# 10 Problems and Solutions to Practice Fluid Dynamics

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## 10 Fluid Dynamics Problems And Solutions Explained In Detail

You can use these 10 problems and solutions to help you learn fluid dynamics and solidify your mastery of this concept. The 10 problems are aimed to help you achieve several important competencies listed below :
• Calculate fluid flow rate
• Using the equation of continuity
• Find the speed of fluid in connected pipes
• Using Torricelli’s principle to solve the leaking tank problem
• Using Bernoulli’s equation to solve some problems
• Understand how venturi meter works
• Calculate the lift force on a plane by using Bernoulli’s principle
Untuk pengalaman terbaik, ada baiknya jika Anda mengakses artikel ini lewat komputer.

### 1. Fluid flow rate problem

A taxi driver is refueling at a gas station. The fuel nozzle delivers 20 liters of fuel in 80 seconds. Assuming the continuity principle applies, determine the flow rate of the fuel.
Solution:
Flow rate simply means “The amount of fluid comes out of the nozzle in one second”. Its symbol is Q and the formula is $Q=\frac{V}{t}$. We can use the formula to solve this problem.
\begin{aligned} Q=&\frac{V}{t}\\ =&\frac{20}{80}\\ =&\frac{1}{4}\;\text{L}/\text{second} \end{aligned}
Our calculation shows that every one second a quarter-liter of fuel fills the taxi tank. Well, the unit liter is not really an international unit. We have to use the international unit of volume which is $m^3$.
1 liter is equivalent to $1000 cm^3$ or $1\times 10^{-3}\; m^{3}$ so we can express it as the following
\begin{aligned} Q=\frac{1}{4}\;\text{L}/\text{seconds}=2.5\times 10^{-4}\;\frac{m^3}{\text{second} } \end{aligned}

### 2. Equation of Continuity Problem

Relook at the previous problem. If the diameter of the nozzle is known to be 2 cm. Determine the speed of the fuel flowing into the car.
Solution:
The flow rate equation which is $Q=\frac{V}{t}$ has another alternative and that is $Q=\frac{V}{t}=Av$. The “A” in the equation is cross-sectional area while the “$v$” is the fluid speed.
From the previous problem we know that $Q=2.5\times 10^{-4}\;\frac{m^3}{\text{seconds}}$. And we know the diameter of the nozzle is 2 cm or $2\times 10^{-2}$ m. This means the radius is $2\times 10^{-2}$ m. Then we should just use the formula $Q=Av$ to find the speed of the fuel entering the taxi tank.
\begin{aligned} Q=&Av\\ Q=&\pi r^2v\\ 2.5\times 10^{-4}=&\pi (1\times 10^{-2})^2v\\ v\approx&\;0.8\;m/s \end{aligned}
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### 3. Connected pipes problem

A 5 cm diameter pipe is connected to a 1 cm diameter pipe. In the 5 cm pipe, water flows at a speed of 2 m/s. Determine the speed of water in the 1 cm pipe.
Solution:
You must understand the principle of continuity to solve this problem. The point is this, the flow rate in pipe 1 must be the same as the flow rate in pipe 2. To put it simply, if 1 liter of fluid is flowing in pipe 1 every one second, then fluid also flowing at the same exact rate in pipe 2. Mathematically we express it as $Q_1=Q_2$ or $A_1v_1=A_2v_2$.
How do we solve the problem? do as follows.
\begin{aligned} A_1v_1=&A_2v_2\\ \pi(2.5\times 10^-2)^2(2)=&\pi(0,5\times 10^-2)^2(v_2)\\ (2,5^2)(2)=&(0,5^2)(v_2)\\ (v_2)=&\frac{(2,5^2)(2)}{(0,5^2)}\\ =&\;50\;m/s \end{aligned}

### 4. The ration between cross-sectional area and fluid speed problem

Look at the pipes above. The cross-sectional areas of A, B, and C have a ratio of 3:1:2. If the speed of the fluid in pipe B is defined as v, find the speed of fluid in pipe A.
Solution:
Similar to the previous problem, this problem also requires us to use the principle of continuity. The flow rate in all 3 pipes is the same. We would solve this problem as the following.
\begin{aligned} \text{Flow Rate in B}=&\text{Flow Rate in C}\\ A_Bv_B=&A_Cv_C\\ A_Bv=&A_Cv_C\\ v_C=&\frac{A_B}{A_C}v\\ v_C=&\frac{1}{2}v \end{aligned}

