QUESTION 12 1 What is the Reynolds number for water flowing

QUESTION 12

1. What is the Reynold\'s number for water flowing in a 24-inch circular pipe at 2. Water is flowing in a 2/2-inch pipe with a discharge of 0.150 cfs. Find the 3. Water is flowing in a 4-inch pipe with a discharge of 4.0 × 10-3 cts. Find the 4. Water is flowing in a 300-mm circular pipe with a discharge of 0.250 m\'ls. a velocity of 6.21 ft/s? Is the flow laminar or turbulent? Reynold\'s number for this flow. Is the flow laminar or turbulent? Reynolds number for this flow. Is the flow laminar or turbulent? Find the Revnold\'s number for this flow. Is the flow laminar or turbulent? 5. What is the velocity head of the flow in problem 1? 6. Water in a pipe has a pressure of 22.1 lb/ft. what is the pressure head? 7. Water in a pipe has a velocity of 4.25 ft/s. What is the velocity head? 8. Water is flowing in an 8-inch-diameter new cast iron pipe with a velocity of 7.49 ft/s. The friction factor is 0.0215. What is the friction head loss over a length of 115 feet? 9. Water is flowing in a 36-inch concrete pipe (finished surface) with a velocity 10. Water is flowing in 2-inch tubing with a velocity of 1.35ft/s. What is the fric- 11. A reservoir of water is connected to an 8-inch-diameter pipe 50 feet long of 12.5 ft/s. What is the friction head loss over a length of 225 feet? tion head loss over a length of 25 ft? and discharging freely. The water surface elevation of the reservoir is 410.0, and the elevation of the center of the pipe at its discharge end is 372.5. Neglecting friction, determine the velocity, v, and discharge, Q, at the end of the pipe. 12. Find the velocity, v, and discharge, Q, for the pipe in problem 9 if the pipe is composed of new cast iron and friction is considered.

Solution

Solution:-

(1) Reynold number = VD/

V =velocity

D = diameter of pipe

=kinematic viscosity = 1.004 *10^(-6) m²/s

= 1.02 *10^-5 feet²/s

Reynold number (Re) = 6.21 *2/(1.02*10^-5)

= 1.21 *10⁶

Re is greater than 4000 so turbulent flow

(2) Re =VD/

           = QD/(A*)

            = Q*D/(/4*D²*)

            = 4 Q/(D)

Re = 4*0.15/(*2.5/12*1.02*10^-5)

        = 89875

Re is greater than 4000 so flow is turbulent flow

(3) Re = 4Q/(D)

           = 4*4*10^-3 /(*4/12*1.02*10^-5)

         = 1497.92

Re is less than 2000 so flow is laminar

(4) Re = 4 *0.250/(*0.3*1.004*10^-6)

       = 1.056 *10⁶

Re is greater than 4000. So Flow is turbulent.

(5)

Velocity head =V²/(2g)

                         = 6.21²/(2*32) = 0.602 feet Answer

(6) Pressure head = P/(pg)

                               = 22.1/(1.94*32) = 0.355 feet

Where p = density of water = 1.94 lb/feet³

(7) Velocity head = V²/(2g)

= (4.25)²/(2*32) = 0.282 feet

(8) Darcy weisback formula

H_L = f LV²/(2gD)

     = 0.0215 *115*(7.49)²/(2*32*8/12)

    = 3.25 feet

(9)   f= 0.011 (for concrete pipe)

H_L = 0.011*225*(12.5)²/(2*32*3)

     = 2.014 feet

(10) Roughness of tube (f) = 0.05

H_L = f L V²/(2gD)

       = 0.05*25*(1.35)²/(2*32*2/12)

      = 0.213 feet