口Q=ACi=A=bh=1.5×1.7=2.55m2=b+2h=1.5+2×1.7=4.9mR===0.52mC===69所以i==0. D出口=D进口－i=51.64m6.15解：=b+h=3.677m=1.5m===0.02678A=(b+b+mh)×=3.56m2=b+h+h=5.177mR==0.688mC===35.09()Q=AC=3.56×35.09×=4.63(m3/s)6.14已知：b=3.4m,梯形m=1.5,i=,渠底至堤顶的高差为3.2m(2)安全超高a=0.5m,Q=67m3/s解：（2）h=3.2-0.5=2.7mA=(b+mh)h=(3.4+1.5×2.7)×2.7=20.12㎡=b+2h=3.4+2×2.7×=13.1mR==1.54mC=查表取n=0.03C==35.82()Q=AC=20.12×35.82×=11.09m3/sv==0.55m/sv不冲=vR×=0.95m/s>v不冲（3）当h=1.5m时A=(b+mh)h=(3.4+1.5×1.5)×1.5=8.48㎡=b+2h=3.4+2×1.5=8.81mR===0.96m 当n=0.03,C===33.1()Q=AC=8.48×33.1×=3.41m3/sv===0.4m/s=0.5m/s>v所以在此条件下会发生淤积'