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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 14

Fluids in Motion - all with Video Answers

Educators

Chapter Questions

01:43

Problem 1

Oil flows through a pipe $8.0 \mathrm{~cm}$ in diameter, at an average speed of $4.0 \mathrm{~m} / \mathrm{s}$. What is the flow rate, $J$, in $\mathrm{m}^{3} / \mathrm{s}$ and $\mathrm{m}^{3} / \mathrm{h}$ ?
$$
\begin{aligned}
J &=A v=\pi(0.040 \mathrm{~m})^{2}(4.0 \mathrm{~m} / \mathrm{s})=0.020 \mathrm{~m}^{3} / \mathrm{s} \\
&=\left(0.020 \mathrm{~m}^{3} / \mathrm{s}\right)(3600 \mathrm{~s} / \mathrm{h})=72 \mathrm{~m}^{3} / \mathrm{h}
\end{aligned}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
02:13

Problem 2

Exactly $250 \mathrm{~mL}$ of fluid flows out of a tube whose inner diameter is $7.0 \mathrm{~mm}$ in a time of $41 \mathrm{~s}$. What is the average speed of the fluid in the tube?
From $J=A v$, since $1 \mathrm{~mL}=10^{-6} \mathrm{~m}^{3}$,
$$
v=\frac{J}{A}=\frac{\left(250 \times 10^{-6} \mathrm{~m}^{3}\right) /(41 \mathrm{~s})}{\pi(0.0035 \mathrm{~m})^{2}}=0.16 \mathrm{~m} / \mathrm{s}
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
02:21

Problem 3

A 14 -cm inner diameter (i.d.) water main furnishes water to a $1.00 \mathrm{~cm}$ i.d. (i.e., inner diameter) faucet pipe. If the average speed in the faucet pipe is $3.0 \mathrm{~cm} / \mathrm{s}$, what will be the average speed it causes in the water main?
The two flows are equal. From the Continuity Equation,
$$J=A_{1} v_{1}=A_{2} v_{2}$$
Letting 1 be the faucet and 2 be the water main, we have
$$
v_{2}=v_{1} \frac{A_{1}}{A_{2}}=v_{1} \frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}=(3.0 \mathrm{~cm} / \mathrm{s})\left(\frac{1}{14}\right)^{2}=0.015 \mathrm{~cm} / \mathrm{s}$$.

Kajal Gautam
Kajal Gautam
Numerade Educator
03:41

Problem 4

How much water will flow in $30.0$ s through $200 \mathrm{~mm}$ of capillary tube of $1.50 \mathrm{~mm}$ i.d., if the pressure differential across the tube is $5.00 \mathrm{~cm}$ of mercury? The viscosity of water is $0.801 \mathrm{cP}$ and $\rho$ for mercury is $13600 \mathrm{~kg} / \mathrm{m}^{3}$.
We shall make use of Poiseuille's Law, $J=\pi r^{4}\left(P_{i}-P_{o}\right) / 8 \eta L$, and therefore,
$$P_{i}-P_{o}=\left(\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.0500 \mathrm{~m})=6660 \mathrm{~N} / \mathrm{m}^{2}\right.$$
The viscosity expressed in $\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}$ is
$$\eta=(0.801 \mathrm{cP})\left(10^{-3} \frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{cP}}\right)=8.01 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}$$
Thus,
$$J=\frac{\pi r^{4}\left(P_{i}-P_{o}\right)}{8 \eta L}=\frac{\pi\left(7.5 \times 10^{-4} \mathrm{~m}\right)^{4}\left(6660 \mathrm{~N} / \mathrm{m}^{2}\right)}{8\left(8.01 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)(0.200 \mathrm{~m})}=5.2 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{s}=5.2 \mathrm{~mL} / \mathrm{s}$$
In $30.0 \mathrm{~s}$, the quantity that would flow out of the tube is $(5.2 \mathrm{~mL} / \mathrm{s})(30 \mathrm{~s})=1.6 \times 10^{2} \mathrm{~mL}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
01:14

