# Material on Static and Dynamic Fluids

# Understanding Static and Dynamic Fluids: Concepts, Equations, and Example Problems

Fluid mechanics is the branch of physics concerned with the behavior of fluids at rest (statics) and in motion (dynamics). It addresses numerous phenomena and practical applications, from the circulation of the blood in the body to the design of complex fluid machinery.

## Static Fluids

Fluid statics, or hydrostatics, involves the study of fluids at rest. The primary principle in fluid statics is Pascal’s Law, which states that pressure exerted anywhere in a confined, incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations remain the same.

### Hydrostatic Pressure
The hydrostatic pressure at any point within a fluid is given by the equation:
$P = P_0 + \rho g h$
Where:
$$P$$ = Pressure at depth $$h$$,
$$P_0$$ = Pressure at the surface,
$$\rho$$ = Density of the fluid,
$$g$$ = Acceleration due to gravity,
$$h$$ = Depth below the surface.

## Dynamic Fluids

Fluid dynamics deals with fluids in motion. The primary equations governing fluid dynamics are the continuity equation and Bernoulli’s equation.

### Continuity Equation
The continuity equation, derived from the conservation of mass, states that for an incompressible, steady flow, the mass flow rate must remain constant from one cross-section of a pipe to another:
$A_1 V_1 = A_2 V_2$
where:
$$A$$ = Cross-sectional area,
$$V$$ = Flow velocity.

### Bernoulli’s Equation
Bernoulli’s equation, which comes from the conservation of energy, can be written as:
$P + \frac{1}{2} \rho V^2 + \rho gh = \text{constant}$
where each term represents the pressure energy, kinetic energy per unit volume, and potential energy per unit volume, respectively.

Now, let’s look at 20 problems that apply these core principles of static and dynamic fluids:

## Problems

1. _Static Fluids Problem_:
What is the pressure at the bottom of a tank that is 3 meters deep and filled with water? (Atmospheric pressure is $$P_0 = 101,325 \text{ Pa}$$, and the density of water is $$\rho = 1000 \text{ kg/m}^3$$).

**Solution**:
$$P = P_0 + \rho g h = 101,325 \text{ Pa} + (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(3 \text{ m}) = 101,325 \text{ Pa} + 29,430 \text{ Pa} = 130,755 \text{ Pa}$$.

2. _Dynamic Fluids Problem_:
Water flows through a pipe with a radius of 0.5 meters at a velocity of 2 m/s. What is the flow rate of the water in $$\text{m}^3/\text{s}$$?

**Solution**:
The flow rate $$Q$$ is $$A \times V$$, where $$A$$ is the cross-sectional area. $$A = \pi r^2 = \pi (0.5 \text{ m})^2 = 0.785 \text{ m}^2$$. The flow rate $$Q$$ is then $$0.785 \text{ m}^2 \times 2 \text{ m/s} = 1.57 \text{ m}^3/\text{s}$$.

3. _Static Fluids Problem_:
If the density of oil is $$800 \text{ kg/m}^3$$, what is the pressure due to the oil column of 2 meters?

**Solution**:
$$P = \rho g h = (800 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(2 \text{ m}) = 15,696 \text{ Pa}$$.

4. _Dynamic Fluids Problem_:
A fluid with a density of $$1000 \text{ kg/m}^3$$ flows steadily from a pipe with an inner diameter of 0.3 meters and velocity 4 m/s into a pipe with an inner diameter of 0.15 meters. What is the velocity in the smaller pipe?

**Solution**:
Using the continuity equation $$A_1 V_1 = A_2 V_2$$, we find $$V_2 = \frac{A_1 V_1}{A_2} = \frac{\pi(0.3/2)^2 \cdot 4}{\pi(0.15/2)^2} = 16 \text{ m/s}$$.

5. _Static Fluids Problem_:
How high can water rise in a barometer if the atmospheric pressure is $$1.013 \times 10^5 \text{ Pa}$$ and the density of mercury is $$13,600 \text{ kg/m}^3$$?

