All Solutions
Page 439: Lesson Check
As dictated by the Equation of Continuity this brings about an increase in the speed of the fluid, since
$$
begin{align*}
v_1 cdot A_1 = v_2 cdot A_2
end{align*}
$$
So we conclude that the speed of the water coming out of the nozzle is greater than the speed in the hose.
This reduction in pressure leads to a net upward force that lifts the paper.
Just as work against friction must be done to move a block on a rough surface, so work must be done to push a real fluid through a pipe.
Since the Equation of Continuity states that:
$$
begin{equation*}
v cdot A = text{const}
end{equation*}
$$
We see that an increase in the speed of the fluid will bring about a decrease in the effective area of the fluid flow. Hence the stream becomes narrower as it falls.
Using the equation of continuity for better understanding:
$$
A_1v_1 = A_2v_2
$$
Where the initial cross sectional area of the flow of water is $A_1$,
Initial speed of water is $v_1$,
Final cross sectional area of the flow of water is $A_2$,
Final speed of water is $v_2$,
Therefore, if velocity of the flow increases then area gets reduced and thus the stream of water becomes narrower.
Since the Equation of Continuity states that:
$$
begin{equation*}
v cdot A = text{const}
end{equation*}
$$
We see that an increase in the area of the flow of the fluid by a factor of 9 will bring about a decrease in the speed by a factor of 9.
Hence the final speed is $frac{v}{9}$
$$
A_1v_1 = A_2v_2
$$
Where the cross sectional area of the hose is $A_1 = pi dfrac{d^2}{4}$,
Speed of water is $v_1 = v$,
New area of the pipe is $A_2 = pi dfrac{(3d)^2}{4}$,
Therefore,
$$
begin{align*}
A_1v_1 &= A_2v_2\
\
v_2 &= dfrac{A_1v_1}{A_2}\
\
v_2 &= dfrac{pi dfrac{d^2}{4} cdot v}{pi dfrac{(3d)^2}{4}}\
\
v_2 &= dfrac{d^2 cdot v}{9d^2}\
\
v_2 &= boxed{dfrac{v}{9}}\
end{align*}
$$
Therefore the new speed of the fluid is equal to $dfrac{v}{9}$, because if the diameter of the pipe is increased 3 times, then its area would increase nine time and according to the equation of continuity, its velocity would decrease nine times.
– The first cross-sectional area of the hose $A_1 = 0.0075text{ m}^2$
– The speed of water through the hose $v_1 = 1.3 ; frac{text{m}}{text{s}}$
– The second cross-sectional area of the hose $A_2 = 0.0033text{ m}^2$
Using the Equation of Continuity we can find the new speed $v_2$:
$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$
Rearranging for $v_2$ we have:
$$
begin{align*}
v_2 = v_1 cdot frac{A_1}{A_2}
end{align*}
$$
Plugging in the values we have:
$$
begin{align*}
v_2 &= 1.3 ; frac{text{m}}{text{s}} cdot frac{0.0075text{ m}^2}{0.0033text{ m}^2} \
v_2 &= 2.95 ; frac{text{m}}{text{s}}
end{align*}
$$
$$
A_{1}v_{1} = A_{2}v_{2}
$$
In the problem, the speed of the water decreased, therefore, the cross-sectional area is expected to increase.
$$
A_{2} = dfrac{A_{1}V_{1}}{V_{2}} = dfrac{(0.0084m^2)(2.5m/s)}{1.1m/s}
$$
$$
A_{2} = 0.019m^2
$$
b. $A_{2} = 0.019m^2$
– The first cross-sectional area of the hose $A_1 = 0.0084 text{ m}^2$
– The speed of water through the hose $v_1 = 2.5 ; frac{text{m}}{text{s}}$
– The new speed of the water $v_2 = 1.1 ; frac{text{m}}{text{s}}$
begin{enumerate}[a)]
item
We know from the Continuity Equation that $v cdot A = text{const}$ \
Since our speed decreases, the area must increase, so $A_2 > 0.0084text{ m}^2$
item
Using the Equation of Continuity we can find the required cross-sectional area of the hose $A_2$:
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
Rearranging for $A_2$ we have:
begin{align*}
A_2 = A_1 cdot frac{v_1}{v_2}
end{align*}
Plugging in the values we have:
begin{align*}
A_2 &= 0.0084text{ m}^2 cdot frac{2.5 ; frac{text{m}}{text{s}}}{1.1 ; frac{text{m}}{text{s}}} \
A_2 &= 0.0191text{ m}^2
end{align*}
end{enumerate}
item
The new cross-sectional area of the hose must increase owing to the Equation of Continuity, so $A_2$ is greater than $0.0084text{ m}^2$
item
$A_2 = 0.0191text{ m}^2$
end{enumerate}
– The diameter of the pipe $d_1 = 27text{ cm}$
– The speed of the fluid through the pipe $v_1 = 1.9 ; frac{text{m}}{text{s}}$
– The final speed of the fluid $v_2 = 3.1 ; frac{text{m}}{text{s}}$
Using the Equation of Continuity we can find the required diameter $d_2$:
$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$
Where the areas are found as follows:
$$
begin{align*}
A_1 &= left(frac{d_1}{2} right)^2 , pi \
A_2 &= left(frac{d_2}{2} right)^2 , pi
end{align*}
$$
So we have:
$$
begin{align*}
v_1 , d_1^2 = v_2 , d_2^2
end{align*}
$$
Rearranging for $d_2$ we have:
$$
begin{align*}
d_2 = d_1 cdot sqrt{frac{v_1}{v_2}}
end{align*}
$$
Plugging in the values we have:
$$
begin{align*}
d_2 &= 2.7text{ cm} cdot sqrt{frac{1.8 ; frac{text{m}}{text{s}}}{3.1 ; frac{text{m}}{text{s}}}} \
d_2 &approx 2.06text{ cm}
end{align*}
$$