Problem 5

An artery in a person has been reduced to half its original inside diameter by deposits on the inner artery wall. By what factor will the blood flow through the artery be reduced if the pressure differential across the artery has remained unchanged?
The relationship governing flow rate, pressure differential, and opening radius is Poiseuille's Law, wherein $J \propto r^{4}$. Therefore,
$$
\frac{J_{\text {final }}}{J_{\text {oniginal }}}=\left(\frac{r_{\text {final }}}{r_{\text {oiginal }}}\right)^{4}=\left(\frac{1}{2}\right)^{4}=0.0625
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
02:45

Problem 6

Under the same pressure differential, compare the flow of water through a pipe to the flow of SAE No. 10 oil. $\eta$ for water is $0.801 \mathrm{cP} ; \eta$ for the oil is $200 \mathrm{cP}$.
From Poiseuille's Law, $J \propto 1 / \eta$. Therefore, since everything else cancels,
$$\frac{J_{\text {water }}}{J_{\text {oil }}}=\frac{200 \mathrm{cP}}{0.801 \mathrm{cP}}=250$$
The flow of water is 250 times as large as that of the oil under the same pressure differential.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
01:57

Problem 7

Calculate the power output of the heart if, in each heartbeat, it pumps $75 \mathrm{~mL}$ of blood at an average pressure of $100 \mathrm{mmHg}$. Assume 65 heartbeats per minute.
The work done by the heart is $P \Delta V$. In one minute, $\Delta V=(65)\left(75 \times 10^{-6} \mathrm{~m}^{3}\right)$. Also
$$P=(100 \mathrm{mmHg}) \frac{1.01 \times 10^{5} \mathrm{~Pa}}{760 \mathrm{mmHg}}=1.33 \times 10^{4} \mathrm{~Pa}$$
consequently Power $=\frac{\text { Work }}{\text { Time }}=\frac{\left(1.33 \times 10^{4} \mathrm{~Pa}\right)\left[(65)\left(75 \times 10^{-6} \mathrm{~m}^{3}\right)\right]}{60 \mathrm{~s}}=1.1 \mathrm{~W}$

Kajal Gautam
Kajal Gautam
Numerade Educator
03:27

Problem 8

What volume of water will escape per minute from an open-top tank through an opening $3.0 \mathrm{~cm}$ in diameter that is $5.0 \mathrm{~m}$ below the water level in the tank? (See Fig. $14-1 .$ )
There is a steady flow of fluid, and therefore we can use Bernoulli's Equation, with 1 representing the top level and 2 the orifice. The pressure at the outlet inside the free jet is atmospheric. Then $P_{1}=P_{2}$ and $h_{1}=5.0 \mathrm{~m}, h_{2}=0 .$
$$\begin{aligned}
P_{1}+\frac{1}{2} \rho v_{1}^{2}+h_{1} \rho g &=P_{2}+\frac{1}{2} \rho v_{2}^{2}+h_{2} \rho g \\
\frac{1}{2} \rho v_{1}^{2}+h_{1} \rho g &=\frac{1}{2} \rho v_{2}^{2}+h_{2} \rho g
\end{aligned}$$
If the tank is large, $v_{1}$ can be approximated as zero. Then, solving for $v_{2}$, we obtain Torricelli's Equation:
$$v_{2}=\sqrt{2 g\left(h_{1}-h_{2}\right)}=\sqrt{2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~m})}=9.9 \mathrm{~m} / \mathrm{s}$$
and the flow is given by
$$
J=v_{2} A_{2}=(9.9 \mathrm{~m} / \mathrm{s}) \pi\left(1.5 \times 10^{-2} \mathrm{~m}\right)^{2}=7.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}=0.42 \mathrm{~m}^{3} / \mathrm{min}.
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
02:16

