Question

A 0.8-km concrete lined canal carries a discharge of 1.20 m³/s. Determine the head loss along the channel if the width is 2 m and the flow depth is at 1.2 m. Assume n = 0.018

Answer 1

Solution:- the values given in the question are as follows:

length of canal(L)=0.8 km , or 800 m

discharge through canal(Q)=1.2 m^3/s

width of canal(B)=2 m

depth of water(y)=1.2 m

manning's canstant(n)=0.018

discharge(Q)=velocity(v)*area(A)

velocity of flow(v)=discharge(Q)/area(A)

velocity of flow(v)=1.2/(B*y)

velocity of flow(v)=1.2/(2*1.2)

velocity of flow(v)=0.5 m/s

According to lacy's theory-

velocity(v)=[(Q*f^2)/140]^(1/6) , [Eq-1]

where, f=friction factor(silt factor)

values put in above equation-(1) and calculate the values of f

0.5=[(1.2*f^2)/140]^(1/6)

0.5^6=[(1.2*f^2)/140]

140*0.5^6=1.2*f^2

f^2=(140*0.5^6)/1.2

f^2=1.8229167

f=1.35015

Calculating head loss due to friction factor-

head loss for canal(H_L)=[(f*L*v^2)/(2g*P)] , [Eq-2]

where, P=wetted perimeter of canal=B+2y

P=2+2*1.2=4.4 m

values put in above equation-(2) and calculate head loss(H_L)

head loss(H_L)=1.35015*800*0.5^2/(2*9.81*4.4)

head loss(H_L)=3.12795 m

[Ans]