**Solution**:
Use the hydrostatic pressure equation $$P = \rho g h$$ to find $$h = \frac{P}{\rho g} = \frac{1.013 \times 10^5 \text{ Pa}}{(13,600 \text{ kg/m}^3)(9.81 \text{ m/s}^2)} = 0.76 \text{ m}$$.

6. _Dynamic Fluids Problem_:
Find the dynamic pressure of the fluid with density $$1.225 \text{ kg/m}^3$$ and velocity $$50 \text{ m/s}$$.

**Solution**:
Dynamic pressure $$p_d$$ is $$\frac{1}{2} \rho V^2 = \frac{1}{2} \cdot 1.225 \text{ kg/m}^3 \cdot (50 \text{ m/s})^2 = 1,531.25 \text{ Pa}$$.

7. _Static Fluids Problem_:
What is the buoyant force on a submerged object with a volume of $$0.1 \text{ m}^3$$ in water?

**Solution**:
Buoyant force $$F_b = \rho V g = (1000 \text{ kg/m}^3)(0.1 \text{ m}^3)(9.81 \text{ m/s}^2) = 981 \text{ N}$$.

8. _Dynamic Fluids Problem_:
Air flows over the wing of an airplane with speed $$180 \text{ m/s}$$ at sea level where the density of the air is $$1.225 \text{ kg/m}^3$$. What is the dynamic pressure on the wing?

**Solution**:
$$p_d = \frac{1}{2} \rho V^2 = \frac{1}{2} \cdot 1.225 \text{ kg/m}^3 \cdot (180 \text{ m/s})^2 = 19,845 \text{ Pa}$$.

9. _Static Fluids Problem_:
The pressure 5 meters underwater in the ocean is how much greater than the pressure at the surface?

**Solution**:
$$\Delta P = \rho g h = (1025 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(5 \text{ m}) = 50,253.125 \text{ Pa}$$.

10. _Dynamic Fluids Problem_:
Water in a river flows at a speed of 3 m/s and has a depth of 2 meters. If the river narrows to half its width, what is the flow speed in the narrower section, assuming depth remains constant?

**Solution**:
Using the continuity equation $$V_1 A_1 = V_2 A_2$$, and given that $$A_2 = \frac{1}{2}A_1$$, the new speed $$V_2 = 2V_1 = 2 \cdot 3 \text{ m/s} = 6 \text{ m/s}$$.

11. _Static Fluids Problem_:
How much pressure is exerted by a column of glycerin 0.5 meters high with a density of $$1260 \text{ kg/m}^3$$?

**Solution**:
$$P = \rho g h = (1260 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(0.5 \text{ m}) = 6,163.8 \text{ Pa}$$.

12. _Dynamic Fluids Problem_:
Determine the kinetic energy per unit volume of water flowing at 5 m/s with a density of $$1000 \text{ kg/m}^3$$.

**Solution**:
Kinetic energy per unit volume $$\frac{1}{2} \rho V^2 = \frac{1}{2} \cdot 1000 \text{ kg/m}^3 \cdot (5 \text{ m/s})^2 = 12,500 \text{ J/m}^3$$.

13. _Static Fluids Problem_:
Calculate the force needed to hold a 2 m x 2 m square gate closed that is submerged in water and has its top edge located 0.5 m below the surface.

**Solution**:
The center of pressure is at the center of the gate, which is 1.5 m below the surface. $$P = \rho g h = 1000 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2 \cdot 1.5 \text{ m} = 14,715 \text{ Pa}$$. The force is $$P \cdot A = 14,715 \text{ Pa} \cdot (2 \text{ m} \times 2 \text{ m}) = 58,860 \text{ N}$$.

14. _Dynamic Fluids Problem_:
In a Venturi meter, the water speed in a 0.5 m diameter section is 2 m/s. If the pressure there is $$1.5 \times 10^5 \text{ Pa}$$, what is the speed in a 0.25 m diameter section, assuming incompressibility and no friction?

