To solve the problem, we need to find the velocity of water in the narrower section of a horizontal pipe when the cross-sectional area reduces to half.
1. Principle of Continuity:
According to the continuity equation for incompressible fluids:
$ A_1 v_1 = A_2 v_2 $
where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.
2. Given Data:
Initial velocity, $v_1 = 2\, m/s$
Cross-sectional area reduces to half, so $A_2 = \frac{1}{2} A_1$
3. Calculate Velocity in Narrower Section:
Using continuity:
$ A_1 v_1 = A_2 v_2 \Rightarrow A_1 \times 2 = \frac{1}{2} A_1 \times v_2 $
Simplify:
$ 2 = \frac{1}{2} v_2 \Rightarrow v_2 = 4\, m/s $
Final Answer:
The velocity of water in the narrower section is $ {4\, m/s} $.
A fixed control volume has four one-dimensional boundary sections (1, 2, 3, and 4). For a steady flow inside the control volume, the flow properties at each section are tabulated below:
The rate of change of energy of the system which occupies the control volume at this instant is \( E \times 10^6 \, {J/s} \). The value of \( E \) (rounded off to 2 decimal places) is ........
A liquid flows under steady and incompressible flow conditions from station 1 to station 4 through pipe sections P, Q, R, and S as shown in the figure. Consider, \( d \), \( V \), and \( h \) represent the diameter, velocity, and head loss, respectively, in each pipe section with subscripts ‘P’, ‘Q’, ‘R’, and ‘S’. \( \Delta h \) represents the head difference between the inlet (station 1) and outlet (station 4). All the pipe sections are placed on the same horizontal plane for which the figure shows the top view.
(Insert diagram here, if possible)