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As far as I know, the scalar velocity should be :
V = (u²+v²)^0.5

where u and v are the velocity components in x and y directions. So if I have u = -0.037 and v = 0, I should have V = 0.037. In a simple rectangular case for the same mentioned values of u and v , V was 0.061 as shown in the attached graph. Is there something I understood wrongly.

Hello,

Looking at your figure, I have the feeling that u (VELOCITY_U) = -0.061, do not you?

Chi-TUan

yes, you are correct, I was wrong, but why I am getting this value, although the giving flow rate is 1.85 m³/s and the cross sectional area is 50 m², so I should have a velocity of 0.037 m/s

Hi
You wrote the cross section area is 50 m2 but did you verify this value in your result?

Regards

It is actually a very simple case of a rectangular channel with bottom width 50 m , you know in a rectangular channel we give just the bottom and the water level , if you look to the figure here you will see the plan of the channel and just after it you will see the level of the bottom and the free surface ( 0 and 1 respectively) which means that the water depth is 1.0 m , also the velocity for the whole domain is shown in addition to the velocity at the upstream, the given value of the flow as mentioned was 1.85 m3/s. Attached also are the steering file, the condim file, and the geo file.