**Solution**:
Using the continuity equation, we get $$V_2 = \frac{A_1}{A_2} V_1 = \frac{\pi (0.5/2)^2}{\pi (0.25/2)^2} \cdot 2 \text{ m/s} = 8 \text{ m/s}$$.

15. _Static Fluids Problem_:
If the atmospheric pressure is $$1.013 \times 10^5 \text{ Pa}$$, what is the absolute pressure at the bottom of a swimming pool that is 3 meters deep?

**Solution**:
Absolute pressure $$P = P_0 + \rho g h = 1.013 \times 10^5 \text{ Pa} + (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(3 \text{ m}) = 1.013 \times 10^5 \text{ Pa} + 29,430 \text{ Pa} = 1.313 \times 10^5 \text{ Pa}$$.

16. _Dynamic Fluids Problem_:
Calculate the decrease in pressure as air flows from a 0.4 m diameter duct at 10 m/s to a 0.2 m diameter duct where the speed increases. Assume air density is $$1.2 \text{ kg/m}^3$$.

**Solution**:
First, find the new speed $$V_2$$ using the continuity equation, then apply Bernoulli’s equation to find the pressure difference:
$$V_2 = \frac{(0.4 \text{ m})^2}{(0.2 \text{ m})^2} \cdot 10 \text{ m/s} = 40 \text{ m/s}$$.
Then $$\Delta P = \frac{1}{2} \rho ((V_2)^2 – (V_1)^2) = \frac{1}{2} \cdot 1.2 \text{ kg/m}^3 \cdot (1600 – 100) \text{ m}^2/\text{s}^2 = 900 \text{ Pa}$$.

17. _Static Fluids Problem_:
Find the pressure difference between the top and bottom of a 10 m high column of ethanol with the density $$789 \text{ kg/m}^3$$.

**Solution**:
$$\Delta P = \rho g h = (789 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(10 \text{ m}) = 77,421.9 \text{ Pa}$$.

18. _Dynamic Fluids Problem_:
A pipe that is 3 cm in diameter carries water at 2 m/s; what is the Reynolds number if the viscosity of water is $$1 \times 10^{-3} \text{ Pa}\cdot\text{s}$$?

**Solution**:
Reynolds number $$Re = \frac{\rho V D}{\mu} = \frac{(1000 \text{ kg/m}^3) \cdot (2 \text{ m/s}) \cdot (0.03 \text{ m})}{1 \times 10^{-3} \text{ Pa}\cdot\text{s}} = 60,000$$.

19. _Static Fluids Problem_:
A submerged object displaces $$0.05 \text{ m}^3$$ of seawater with a density of $$1025 \text{ kg/m}^3$$. What is the buoyant force?

**Solution**:
$$F_b = \rho V g = (1025 \text{ kg/m}^3)(0.05 \text{ m}^3)(9.81 \text{ m/s}^2) = 502.9125 \text{ N}$$.

20. _Dynamic Fluids Problem_:
Water flows into a tank with a flat top at a speed of $$5 \text{ m/s}$$ and exits through a hole at the bottom with a speed of $$20 \text{ m/s}$$. Assuming no height difference between the entrance and exit, what is the pressure difference?

**Solution**:
Using Bernoulli’s equation for the entrance ($$1$$) and exit ($$2$$), with $$y_1 = y_2$$:
$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$
$\Delta P = P_1 – P_2 = \frac{1}{2} \rho (V_2^2 – V_1^2)$
$\Delta P = \frac{1}{2} \cdot 1000 \text{ kg/m}^3 \cdot (400 – 25) \text{ m}^2/\text{s}^2$
$\Delta P = 187,500 \text{ Pa}$

These problems illustrate how the principles of static and dynamic fluid mechanics can be applied to solve a range of real-world problems. Fluid mechanics engages concepts and skills that are vital across many scientific disciplines and engineering applications.

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