Question

At what speed, the velocity head of a stream of water be equal to 40 cm of Hg?
A. 1032.6 \[cm{s^{ - 1}}\]
B. 432.6 \[cm{s^{ - 1}}\]
C. 632.6 \[cm{s^{ - 1}}\]
D. 832.6 \[cm{s^{ - 1}}\]

Answer 1

Hint:When the area of cross-section is same, then the mass of fluid is in the ratio of product of the height of the fluid and the density of the fluid. The velocity head is defined as the height through which the fluid should have fallen to attain the required velocity.

Formula used:
\[{v_h} = \dfrac{{{v^2}}}{{2g}}\]
where \[{v_h}\] is the velocity head, v is the velocity and g is the acceleration due to gravity.

Complete step by step solution:
The density of Hg is 13600 \[kg/{m^3}\] and the density of water is 1000 \[kg/{m^3}\]. If the area of cross-section is A and the height is h then the volume is,
\[V = Ah\]
And the mass is the product of the density and the volume,
\[m = \rho Ah\]

As it is to find the equivalent height of the water equivalent to 40 cm of Hg, the mass of water is equal to the mass of the Hg,
\[{\rho _W}A{h_W} = {\rho _{Hg}}A{h_{Hg}} \\ \]
\[\Rightarrow {h_W} = \dfrac{{{\rho _{Hg}}{h_{Hg}}}}{{{\rho _W}}}\]

Putting the values, we get
\[{h_W} = \dfrac{{13600 \times 40cm}}{{1000}} \\ \]
\[\Rightarrow {h_W} = 544\,cm \\ \]
\[\Rightarrow {h_W} = 5.44\,m\]
Using the velocity head formula, we get
\[v = \sqrt {2gh} \]
Putting the values, we get the value of required speed of the streamline of water;
\[v = \sqrt {2 \times 9.8 \times 5.44} m/s \\ \]
\[\Rightarrow v = 10.326\,m/s\]

As the numerical values for the speed of the streamline of water is given in the unit of cm per seconds, so we need to change the unit of obtained value of speed in respective unit,
\[v = 10.326 \times 100\,cm{s^{ - 1}} \\ \]
\[\therefore v = 1032.6\,cm{s^{ - 1}}\]
Therefore, when the water streamline is flowing with speed of 1032.6 cm per sec, the velocity head of water is equal to the velocity head of 40 cm Hg.

Therefore, the correct option is A.

Note: When the pipe through which the fluid is flowing is not frictionless then due resistance there is decrease in the velocity of the fluid. As the velocity head is proportional to the square of the velocity, so the velocity head decreases and it is called the velocity head loss.