1. The blacktip shark shown below (nature.org) is swimming very slowly 1m above the top of the coral reef. The vertical distance from its nose to the water surface is 2m. Set the top of the coral reef as the datum, assume the shark is stationary, and calculate: a. The pressure head at the shark's nose. b. The elevation head at the shark's nose. c. The total head at the shark's nose. d. The pressure at the shark's nose.

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### Fluid Mechanics Problem: Calculating Pressure and Head

#### Scenario:
The blacktip shark shown below (nature.org) is swimming very slowly 1m above the top of the coral reef. The vertical distance from its nose to the water surface is 2m. Set the top of the coral reef as the datum, assume the shark is stationary, and calculate:

a. The pressure head at the shark’s nose.  
b. The elevation head at the shark’s nose.  
c. The total head at the shark’s nose.  
d. The pressure at the shark’s nose.

[Image: Blacktip shark swimming above a coral reef]

---

#### Explanation of the Concepts:

1. **Pressure Head:**
   - Pressure head is the height equivalent of the pressure exerted by the fluid at a given point. It can be expressed in terms of meters of fluid column.
   - Formula: \( \text{Pressure Head} = \frac{P}{\rho g} \)
     where \( P \) is the pressure, \( \rho \) is the density of the fluid, and \( g \) is the acceleration due to gravity.

2. **Elevation Head:**
   - Elevation head is the height of the point above a reference level (datum). In this scenario, set the top of the coral reef as the datum.
   - Formula: \( \text{Elevation Head} = z \)
     where \( z \) is the height above the datum. 

3. **Total Head:**
   - Total head is the sum of the pressure head and the elevation head (since the shark is stationary, the velocity head is negligible).
   - Formula: \( \text{Total Head} = \text{Pressure Head} + \text{Elevation Head} \)

4. **Pressure:**
   - Pressure is the force exerted perpendicular to the surface of an object per unit area over which that force is distributed. In fluid statics, it can be found using the depth below the surface.
   - Formula: \( P = \rho gh \)
     where \( h \) is the depth of fluid above the point of interest.

By understanding and applying these concepts, one can solve the given problem related to the blacktip shark's positioning in water.

---
Transcribed Image Text:### Fluid Mechanics Problem: Calculating Pressure and Head #### Scenario: The blacktip shark shown below (nature.org) is swimming very slowly 1m above the top of the coral reef. The vertical distance from its nose to the water surface is 2m. Set the top of the coral reef as the datum, assume the shark is stationary, and calculate: a. The pressure head at the shark’s nose. b. The elevation head at the shark’s nose. c. The total head at the shark’s nose. d. The pressure at the shark’s nose. [Image: Blacktip shark swimming above a coral reef] --- #### Explanation of the Concepts: 1. **Pressure Head:** - Pressure head is the height equivalent of the pressure exerted by the fluid at a given point. It can be expressed in terms of meters of fluid column. - Formula: \( \text{Pressure Head} = \frac{P}{\rho g} \) where \( P \) is the pressure, \( \rho \) is the density of the fluid, and \( g \) is the acceleration due to gravity. 2. **Elevation Head:** - Elevation head is the height of the point above a reference level (datum). In this scenario, set the top of the coral reef as the datum. - Formula: \( \text{Elevation Head} = z \) where \( z \) is the height above the datum. 3. **Total Head:** - Total head is the sum of the pressure head and the elevation head (since the shark is stationary, the velocity head is negligible). - Formula: \( \text{Total Head} = \text{Pressure Head} + \text{Elevation Head} \) 4. **Pressure:** - Pressure is the force exerted perpendicular to the surface of an object per unit area over which that force is distributed. In fluid statics, it can be found using the depth below the surface. - Formula: \( P = \rho gh \) where \( h \) is the depth of fluid above the point of interest. By understanding and applying these concepts, one can solve the given problem related to the blacktip shark's positioning in water. ---
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