### 5. Leaking tank problem

Look at the image above. A giant glass is 3.4 meters high. There is a leak located at 1.8 meters from the bottom of the glass. Calculate the speed of the water leak and calculate the distance at which the water leak hits the ground
Solution:
You’ll have to understand Torricelli’s principle in order to solve this problem. To put it simply, if there is a small hole in a tank, the speed of water coming out of the small hole can be calculated using this equation $v=\sqrt{2gh_1}$.
What is $h_1$? $h_1$ is the distance between the leak point and the water surface in the glass(160 cm). Meanwhile, $h_2$ is the distance between the bottom of the glass and the leak point (180 cm).
Then we use the equation.
\begin{aligned} v=&\sqrt{2gh_1}\\ =&\sqrt{2(10)(1,6)}\\ =&\sqrt{32}\\ =&4\sqrt{2}\;m/s \end{aligned}
We’ve got water leak speed. Then we have to find where the water would hit the ground. Watch the leak, it floats like a projectile motion, right? Now for a projectile motion, we have the equation for vertical position, and that is $y=v_{oy}+1/2gt^2$. In this case y is $h_2$ and $v_{oy}=0$ m/s. Use the equation as follows.
\begin{aligned} y=h_2=&v_{\circ y}t+\frac{1}{2}gt^2\\ -1.8=&0+\frac{1}{2}(-10)t^2\\ t^2=&0.36\\ t=&0.6\;\text{second} \end{aligned}
$y=-1.8$ means water falls 1.8 meters down.
We have t=0.6 seconds, this means the water will reach the ground in 0.6 seconds. If we already know t, we can use the projectile distance equation which is$x={v_x}t$. The $v_x$ is the speed we got from Torricelli’s formula earlier.
\begin{aligned} x=&v_xt\\ =&(4\sqrt{2})(0,6)\\ =&2.4\sqrt{2}\\ \approx&3,4\;meter \end{aligned}
Finally, we get the position where the water leak touches the ground. It is 3.4 meters horizontally from the glass.

### 6. Leaking tank problem II

A water tank has a leak. The location of the leak point is 60 cm from the bottom of the tank. The water leak hits the ground at a distance of $\frac{2}{5}\sqrt{15}$ meters from the tank. Determine the speed of the water leak and find the distance between the leak point and the surface of the water.
Solution:
This problem also requires us to use Torricelli’s principle.
I tell you. There’s actually a quick formula to find the distance at which the water hits the ground. The formula is $x=2\sqrt{h_1h_2}$. Try using this formula to work on the previous problem, the result would be the same. I intentionally didn’t tell you, so that you could remember the concept of projectile motion.
Okay, let’s now use the formula.
\begin{aligned} x=&2\sqrt{h_1h_2}\\ \frac{2}{5}\sqrt{15}=&2\sqrt{h_1(0,6)}\\ \frac{4}{25}(15)=&4(0,6)(h_1)\\ h_1=&1\;m \end{aligned}
We’ve calculated $h_1$. Then we should calculate the speed of the water leak.
\begin{aligned} v=&\sqrt{2gh_1}\\ =&\sqrt{2(10)(1)}\\ =&2\sqrt{5}\;m/s \end{aligned}
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### 7.Find the speed of flowing water with a venturimeter

Water flows in a venturimeter as shown above. In pipe 1 the speed of the water is 4 m/s. If g=10 m/s, find the speed of the water in pipe 2.
Solution:
This is a bit tricky. You should understand Bernoulli’s principle. Bernoulli’s principle is like the principle of conservation of mechanical energy but for fluids. Bernoulli’s equation is written below.
\begin{aligned} P+\frac{1}{2}\rho{v^2}+\rho{gh}=constant \end{aligned}
Then we use the equation to solve this problem.
$P_1+\frac{1}{2}\rho {v_1}^2+\rho gh_1=P_2+\frac{1}{2}\rho {v_2}^2+\rho gh_2\\$
Look at both pipes. There is no significant difference in height between pipe 1 and pipe 2. So $\rho gh_1=\rho gh_2$ and we cancel them out so they are not in the equation at all. The equation becomes as follows.
$P_1-P_2=\frac{1}{2}\rho ({v_2}^2-{v_1}^2)\\$
Now, what is $P_1-P_2$? Look at the tubes installed above each pipe. Both tubes are used to measure the pressure difference between pipe 1 and pipe 2 aka $P_1-P_2$. We can see from both tubes that there is a height difference of 20 cm or 0.2 m. This means that the pressure difference in pipes 1 and 2 is $\rho g(\Delta h_{\text{venturimeter}})=\rho g (0.2\;m)$. Then we continue the mathematical calculations below.
\begin{aligned} P_1+\frac{1}{2}\rho {v_1}^2+\rho gh_1=&P_2+\frac{1}{2}\rho {v_2}^2+\rho gh_2\\ P_1-P_2=&\frac{1}{2}\rho ({v_2}^2-{v_1}^2)\\ \rho g(\Delta h_{\text{venturimeter}})=&\frac{1}{2}\rho ({v_2}^2-{v_1}^2)\\ 1000(10)(0,2)=&\frac{1}{2}(1000)({v_2}^2-4^2)\\ {v_2}^2=&20\\ v_2=&2\sqrt{5}\;m/s \end{aligned}
We have found the speed of water in pipe 2 which is $2\sqrt{5}\;m/s$