Problem 9

An open water tank in air springs a leak at position- 2 in Fig. $14-2$, where the pressure due to the water at position- 1 is $500 \mathrm{kPa}$. What is the velocity of escape of the water through the hole? The pressure at position- 2 in the free jet is atmospheric. We use Bernoulli's Equation with $P_{1}-P_{2}=5.00 \times 10^{5}$ $\mathrm{N} / \mathrm{m}^{2}, h_{1}=h_{2}$, and the approximation $v_{1}=0 .$ Then $$\left(P_{1}-P_{2}\right)+\left(h_{1}-h_{2}\right) \rho g=\frac{1}{2} \rho v_{2}^{2}$$
whence
$$v_{2}=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}}=\sqrt{\frac{2\left(5.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)}{1000 \mathrm{~kg} / \mathrm{m}^{3}}}=31.6 \mathrm{~m} / \mathrm{s}.
$$

Kajal Gautam
Kajal Gautam
Numerade Educator
05:34

Problem 10

Water flows at the rate of $30 \mathrm{~mL} / \mathrm{s}$ through an opening at the bottom of a large tank in which the water is $4.0 \mathrm{~m}$ deep. Calculate the rate of escape of the water if an added pressure of $50 \mathrm{kPa}$ is applied to the top of the water.
Take position- 1 at the liquid surface at the top of the tank, and position- 2 at the opening. From Bernoulli's Equation where $v_{1}$ is essentially zero,
$$\left(P_{1}-P_{2}\right)+\left(h_{1}-h_{2}\right) \rho g=\frac{1}{2} \rho v_{2}^{2}$$
We can apply this expression twice, before the pressure is added and after.
$$\begin{aligned}
\left(P_{1}-P_{2}\right)_{\text {bcfore }}+\left(h_{1}-h_{2}\right) \rho g &=\frac{1}{2} \rho\left(v_{2}^{2}\right)_{\text {beffore }} \\
\left(P_{1}-P_{2}\right)_{\text {before }}+5 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}+\left(h_{1}-h_{2}\right) \rho g &=\frac{1}{2} \rho\left(v_{2}^{2}\right)_{\text {after }}
\end{aligned}$$
If the opening and the top of the tank are originally at atmospheric pressure, then
$$\left(P_{1}-P_{2}\right)_{\text {beffore }}=0$$
and division of the second equation by the first gives
$$\frac{\left(v_{2}^{2}\right)_{\text {atter }}}{\left(v_{2}^{2}\right)_{\text {before }}}=\frac{5 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}+\left(h_{1}-h_{2}\right) \rho g}{\left(h_{1}-h_{2}\right) \rho g}$$
But $\quad\left(h_{1}-h_{2}\right) \rho g=(4.0 \mathrm{~m})\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=3.9 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$
Therefore,
$$\frac{\left(v_{2}\right)_{\text {after }}}{\left(v_{2}\right)_{\text {before }}}=\sqrt{\frac{8.9 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}}{3.9 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}}}=1.51$$
Since $J=A v$, this can be written as
$$\frac{J_{\text {after }}}{J_{\text {beffore }}}=1.51 \quad \text { or } \quad J_{\text {after }}=(30 \mathrm{~mL} / \mathrm{s})(1.51)=45 \mathrm{~mL} / \mathrm{s}.$$

Kajal Gautam
Kajal Gautam
Numerade Educator
View

Problem 11

How much work $W$ is done by a pump in raising $5.00 \mathrm{~m}^{3}$ of water $20.0 \mathrm{~m}$ and forcing it into $a$ main at a gauge pressure of $150 \mathrm{kPa}$ ?
$W=($ Work to raise water $)+($ Work to push it in $)=m g h+P \Delta V$ $W=\left(5.00 \mathrm{~m}^{3}\right)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20.0 \mathrm{~m})+\left(1.50 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(5.00 \mathrm{~m}^{3}\right)=1.73 \times 10^{6} \mathrm{~J}$

