Open Channel Flow - 7 [How to apply momentum equation in open channel flow]

Unit 5 part 7
Topics covered in this lecture are
1. how to apply momentum equation in open channel flow to get expression for resistance force.

[@priyaverma5189]

Saw alot of vedios but this was the best

[@abhijeet246]

Thanks priya

[@rashmijichkar6974]

You are best sir 😀

[@asziolaasyraflinazhif4504]

Excuse me Sir, why do these equations have different units? (N/m = N; So does the Q become q); Correct me if i am wrong. Thx

[@abhijeet246]

The equation is Dimensionally homogenous. i.e.the dimension on LHS = dimension of RHS LHS = F now all F are force terms, if you see the unit of pressure force it will come out as Newton itself N. RHS = beta * rho * Q * V if you write dimension you will get Beta = Dimensionless Rho = kg /m3 Q = m3/s V = m /s Therefore RHS = kg /m3 * m3/s * m /s RHS = kg m /s2 that is dimension of 1N (1N = 1kg * 1 m/s2) Therefore dimension of both sides of the equation is N.

[@abhijeet246]

The equation has same units on both sides. Newton N. I have derived the dimension below.. Please have a look.. If you have any doubts please feel free to comment. I will try to solve them. Thanks

[@Santon-Motho]

@@abhijeet246 Hello, I believe the confusion is that the LHS pressure forces you derived were only for a unit width of the channel, i.e. to get the complete pressure force you have to multiply the area of the pressure distribution (the triangle) by the width of the channel to get dimensional homogeneity. Or use q = m3/s/m instead of Q=m3/s on the RHS. I think that is the point Asziola was trying to make

[@abhijeet246]

@@Santon-Motho Thanks, I got the point. On the LHS, the Pressure Force P will come in N anyways, because as you rightly said we will multiply Pressure intensity x Area ( considering complete channel width). So on R H S we have considered Q and not q. So we will get N on both sides. I should have drawn the perspective of pressure distribution for better clarity.