下射式閘門水躍(Undershoot gate water jump)

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 明渠水理計算_曼寧公式

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例題

已知條件:y0=1m，y1=0.1m

連續方程式(C.E.)：V0*y0=V1*y1

        ==> V0=V1*y1/y0=0.1*V1 

        ==> V0^2=0.01*V1^2 ...............................(1)                                         

能量方程式(E.E.)：V0^2/(2g)+y0=V1^2/(2g)+y1

        ==>(V1^2-V0^2)/(2g)=y0-y1.....................(2)

        由(1)式可知，V0^2為V1^2之0.01，可省略

不計，故(2)式可簡化為

        ==>V1^2/(2g)=y0-y1=1-0.1=0.9

        ==>V1=SQRT(2*9.81*0.9)=4.202(m/s)

依水躍共軛水深公式：

y2=0.5*y1*{-1+SQRT[1+8*V1^2/(g*y1)]}

    =0.5*0.1*{-1+SQRT[1+8*4.202^2/(9.81*0.1)]}

    =0.552(m)

水躍能量損失:

△ E=(y2-y1)^3/(4*y1*y2)

      =(0.552-0.1)^3/(4*0.1*0.552)

      =0.418(m)

 

Undershoot gate water jump

example

Known conditions: y0=1m, y1=0.1m

Continuity Equation (CE): V0*y0=V1*y1

        ==> V0=V1*y1/y0=0.1*V1 

        ==> V0^2=0.01*V1^2 ............................(1)                                         

Energy Equation (EE): V0^2/(2g)+y0=V1^2/(2g)+y1

        ==>(V1^2-V0^2)/(2g)=y0-y1..........(2)

        It can be seen from formula (1) that V0^2 is 0.01 of V1^2 and can be omitted

Neglected, so (2) can be simplified as

        ==>V1^2/(2g)=y0-y1=1-0.1=0.9

        ==>V1=SQRT(2*9.81*0.9)=4.202(m/s)

According to the water jump conjugate water depth formula:

y2=0.5*y1*{-1+SQRT[1+8*V1^2/(g*y1)]}

    =0.5*0.1*{-1+SQRT[1+8*4.202^2/(9.81*0.1)]}

    =0.552(m)

Hydraulic jump energy loss:

△ E=(y2-y1)^3/(4*y1*y2)

      =(0.552-0.1)^3/(4*0.1*0.552)

      =0.418(m)