Question
water with a density of 1 000 kg/m^3flowing through a tapered pipe.
The larger diameter of the pipe is 130 mm, while the smaller diameter is 105 mm. For the smaller diameter, the water pressure reading is 125 kN/m^2; for the larger diameter, the reading is 150 kN/m^2. Calculate the flow rate in l/s.
The larger diameter of the pipe is 130 mm, while the smaller diameter is 105 mm. For the smaller diameter, the water pressure reading is 125 kN/m^2; for the larger diameter, the reading is 150 kN/m^2. Calculate the flow rate in l/s.
Answers
GPT-4o mini
To calculate the flow rate of water in a tapered pipe given the diameters, pressure readings, and density, we can use the principles of fluid dynamics, specifically the Bernoulli equation and the continuity equation.
### Step 1: Parameters and Conversions
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Larger diameter, \( D_1 = 130 \, \text{mm} = 0.130 \, \text{m} \)
- Smaller diameter, \( D_2 = 105 \, \text{mm} = 0.105 \, \text{m} \)
- Pressure at larger diameter, \( P_1 = 150 \, \text{kN/m}^2 = 150,000 \, \text{Pa} \)
- Pressure at smaller diameter, \( P_2 = 125 \, \text{kN/m}^2 = 125,000 \, \text{Pa} \)
### Step 2: Areas
Calculate the cross-sectional areas of both sections of the pipe:
\[
A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.130)^2}{4} \approx 0.01327 \, \text{m}^2
\]
\[
A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.105)^2}{4} \approx 0.00867 \, \text{m}^2
\]
### Step 3: Apply Bernoulli's Equation
Using Bernoulli's equation between the two points, we have:
\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g z_2
\]
Assuming the height (\( z \)) of both points is the same (we can ignore the height), the equation simplifies to:
\[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
\]
Rearranging gives:
\[
P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2
\]
Or:
\[
P_1 - P_2 = \frac{\rho}{2} (v_2^2 - v_1^2)
\]
### Step 4: Express Velocities in Terms of Each Other
From the continuity equation \( A_1 v_1 = A_2 v_2 \):
\[
v_2 = \frac{A_1}{A_2} v_1
\]
Substituting this into the Bernoulli equation gives:
\[
P_1 - P_2 = \frac{\rho}{2} \left( \left( \frac{A_1}{A_2} v_1 \right)^2 - v_1^2 \right)
\]
### Step 5: Solve for \( v_1 \)
Let's denote \( \frac{A_1}{A_2} = k \):
\[
k = \frac{0.01327}{0.00867} \approx 1.53
\]
The equation now becomes:
\[
P_1 - P_2 = \frac{\rho}{2} ( (k v_1)^2 - v_1^2 )
\]
\[
25,000 = \frac{1000}{2} ( (1.53 v_1)^2 - v_1^2 )
\]
Expanding and simplifying:
\[
25,000 = 500 \left( 2.3409 v_1^2 - v_1^2 \right)
\]
\[
25,000 = 500 (1.3409 v_1^2)
\]
Dividing both sides by 500:
\[
50 = 1.3409 v_1^2
\]
Solving for \( v_1^2 \):
\[
v_1^2 = \frac{50}{1.3409} \approx 37.24
\]
\[
v_1 \approx \sqrt{37.24} \approx 6.1 \, \text{m/s}
\]
### Step 6: Calculate \( v_2 \)
Using \( v_2 = k v_1 \):
\[
v_2 = 1.53 \times 6.1 \approx 9.33 \, \text{m/s}
\]
### Step 7: Calculate Flow Rate
The flow rate \( Q \) can be calculated from:
\[
Q = A_2 v_2
\]
Converting to liters per second:
\[
Q = 0.00867 \times 9.33 \approx 0.08093 \, \text{m}^3/s \quad (\text{in m}^3/s)
\]
\[
Q \approx 0.08093 \times 1000 \approx 80.93 \, \text{l/s}
\]
Thus, the flow rate of water in the pipe is approximately **80.93 l/s**.
### Step 1: Parameters and Conversions
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Larger diameter, \( D_1 = 130 \, \text{mm} = 0.130 \, \text{m} \)
- Smaller diameter, \( D_2 = 105 \, \text{mm} = 0.105 \, \text{m} \)
- Pressure at larger diameter, \( P_1 = 150 \, \text{kN/m}^2 = 150,000 \, \text{Pa} \)
- Pressure at smaller diameter, \( P_2 = 125 \, \text{kN/m}^2 = 125,000 \, \text{Pa} \)
### Step 2: Areas
Calculate the cross-sectional areas of both sections of the pipe:
\[
A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.130)^2}{4} \approx 0.01327 \, \text{m}^2
\]
\[
A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.105)^2}{4} \approx 0.00867 \, \text{m}^2
\]
### Step 3: Apply Bernoulli's Equation
Using Bernoulli's equation between the two points, we have:
\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g z_2
\]
Assuming the height (\( z \)) of both points is the same (we can ignore the height), the equation simplifies to:
\[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
\]
Rearranging gives:
\[
P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2
\]
Or:
\[
P_1 - P_2 = \frac{\rho}{2} (v_2^2 - v_1^2)
\]
### Step 4: Express Velocities in Terms of Each Other
From the continuity equation \( A_1 v_1 = A_2 v_2 \):
\[
v_2 = \frac{A_1}{A_2} v_1
\]
Substituting this into the Bernoulli equation gives:
\[
P_1 - P_2 = \frac{\rho}{2} \left( \left( \frac{A_1}{A_2} v_1 \right)^2 - v_1^2 \right)
\]
### Step 5: Solve for \( v_1 \)
Let's denote \( \frac{A_1}{A_2} = k \):
\[
k = \frac{0.01327}{0.00867} \approx 1.53
\]
The equation now becomes:
\[
P_1 - P_2 = \frac{\rho}{2} ( (k v_1)^2 - v_1^2 )
\]
\[
25,000 = \frac{1000}{2} ( (1.53 v_1)^2 - v_1^2 )
\]
Expanding and simplifying:
\[
25,000 = 500 \left( 2.3409 v_1^2 - v_1^2 \right)
\]
\[
25,000 = 500 (1.3409 v_1^2)
\]
Dividing both sides by 500:
\[
50 = 1.3409 v_1^2
\]
Solving for \( v_1^2 \):
\[
v_1^2 = \frac{50}{1.3409} \approx 37.24
\]
\[
v_1 \approx \sqrt{37.24} \approx 6.1 \, \text{m/s}
\]
### Step 6: Calculate \( v_2 \)
Using \( v_2 = k v_1 \):
\[
v_2 = 1.53 \times 6.1 \approx 9.33 \, \text{m/s}
\]
### Step 7: Calculate Flow Rate
The flow rate \( Q \) can be calculated from:
\[
Q = A_2 v_2
\]
Converting to liters per second:
\[
Q = 0.00867 \times 9.33 \approx 0.08093 \, \text{m}^3/s \quad (\text{in m}^3/s)
\]
\[
Q \approx 0.08093 \times 1000 \approx 80.93 \, \text{l/s}
\]
Thus, the flow rate of water in the pipe is approximately **80.93 l/s**.