14.6 Bernoulli’s equation  (Page 4/8)

becomes

Thus pressure $p_2$ over the second opening is reduced by $\frac{1}{2}\rho v_2^2$ , so the fluid in the manometer rises by h on the side connected to the second opening, where

(Recall that the symbol $\propto$ means “proportional to.”) Solving for $v_2$ , we see that

Part (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air-speed indicators in aircraft.

A fire hose

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.

Calculating pressure: a fire hose nozzle

Fire hoses used in major structural fires have an inside diameter of 6.40 cm ( [link] ). Suppose such a hose carries a flow of 40.0 L/s, starting at a gauge pressure of $1.62\times {10}^6{\text{N/m}}^2$ . The hose rises up 10.0 m along a ladder to a nozzle having an inside diameter of 3.00 cm. What is the pressure in the nozzle?

Strategy

We must use Bernoulli’s equation to solve for the pressure, since depth is not constant.

Solution

Bernoulli’s equation is

$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds $v_1$ and $v_2$ . Since $Q=A_1{v}_1$ , we get

$v_1=\frac{Q}{A_1}=\frac{40.0\times {10}^{\mathrm{-3}}{\text{m}}^3\text{/}\text{s}}{\pi {(3.20\times {10}^{\mathrm{-2}}\text{m)}}^2}=12.4\text{m/s}.$

Similarly, we find

$v_2=56.6\text{m/s}.$

This rather large speed is helpful in reaching the fire. Now, taking $h_1$ to be zero, we solve Bernoulli’s equation for $p_2$ :

$p_2=p_1+\frac{1}{2}\rho (v_1^2-v_2^2)-\rho g{h}_2.$

Substituting known values yields

$\begin{array}{cc}\hfill p_2& =1.62\times {10}^6{\text{N/m}}^2+\frac{1}{2}(1000{\text{kg/m}}^3)[{\text{(12.4 m/s)}}^2-{\text{(56.6 m/s)}}^2]\hfill \\ & -(1000{\text{kg/m}}^3)(9.80{\text{m/s}}^2)(10.0\text{m})\hfill \\ & =0.\hfill \end{array}$

Significance

This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus, the nozzle pressure equals atmospheric pressure as it must, because the water exits into the atmosphere without changes in its conditions.

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Summary

• Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
  
  $p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2.$
• Bernoulli’s principle is Bernoulli’s equation applied to situations in which the height of the fluid is constant. The terms involving depth (or height h ) subtract out, yielding
  
  $p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2.$
• Bernoulli’s principle has many applications, including entrainment and velocity measurement.