Ankur S
Ankur S
Numerade Educator
02:57

Problem 12

A horizontal pipe has a constriction in it, as shown in Fig. 14-3. At point- 1 the diameter is $6.0 \mathrm{~cm}$, while at point- 2 it is only $2.0 \mathrm{~cm}$. At point- $1, v_{1}=2.0 \mathrm{~m} / \mathrm{s}$ and $P_{1}=180 \mathrm{kPa}$. Calculate $v_{2}$ and $P_{2}$.
We have two unknowns and will need two equations. Using Bernoulli's Equation with $h_{1}=h_{2}$, we have
$$P_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\frac{1}{2} \rho v_{2}^{2} \quad \text { or } \quad P_{1}+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)=P_{2}$$
Furthermore, $v_{1}=2.0 \mathrm{~m} / \mathrm{s}$, and the equation of continuity tells us that
$$v_{2}=v_{1} \frac{A_{1}}{A_{2}}=(2.0 \mathrm{~m} / \mathrm{s})\left(\frac{r_{1}}{r_{2}}\right)^{2}=(2.0 \mathrm{~m} / \mathrm{s})(9.0)=18 \mathrm{~m} / \mathrm{s}$$
Substituting then gives
$$1.80 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\frac{1}{2}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left[(2.0 \mathrm{~m} / \mathrm{s})^{2}-\left(18 \mathrm{~m} / \mathrm{s}^{2}\right)\right]=P_{2}$$
from which $P_{2}=0.20 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=20 \mathrm{kPa}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
02:07

Problem 13

We have two unknowns and will need two equations. Using Bernoulli's Equation with $h_{1}=h_{2}$, we have
$$P_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\frac{1}{2} \rho v_{2}^{2} \quad \text { or } \quad P_{1}+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)=P_{2}$$
Furthermore, $v_{1}=2.0 \mathrm{~m} / \mathrm{s}$, and the equation of continuity tells us that
$$v_{2}=v_{1} \frac{A_{1}}{A_{2}}=(2.0 \mathrm{~m} / \mathrm{s})\left(\frac{r_{1}}{r_{2}}\right)^{2}=(2.0 \mathrm{~m} / \mathrm{s})(9.0)=18 \mathrm{~m} / \mathrm{s}$$
Substituting then gives
$$1.80 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\frac{1}{2}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left[(2.0 \mathrm{~m} / \mathrm{s})^{2}-\left(18 \mathrm{~m} / \mathrm{s}^{2}\right)\right]=P_{2}$$
from which $P_{2}=0.20 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=20 \mathrm{kPa}$.
Since $P_{\text {out }}=P_{A}$, this is the gauge pressure inside the hose. How could you obtain this latter equation directly from Torricelli's Theorem?

Kajal Gautam
Kajal Gautam
Numerade Educator
02:32

Problem 14

At what rate does water flow from a $0.80 \mathrm{~cm}$ i.d. faucet if the water (or gauge) pressure is $200 \mathrm{kPa}$ ?
We apply Bernoulli's Equation for points just inside and outside the faucet (using absolute pressure):
$$P_{\text {in }}+\frac{1}{2} \rho v_{\text {in }}^{2}+h_{\text {in }} \rho g=P_{\text {out }}+\frac{1}{2} \rho v_{\text {out }}^{2}+h_{\text {out }} \rho g$$
Note that the pressure inside due only to the water is $200 \mathrm{kPa}$, and therefore $P_{\text {in }}=P_{\text {out }}=200 \mathrm{kPa}$ since $P_{\text {out }}=P_{A}$. Taking $h_{\text {out }}=h_{\text {in }}$
$$v_{\text {out }}^{2}-v_{\text {in }}^{2}=\left(200 \times 10^{3} \text { Pa }\right) \frac{2}{\rho}$$
Assuming $v_{\text {in }}^{2} \ll v_{\text {out }}^{2}$, we solve to obtain $v_{\text {out }}=20 \mathrm{~m} / \mathrm{s}$. The flow rate is then
$$J=v A=(20 \mathrm{~m} / \mathrm{s})\left(\pi r^{2}\right)=(20 \mathrm{~m} / \mathrm{s})(\pi)\left(0.16 \times 10^{-4} \mathrm{~m}^{2}\right)=1.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}$$

