Assume ideal fluid in a frictionless pipe at same horizontal level. If we apply an external force and then remove it.
1.Will the fluid keep flowing with the same velocity?
Please explain in simple words. I am an 11th grader.Please help me.
1)yes, just like for bigger objects.
2)Pressure at a point is defined to be the
$$\sum_{i=0}^n F_i/A_i$$
where $F_i$ is some external force(usually excluding gravity). You can imagine this force as a person squeezing the pipe. Additionally, When we're studying fluid flows through pipes(especially in 11th grade), we assume zero velocity in the radial direction of the pipe. And as for your twin question, No that is wrong. The pressure is the same along a level only in fluid statics. This is just Newton's second law, Consider a small cylinder in the fluid. Assuming the fluid is at rest, apply Newton's second law to this piece of fluid. We get
$$P_1S – P_2S = 0$$
$$P_1 = P_2$$
where S is the area of the cross-section of the cylinder and $P_1$ and $P_2$ are the pressures at two ends of the cylinder. You can see that in case the fluid is not static, we wouldn't get 0 on the right-hand side of the equation above. We would then have to equate that with the time derivative of velocity, you would get the Navier-Stokes equation conserving momentum.
$$\nabla p = \partial (\rho v)/ \partial t + \rho g$$
applying your laminar flow conditions, i.e no viscosity, constant force ($F_i$), and most importantly $$\partial \rho / \partial t = 0$$
Is this satisfactory or do you need me to derive the Navier-stokes and Bernoulli's equation too?
we obtain the fundamental fluid equation in the form of an 11th grader (Bernoulli's equation)
$$P_1 + \rho {v_1}^2/2 + \rho g{h_1} = P_2 + \rho {v_2}^2/2 + \rho g{h_2}$$
