Water is flowing smoothly through a closed-pipe system. At one point the speed of the water is `3.0m//s`. While at another point `1.0m` higher the speed is `4.0m//s`. If the pressure is `20 Kpa` at the lower point, what is the pressure at the upper point? What would the pressure at the upper point be if the water were to stop flowing and the pressure at the lower point were `18 kpa`?

Explanation

To solve the problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a closed system. The equation is given by:

\[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g H_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g H_2 \]

Where:
- \( P \) is the pressure,
- \( \rho \) is the density of the fluid (for water, approximately \( 1000 \, \text{kg/m}^3 \)),
- \( V \) is the velocity of the fluid,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( H \) is the height.

### Step 1: Identify the known values
- At the lower point (Point 1):
- \( V_1 = 3.0 \, \text{m/s} \)
- \( P_1 = 20 \, \text{kPa} = 20 \times 10^3 \, \text{Pa} \)
- \( H_1 = 0 \, \text{m} \) (reference height)

- At the upper point (Point 2):
- \( V_2 = 4.0 \, \text{m/s} \)
- \( H_2 = 1.0 \, \text{m} \)

### Step 2: Substitute the known values into Bernoulli's equation
Using the values we have:

\[ 20 \times 10^3 + \frac{1}{2} \times 1000 \times (3.0)^2 + 1000 \times 9.81 \times 0 = P_2 + \frac{1}{2} \times 1000 \times (4.0)^2 + 1000 \times 9.81 \times 1 \]

### Step 3: Calculate each term
- Calculate \( \frac{1}{2} \times 1000 \times (3.0)^2 \):
\[ = \frac{1}{2} \times 1000 \times 9 = 4500 \, \text{Pa} \]

- Calculate \( \frac{1}{2} \times 1000 \times (4.0)^2 \):
\[ = \frac{1}{2} \times 1000 \times 16 = 8000 \, \text{Pa} \]

- Calculate \( 1000 \times 9.81 \times 1 \):
\[ = 9810 \, \text{Pa} \]

### Step 4: Substitute back into the equation
Now substituting these values back into the equation:

\[ 20 \times 10^3 + 4500 = P_2 + 8000 + 9810 \]

### Step 5: Simplify and solve for \( P_2 \)
\[ 20000 + 4500 = P_2 + 8000 + 9810 \]
\[ 24500 = P_2 + 17810 \]
\[ P_2 = 24500 - 17810 = 6680 \, \text{Pa} = 6.68 \, \text{kPa} \]

### Step 6: Answer for the first part
The pressure at the upper point when water is flowing is approximately \( 6.68 \, \text{kPa} \).

### Step 7: Second part of the question when water stops flowing
When the water stops flowing, \( V_1 = V_2 = 0 \). The Bernoulli's equation simplifies to:

\[ P_1 + \rho g H_1 = P_2 + \rho g H_2 \]

Substituting the known values:

\[ 18 \times 10^3 + 0 = P_2 + 1000 \times 9.81 \times 1 \]

Calculating \( 1000 \times 9.81 \):

\[ P_2 + 9810 = 18000 \]
\[ P_2 = 18000 - 9810 = 8180 \, \text{Pa} = 8.18 \, \text{kPa} \]

### Final Answer
The pressure at the upper point when the water stops flowing is approximately \( 8.18 \, \text{kPa} \).