Kajal Gautam
Kajal Gautam
Numerade Educator
03:33

Problem 15

The pipe shown in Fig. 14-4 has a diameter of $16 \mathrm{~cm}$ at section- 1 and $10 \mathrm{~cm}$ at section-2. At section-1 the pressure is $200 \mathrm{kPa}$. Point- 2 is $6.0 \mathrm{~m}$ higher than point- $1 .$ When oil of density $800 \mathrm{~kg} / \mathrm{m}^{3}$ flows at a rate of $0.030 \mathrm{~m}^{3} / \mathrm{s}$, find the pressure at point- 2 if viscous effects are negligible.
From $J=v_{1} A_{1}=v_{2} A_{2}$
$$\begin{array}{l}
v_{1}=\frac{J}{A_{1}}=\frac{0.030 \mathrm{~m}^{3} / \mathrm{s}}{\pi\left(8.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=1.49 \mathrm{~m} / \mathrm{s} \\
v_{2}=\frac{J}{A_{2}}=\frac{0.030 \mathrm{~m}^{3} / \mathrm{s}}{\pi\left(5.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=3.82 \mathrm{~m} / \mathrm{s}
\end{array}$$
We can now use Bernoulli's Equation:
$$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g\left(h_{1}-h_{2}\right)=P_{2}+\frac{1}{2} \rho v_{2}^{2}$$
Setting $P_{1}=2.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}, h_{2}-h_{1}=6 \mathrm{~m}$ and $\rho=800 \mathrm{~kg} / \mathrm{m}^{3}$ result in
$\begin{aligned} P_{2} &=2.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{3}+\frac{1}{2}\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left[(1.49 \mathrm{~m} / \mathrm{s})^{2}-(3.82 \mathrm{~m} / \mathrm{s})^{2}\right]-\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(6.0 \mathrm{~m}) \\ &=1.48 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.5 \times 10^{5} \mathrm{kPa} . \end{aligned}$

Kajal Gautam
Kajal Gautam
Numerade Educator
04:18

Problem 16

A venturi meter equipped with a differential mercury manometer is shown in Fig. $14-5$. At the inlet, point- 1 , the diameter is $12 \mathrm{~cm}$, while at the throat, point- 2 , the diameter is $6.0 \mathrm{~cm}$. What is the flow $J$ of water through the meter if the mercury manometer reading is $22 \mathrm{~cm}$ ? The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3}$
From the manometer reading (remembering that $1 \mathrm{~g} / \mathrm{cm}^{3}=1000 \mathrm{~kg} / \mathrm{m}^{3}$ ):
$$P_{1}-P_{2}=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.22 \mathrm{~m})=2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$$
Since $J=v_{1} A_{1}=v_{2} A_{2}$, we have $v_{1}=J / A_{1}$ and $v_{2}=J / A_{2}$. Using Bernoulli's Equation with $h_{1}-h_{2}=0$ gives
$$\begin{array}{c}
\left(P_{1}-P_{2}\right)+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)=0 \\
2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}+\frac{1}{2}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(\frac{1}{A_{1}^{2}}-\frac{1}{A_{2}^{2}}\right) J^{2}=0
\end{array}$$
where $A_{1}=\pi r_{1}^{2}=\pi(0.060)^{2} \mathrm{~m}^{2}=0.01131 \mathrm{~m}^{2} \quad$ and $\quad A_{2}=\pi r_{2}^{2}=\pi(0.030)^{2} \mathrm{~m}^{2}=0.0028 \mathrm{~m}^{2}$
Substitution then gives $J=0.022 \mathrm{~m}^{3} / \mathrm{s}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
02:33

