# 14.6 Bernoulli’s equation  (Page 4/8)

 Page 4 / 8
${p}_{1}+\frac{1}{2}\rho {v}_{1}^{2}={p}_{2}+\frac{1}{2}\rho {v}_{2}^{2}$

becomes

${p}_{1}={p}_{2}+\frac{1}{2}\rho {v}_{2}^{2}.$

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

$h\propto \frac{1}{2}\rho {v}_{2}^{2}.$

(Recall that the symbol $\propto$ means “proportional to.”) Solving for ${v}_{2}$ , we see that

${v}_{2}\propto \sqrt{h}.$

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 $1.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}{\text{N/m}}^{2}$ . 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

${v}_{1}=\frac{Q}{{A}_{1}}=\frac{40.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\text{m}}^{3}\text{/}\text{s}}{\pi {\left(3.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{m)}}^{2}}=12.4\text{m/s}.$

Similarly, we find

${v}_{2}=56.6\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

This rather large speed is helpful in reaching the fire. Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${p}_{2}$ :

${p}_{2}={p}_{1}+\frac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right)-\rho g{h}_{2}.$

Substituting known values yields

$\begin{array}{cc}\hfill {p}_{2}& =1.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}+\frac{1}{2}\left(1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left[{\text{(12.4 m/s)}}^{2}-{\text{(56.6 m/s)}}^{2}\right]\hfill \\ & \phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}\left(1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)\hfill \\ & =0.\hfill \end{array}$

## 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.

## 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.

What is a volt equal to?
list and explain the 3 ways of charging a conductor
conduction convention rubbing
Asdesaw
formula of magnetic field
Integral of a vector
define surface integral of a vector?
Rahat
the number of degree freedom of a rigid body in2-dimantion is:
1
Nathan
A block (A) of weight 5 kN is to be raised by means of a 20° wedge (B) by the application of a horizontal force (P) as shown in Fig.1. The block A is constrained to move vertically by the application of a horizontal force (S). Find the magnitude of the forces F and S, when the coefficient of fricti
Danilo
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Previous
twenty four square plus thirty-five square minus to multiple thirty five twenty four and equal answer number square Via this equation defined Total Total impulse
Cemal
why simple pendulum do not vibrate indefinitely?
Zirmal
define integral vector
Rahat
what is matar
define surface integral vector?
Rahat
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
hi
Asdesaw
yes
Cemal
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
hi
Abdulrahman
mola mass
Abdulrahman
hi
Asdesaw
what exactly is a transverse wave