### 8. The relation between fluid speed and pressure

Look at the image above. 3 pipes with different cross-sectional areas have water flowing in it.
Swipe left the picture and you will see 3 possible outcomes if all three pipes are installed with venturimeter tube. Out of the three options, choose the one that makes sense (according to Bernoulli’s principle).
Solution:
Try think it by yourself and make a guess.
Based on the continuity principle we know that the smaller the cross-sectional area, the faster the fluid moves inside the pipe.
Then from Bernoulli’s principle, we know that the faster the fluid moves, the lower the pressure.
We can conclude that the pressure in pipe 1 is the highest. Pipe 3 has the lowest pressure.
We also need to understand that the height of water in the venturimeter tube reflects the pressure of the pipe underneath. If the water is high, it means that the pressure in the pipe below is also high.
Now you look at option A, option B, and option C. Think for yourself which one do you think is the correct one.
The correct answer is option A.

### 9. Bernoulli's principle Problem.

Pipe 1 which has a diameter of 6 cm is connected to pipe 2 which has a diameter of 4 cm. Pipe 2 is at a height of 2 meters above pipe 1. Water flows at a speed of 4 m/s in pipe 1 with a pressure of 145 KPa. Determine the pressure in pipe 2.
Solution:
OK, here we go. This problem is a bit difficult, but we can get through this.
How do we solve this is similar to problem number 7. But in this problem $\rho gh_1\neq \rho gh_2$. So we can not cancel them out. We have put them in our equation.
We go straight to Bernoulli’s equation and put in the numbers.
\begin{aligned} 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\\ P_1-P_2=&\frac{1}{2}\rho ({v_2}^2-{v_1}^2)+\rho g (h_2-h_1)\\ =&\frac{1}{2}(1000) ({9}^2-{4}^2)+(1000)(10)(2)\\ =&500(65)+20\times{10}^3\\ =&32.5\times{10}^3 \end{aligned}
\begin{aligned} 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\\ \end{aligned}
\begin{aligned}P_1-P_2=&\frac{1}{2}\rho ({v_2}^2-{v_1}^2)+\rho g (h_2-h_1)\\ =&\frac{1}{2}(1000) ({9}^2-{4}^2)+(1000)(10)(2)\\ =&500(65)+20\times{10}^3\\ =&32.5\times{10}^3 \end{aligned}
We know that the pressure at $P_1$ is 145 KPa, so we put that in our equation.
\begin{aligned} P_1-P_2=&32.5\times{10}^3\\ 145\times{10}^3-P_2=&32.5\times{10}^3\\ P_2=&145\times{10}^3-32.5\times{10}^3\\ P_2=&112.5\;KPa \end{aligned}
There we go, we have calculated the pressure in pipe 2 which is 112.5 KPa. If you think about it, our calculation result does make sense, because fluid by the law of physics moves from high pressure to low pressure.

### 10. The lift force of airplane

The speed of the air flowing at the top of the airplane wing is 75 m/s. The pressure difference between the top and bottom of the wing is 240 Pa. Determine the speed of the air flowing below the wing. $\Rho_{air}=1.3\;\frac{Kg}{m^3}$
Solution:
So how do wings exerts an upward force on an airplane? There is a pressure difference between the underside of the plane’s wing and the top of the plane’s wing. The shape of the wing makes air move faster at the topside of the wing than at the underside of the wing. According to Bernoulli’s principle, this means that the fluid pressure at the underside is greater than at the topside. Because the underside pressure is greater, it lift the plane up.
The equation for pressure difference between the underside of the plane and the topside of the plane is $P_1-P_2=\frac{1}{2}\rho ({v_2}^2-{v_1}^2)$.
$P_1$ is the underside pressure while $P_2$ is the pressure at the topside of the wing. $v_1$ is the air speed at the underside and$v_2$ is the airspeed at the topside.
Then we put in them numbers to the equation.
\begin{aligned} P_1-P_2=&\frac{1}{2}\rho ({v_2}^2-{v_1}^2)\\ 240=&\frac{1}{2}(1,3)(75^2-{v_1}^2)\\ \frac{480}{1,3}=&5625-{v_1}^2\\ {v_1}^2=&5255,77\\ v_1=&\sqrt{5255,77}\\ \approx&72.5\;m/s \end{aligned}
There you go. The speed of air at the underside of the wing is 72.5 m/s. And it lived up to our expectations. The airspeed at the underside of the wing of an airplane must always be lower than above.
Okay it’s over. I hope my writing helps you learn fluid dynamics. Follow us on Instagram @creaticalsblog or twitter @creatical3.
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