Question

Water is flowing through a horizontal tube 4 km in length and 4 cm in radius at a rate of 20 litre calculate the pressure required to maintain the flow in terms of the height of mercury column

Answer 1

Explanation:

Volume of the tube = pi r^2 L = pi * 4/100 * 4/100 * 4000 m^3

Volume of water flown per second = 20 ltr = 20 * 10^-3 m^3 /s

So time of flow = pi * 4/100 * 4/100 * 4000 / 20 * 10^-3

==> 320 pi second

So velocity of rush = 4000 / 320 pi = 25/2pi m/s

Hence kinetic energy of liquid going out 1/2 * ρ * V * v^2

Force = Pressure * area

Work performed to push water through length of tube = Pressure * area * length = pressure * volume = P * V

By the conservation of energy P V = 1/2 * ρ  * V * v^2

Cancelling and plugging v you can get P = 125 * 625 / pi^2 Pa

Answer 2

The pressure required to maintain the flow is 32 k pa

Explanation:

Given as :

Radius of tube = 4 cm = 0.04 m

The length of the tube = l = 4 km = 4000 m

The rate of water discharge = 20 l/sec = 20000 ml/s

Coefficient of viscosity of water = 0.001 pa s

Let The pressure require to maintain the flow = P pascal

According to question

Volume of tube = V = π × r² × l

                                = 3.14 × (0.04)² × 4000

                                = 20.096  m³

∵  rate of water discharge = 20000  ml/s

So, Time of flow = $\dfrac{volume}{rate}$

                           = $\dfrac{20.096}{20000}$

                          = 1 milli sec

So, Velocity of rush = $\dfrac{length of tube}{tme of flow}$

                                = $\dfrac{4000}{10^{-3} }$

                                = 4 × $10^{6}$  m/s = 4000 km/s

Again'

work perform to push water through length of tube = P ×V

By conservation of energy

P V = $\dfrac{1}{2}$ × $\rho$ × V × v²

Or, P = 0.5 × 0.001 × ( 4000  )²

       = 32000

       = 32 k pa

So, The pressure required to maintain the flow = 32 k pa

Hence, The pressure required to maintain the flow is 32 k pa   Answer