Problem 17

A wind tunnel is to be used with a $20-\mathrm{cm}$ -high model car to approximately reproduce the situation in which a $550-\mathrm{cm}$ -high car is moving at $15 \mathrm{~m} / \mathrm{s}$. What should be the wind speed in the tunnel? Is the flow likely to be turbulent?
We want the Reynolds number $N_{R}$ to be the same in both cases, so that the situations will be similar. That is,
$$N_{R}=\left(\frac{\rho v D}{\eta}\right)_{\text {tunnel }}=\left(\frac{\rho v D}{\eta}\right)_{\text {air }}$$
Both $\rho$ and $\eta$ are the same in the two cases, hence
$$v_{t} D_{t}=v_{a} D_{a} \quad \text { from which } \quad v_{t}=v_{a} \frac{D_{a}}{D_{t}}=(15 \mathrm{~m} / \mathrm{s})(550 / 20)=0.41 \mathrm{~km} / \mathrm{s}$$
To investigate turbulence, evaluate $N_{R}$ using $\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}$ and $\eta=1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}$ for air. Consequently $N_{R}=5.9 \times 10^{6}$, a value far in excess of that required for turbulent flow. The flow will certainly be turbulent.

Kajal Gautam
Kajal Gautam
Numerade Educator
01:33

Problem 18

Oil flows through a $4.0-\mathrm{cm}$ -i.d. pipe at an average speed of $2.5 \mathrm{~m} / \mathrm{s}$. Find the flow in $\mathrm{m}^{3} / \mathrm{s}$ and $\mathrm{cm}^{3} / \mathrm{s}$.

Kajal Gautam
Kajal Gautam
Numerade Educator
01:45

Problem 19

Compute the average speed of water in a pipe having an i.d. of $5.0 \mathrm{~cm}$ and delivering $2.5 \mathrm{~m}^{3}$ of water per hour.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:09

Problem 20

The speed of glycerin flowing in a 5.0-cm-i.d. pipe is $0.54 \mathrm{~m} / \mathrm{s}$. Find the fluid's speed in a 3.0-cm-i.d. pipe that connects with it, both pipes flowing full.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:45

Problem 21

How long will it take for $500 \mathrm{~mL}$ of water to flow through a $15-\mathrm{cm}$ -long, $3.0$ -mm-i.d. pipe, if the pressure differential across the pipe is $4.0 \mathrm{kPa}$ ? The viscosity of water is $0.80 \mathrm{cP}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:50

Problem 22

A molten plastic flows out of a tube that is $8.0 \mathrm{~cm}$ long at a rate of $13 \mathrm{~cm}^{3} / \mathrm{min}$ when the pressure differential between the two ends of the tube is $18 \mathrm{~cm}$ of mercury. Find the viscosity of the plastic. The i.d. of the tube is $1.30 \mathrm{~mm} .$ The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3} .$

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:14

Problem 23

In a horizontal pipe system, a pipe (i.d. $4.0 \mathrm{~mm}$ ) that is $20 \mathrm{~cm}$ long connects in line to a pipe (i.d. $5.0 \mathrm{~mm}$ ) that is $30 \mathrm{~cm}$ long. When a viscous fluid is being pushed through the pipes at a steady rate, what is the ratio of the pressure difference across the $20-\mathrm{cm}$ pipe to that across the $30-\mathrm{cm}$ pipe?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
04:02

Problem 24

A hypodermic needle of length $3.0 \mathrm{~cm}$ and i.d. $0.45 \mathrm{~mm}$ is used to draw blood $(\eta=4.0 \mathrm{mPl})$. Assuming the pressure differential across the needle is $80 \mathrm{cmHg}$, how long does it take to draw $15 \mathrm{~mL}$ ?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
05:23

