Question

Water flows from a large 4m detention basin (100m by 50m) along a 200mm diameter pipe for a distance of 100m at a slope of 1:200. The pipe centerline is 300mm above the base of the basin and it drains freely to a nearby channel. Calculate the steady flow rate if: (a) the flow is assumed ideal, (b) the pipe is considered smooth for the purposes of the Moody Diagram. (c) the pipe has a relative roughness of 0.006. (d) the time to empty the basin for case (c).

Answer 1

Ans) Given

Length of tank = 100 m

Width of tank = 50 m

Depth of tank = 4 m

Diameter of pipe = 200 mm or 0.20 m

a) Since, flow is ideal, no head loss is present

Apply Bernoulli equation between point 1 at water surface of tank and point 2 just outside the pipe,

P₁/g + V₁²/2g + Z1 =   P₂/g + V₂²/2g + Z2

where, P1 and P2 are pressures at point 1 and 2 respectively

V1 and V2 are velocity at point 1 and 2 respectively

Z1 and Z2 are elevation from datum

Since point 1 and 2 are open to atmpsphere , pressure present is only atmospheric pressure , hence gauge pressure P1 =P2 = 0

Also, velocity at surface is zero , therefore V1 = 0

Let pipe centerline be datum, therefore, Z1 = 4 + 0.30 = 4.30 m and Z2= 0

=> 0 + 0 + 4.30 = 0 + V₂²/2g + 0

=> V₂ = ( 2 x 9.81 x 4.30)^1/2

= 9.18 m/s

We know, flow rate(Q) = A x V

=> Q = (/4) x 0.20 x 0.20 x 9.18

=> Q = 0.287 m³/s

b) Now, consider minor loss at entry and exit

P₁/g + V₁²/2g + Z1 =   P₂/g + V₂²/2g + Z2 + H_m

0 + 0 + 4.30 = 0 + V₂²/2g + 0 + 0.5V₂²/2g + V₂²/2g

=> V₂ = (2 x 9.81 x 4.30/2.5)^1/2

=> V₂ = 5.81 m/s

=> Q = (/4) x 0.20 x 0.20 x 5.81

=> Q = 0.182 m³/s

c) Now consider friction loss as well

P₁/g + V₁²/2g + Z1 =   P₂/g + V₂²/2g + Z2 + H_f + H_m

    0 + 0 + 4.30 = 0 + V₂²/2g + 0 + fLV₂²/ 2gD + 0.5V₂²/2g + V₂²/2g

To determine Darcy friction factor 'f' we have to first calculate Reynold number (Re),

Re = V D/

where, = dynamic viscosity of water ( 10^-3 Ns/m²)

Find approximate Re by using V = 5.81 m/s from part(b),

Re = 1000 x 5.81 x 0.20 / 0.001

Re = 1162000

Using Moody diagram, we can find friction factor 'f' for given value of relative roughness (e/D) and Re ,

For e/D = 0.006 and Re = 1162000 , f = 0.031

=> 4.30 = 2.5 V₂²/2g + 0.031(100)(V₂²)/ 0.20 x 2g

=> 4.30 = 2.5 V₂²/2g + 15.50 V₂²/2g

=> V₂ = (2 x 9.81 x 4.30 /18)^1/2

=> V₂ = 2.165 m/s

Therefore, Q = (/4) x 0.20 x 0.20 x 2.165

Q = 0.068 m³/s

d) Volume of water in tank = 4 x 100 x 50 = 20,000 m³

Discharge = 0.068 m³/s

Time required to empty the basin = 20000/0.068 sec

   = 294117.6 sec or 4902 min or 81.7 hr or 3.40 days