Question

Water is conveyed thourth a horizontal tube 8 cm in diameter and 4 kilometer in length at the rate of 20 litre/s Assumming only viscous resistance , callcalute the pressure required to maintain the flow . Coefficient of viscosity of water is 0.001 pa s

Answer 1

The pressure required to maintain the flow is 32 k pa  

Explanation:

Given as :

The diameter of tube = 8 cm = 0.08 m

Radius of tube = 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

Answer 2

Water is conveyed thourth a horizontal tube 8 cm in diameter and 4 kilometer in length at the rate of 20 litre/s . Coefficient of viscosity of water is 0.001 pa s

Stepwise explanation is given below:

- It is given that,

The diameter of tube = 8 cm = 0.08 m

So, the Radius of tube = 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 = volume/rate

= 20.096 /20000

= 1milli sec

- So, velocity of rush = length of tube velocity / time od flow

4000/(10)^-3

=4 (10)^6. m/s =4000 km /s

- Now,

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

- By conservation of energy

P V = × × 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