Problem 25

In a blood transfusion, blood flows from a bottle at atmospheric pressure into a patient's vein in which the pressure is $20 \mathrm{mmHg}$ higher than atmospheric. The bottle is $95 \mathrm{~cm}$ higher than the vein, and the needle into the vein has a length of $3.0 \mathrm{~cm}$ and an i.d. of $0.45 \mathrm{~mm}$. How much blood flows into the vein each minute? For blood, $\eta=0.0040 \mathrm{~Pa} \cdot \mathrm{s}$ and $\rho=1005 \mathrm{~kg} / \mathrm{m}^{3}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
01:35

Problem 26

How much work does the piston in a hydraulic system do during one $2.0-\mathrm{cm}$ stroke if the end area of the piston is $0.75 \mathrm{~cm}^{2}$ and the pressure in the hydraulic fluid is $50 \mathrm{kPa}$ ?

Kajal Gautam
Kajal Gautam
Numerade Educator
02:37

Problem 27

A large tank of nonviscous liquid, which is open to the surrounding air, springs a leak $4.5 \mathrm{~m}$ below the top of the liquid. What is the theoretical velocity of outflow from the hole? If the area of the hole is $0.25 \mathrm{~cm}^{2}$, how much liquid will escape in exactly 1 minute?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:23

Problem 28

Find the flow in liters/s of a nonviscous liquid through an opening $0.50 \mathrm{~cm}^{2}$ in area and $2.5 \mathrm{~m}$ below the level of the liquid in an open tank surrounded by air.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:32

Problem 29

Calculate the theoretical velocity of efflux of water, into the surrounding air, from an aperture that is $8.0 \mathrm{~m}$ below the surface of water in a large tank, if an added pressure of $140 \mathrm{kPa}$ is applied to the surface of the water.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:01

Problem 30

What horsepower is required to force $8.0 \mathrm{~m}^{3}$ of water per minute into a water main at a pressure of $220 \mathrm{kPa}$ ?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
04:58

Problem 31

A pump lifts water at the rate of $9.0$ liters/s from a lake through a $5.0$ -cm-i.d. pipe and discharges it into the air at a point $16 \mathrm{~m}$ above the level of the water in the lake. What are the theoretical $(a)$ velocity of the water at the point of discharge and $(b)$ power delivered by the pump.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:31

Problem 32

Water flows steadily through a horizontal pipe of varying cross section. At one place the pressure is $130 \mathrm{kPa}$ and the speed is $0.60 \mathrm{~m} / \mathrm{s}$. Determine the pressure at another place in the same pipe where the speed is $9.0 \mathrm{~m} / \mathrm{s}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
04:47

Problem 33

A pipe of varying inner diameter carries water. At point- 1 the diameter is $20 \mathrm{~cm}$ and the pressure is $130 \mathrm{kPa}$. At point- 2 , which is $4.0 \mathrm{~m}$ higher than point- 1 , the diameter is $30 \mathrm{~cm}$. If the flow is $0.080 \mathrm{~m}^{3} / \mathrm{s}$, what is the pressure at the second point?

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
05:27

Problem 34

Fuel oil of density $820 \mathrm{~kg} / \mathrm{m}^{3}$ flows through a venturi meter having a throat diameter of $4.0 \mathrm{~cm}$ and an entrance diameter of $8.0 \mathrm{~cm}$. The pressure drop between entrance and throat is $16 \mathrm{~cm}$ of mercury. Find the flow. The density of mercury is $13600 \mathrm{~kg} / \mathrm{m}^{3}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:30

Problem 35

Find the maximum amount of water that can flow through a 3.0-cm-i.d. pipe per minute without turbulence. Take the maximum Reynolds number for nonturbulent flow to be 2000 . For water at $20^{\circ} \mathrm{C}, \eta=1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
02:33

Problem 36

How fast can a raindrop $(r=1.5 \mathrm{~mm})$ fall through air if the flow around it is to be close to turbulent- that is, for $N_{R}$ close to $10 ?$ For air, $\eta=1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}$ and $\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}$.

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator