A horizontal pipe carries a smoothly flowing liquid with density 1.19 A1 103 kg/m3. At locations 1 and 2 along the pipe the diameters are 6.41 cm and 2.61 cm, respectively. Is the flow speed at location 2 higher or lower than at location 1?
The flow speed at location 1 is 2.03 m/s. What is the pressure difference between location 2 and location 1 (including its sign)? Ignore viscosity.
The Correct Answer and Explanation is :
The question asks for the comparison of flow speed at two locations in a pipe and the pressure difference between those two locations. Let’s break this down step by step using principles from fluid dynamics, specifically the continuity equation and Bernoulli’s equation.
Step 1: Continuity Equation
The continuity equation ensures mass conservation in a flowing fluid. It is given by:
[
A_1 v_1 = A_2 v_2
]
Where:
- (A_1) and (A_2) are the cross-sectional areas at locations 1 and 2.
- (v_1) and (v_2) are the flow speeds at locations 1 and 2.
The area of the pipe is related to its diameter by the equation:
[
A = \pi \left( \frac{D}{2} \right)^2
]
Since the pipe narrows from location 1 to location 2 (the diameter decreases), the area at location 2 will be smaller than at location 1. According to the continuity equation, if the area decreases, the flow speed must increase to maintain the same volumetric flow rate.
Thus, the flow speed at location 2 will be higher than at location 1.
Step 2: Bernoulli’s Equation
Bernoulli’s equation relates the pressure, velocity, and elevation of a fluid in steady flow. It is expressed as:
[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
]
Where:
- (P_1) and (P_2) are the pressures at locations 1 and 2.
- (\rho) is the fluid density.
- (v_1) and (v_2) are the flow velocities at locations 1 and 2.
Rearranging the equation to find the pressure difference:
[
P_2 – P_1 = \frac{1}{2} \rho \left( v_1^2 – v_2^2 \right)
]
Given that (v_2 > v_1) (as established earlier), it follows that (v_2^2 > v_1^2), so:
[
P_2 – P_1 < 0
]
Thus, the pressure at location 2 will be lower than at location 1.
Step 3: Calculate the Pressure Difference
We are given:
- (\rho = 1.19 \times 10^3 \, \text{kg/m}^3)
- (v_1 = 2.03 \, \text{m/s})
- (D_1 = 6.41 \, \text{cm} = 0.0641 \, \text{m})
- (D_2 = 2.61 \, \text{cm} = 0.0261 \, \text{m})
First, find the areas (A_1) and (A_2):
[
A_1 = \pi \left( \frac{D_1}{2} \right)^2 = \pi \left( \frac{0.0641}{2} \right)^2 \approx 3.23 \times 10^{-3} \, \text{m}^2
]
[
A_2 = \pi \left( \frac{D_2}{2} \right)^2 = \pi \left( \frac{0.0261}{2} \right)^2 \approx 5.38 \times 10^{-4} \, \text{m}^2
]
Using the continuity equation, we can solve for (v_2):
[
A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad v_2 = \frac{A_1 v_1}{A_2} = \frac{3.23 \times 10^{-3} \times 2.03}{5.38 \times 10^{-4}} \approx 12.2 \, \text{m/s}
]
Now, substitute the values into Bernoulli’s equation to find the pressure difference:
[
P_2 – P_1 = \frac{1}{2} \times 1.19 \times 10^3 \times \left( 2.03^2 – 12.2^2 \right)
]
[
P_2 – P_1 = \frac{1}{2} \times 1.19 \times 10^3 \times \left( 4.1209 – 148.84 \right)
]
[
P_2 – P_1 = \frac{1}{2} \times 1.19 \times 10^3 \times (-144.7191)
]
[
P_2 – P_1 \approx -86122.3 \, \text{Pa}
]
So, the pressure difference is approximately -86.1 kPa, indicating that the pressure at location 2 is lower than at location 1.
Conclusion:
The flow speed at location 2 is higher than at location 1, and the pressure at location 2 is lower than at location 1, with a pressure difference of approximately -86.1 kPa.