A tank has a nizzle of exit diameter `D_1` at a depth `H_1` below the free surface. At the side opposite to that of nozzle 1, another nozzle is proposed at a depth `H_1/2`. What should be the diameter `D_2` in terms of diameter `D_1` so that the net horizontal force on the tank is zero?

SOLUTION:

Let,

`F_1` be the force due to nozzle 1

`F_2` be the force due to nozzle 2

For `D_1`,

`F_1=d/(dt)(mV)`

`=rho*Q*V`

`=rho*A_1*(V_1)^2`

`=(pi rho)/4*(D_1)^2*(V_1)^2`... (i)

Similarly the force due to nozzle 2 is,

`F_2=(pi rho)/4*(D_2)^2*(V_2)^2`... (ii)

Also,

`V_1=root ()(2gH_1)`

And `V_2=root ()(2gH_2)`

Since `H_2=H_1/2`,

`V_2=root ()(gH_1)`

According to questions,

`F.... Show More

SOLUTION:

Let,

`F_1` be the force due to nozzle 1

`F_2` be the force due to nozzle 2

For `D_1`,

`F_1=d/(dt)(mV)`

`=rho*Q*V`

`=rho*A_1*(V_1)^2`

`=(pi rho)/4*(D_1)^2*(V_1)^2`... (i)

Similarly the force due to nozzle 2 is,

`F_2=(pi rho)/4*(D_2)^2*(V_2)^2`... (ii)

Also,

`V_1=root ()(2gH_1)`

And `V_2=root ()(2gH_2)`

Since `H_2=H_1/2`,

`V_2=root ()(gH_1)`

According to questions,

`F_1=F_2`

`(D_1)^2*(V_1)^2=(D_2)^2*(V_2)^2`

`(D_1)^2*(2gH_1)=(D_2)^2*(gH_1)`

`D_2=root ()(2)*D_1`

`D_2=1.414D_1`