Question

Air flows through a pipe at a rate of 124 L/s. The pipe consists of two sections of diameters 22 cm and 10 cm with a smooth reducing section that connects them. The pressure difference between the two pipe sections is measured by a water manometer.

Neglecting frictional effects, determine the differential height of water between the two pipe sections. Take the air density to be 1.20 kg/m^3.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The differential height is
`h= 1.458cm`

Step-by-step explanation:

The schematic diagram of the venturimeter is shown on the second uploaded image

The continuity equation is mathematically given as

`\r V = Av`

Where A is the area

`\r V` is the flow rate

`v_1` is the velocity

At the inlet making v the subject to obtain the inlet velocity we have

`v_1 = (\r V)/(A_1)`

Substituting 124 L/s =
`0.124 m^3 /s` for
`\r V` and
`A_1 = (\pi)/(4 ) *d^2 = (\pi)/(4) * 0.22^2` given that

`d =22cm= (22)/(100) = 0.22m`

So
`v_1 =(0.124)/((\pi)/(4) * 0.22^2 )`

`=3.26m/s`

At the exit point the velocity is

`v_2 = (\r V)/(A_2)`

Where
`A_ 2 = (\pi)/(4)d^2 = (\pi)/(4) * 0.10^2` given that
`d =10cm= (10)/(100) = 0.10m`

So
`v_2 = (0.124)/((\pi)/(4) *(0.10)^2)`

`= 15.78m/s`

The Bernoulli's flow equation between the inlet and exist is mathematically given as

`(P_1)/(\rho_o ) + (v_1^2)/(2) = (P_2)/(\rho_0) +(v_2^2)/(2) +gz_2 ---(1)`

And

`P_1 - P_2 = \rho_o [(v_2^2)/(2) + (v_1^2)/(2) ] = \rho_(water) gh ---(2)`

Where
`P_1` is the pressure at inlet

`P_2` is the pressure at exist

`\rho_(water)` is the density of water with value of
`1000kg/m^3`

g is acceleration due to gravity

h is the height of the water column

making h the subject in the equation 2

`h = \rho _o (v_2^2 - v_1^2)/(2g\rho_(water))`

where
`\rho _o` is the density of air given in the question

Substituting value

`h = (1.20) (15.78^2 - 3.26^2)/(2 (9.81 )(1000))`

`h= 1.458cm`