state and prove bernoulli's theorem for a liquid in streamline flow
Answers
P^2/rhoxg +v^2/2g +z = contant
Bernoulli's theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar.
Let us consider two different regions in the above diagram. Let us name the first region as BC and the second region as DE. Now consider the fluid was previously present in between B and D. However, this fluid will move in a minute (infinitesimal) interval of time (∆t).
If the speed of fluid at point B is v1 and at point D is v2. Therefore, if the fluid initially at B moves to C then the distance is v1∆t. However, v1∆t is very small and we can consider it constant across the cross-section in the region BC.
Similarly, during the same interval of time ∆t the fluid which was previously present in the point D is now at E. Thus, the distance covered is v2∆t. Pressures, P1 and P2, will act in the two regions, A1 and A2, thereby binding the two parts. The entire diagram will look something like the figure given below.
Finding the Work Done
First, we will calculate the work done (W1) on the fluid in the region BC. Work done is
W1 = P1A1 (v1∆t) = P1∆V
Moreover, if we consider the equation of continuity, the same volume of fluid will pass through BC and DE. Therefore, work done by the fluid on the right-hand side of the pipe or DE region is
W2 = P2A2 (v2∆t) = P2∆V
Thus, we can consider the work done on the fluid as – P2∆V. Therefore, the total work done on the fluid is
W1 – W2 = (P1 − P2) ∆V
The total work done helps to convert the gravitational potential energy and kinetic energy of the fluid. Now, consider the fluid density as ρ and the mass passing through the pipe as ∆m in the ∆t interval of time.
Hence, ∆m = ρA1 v1∆t = ρ∆V
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