A horizontal pipe carries a smoothly flowing liquid of density 1110 kg/m^3. At Locations 1 and 2 along the pipe, the diameters are d1 = 5.45 cm and d2 = 2.83 cm, respectively. The flow speed at Location 1 is 2.83 m/s. What is the pressure difference ΔP between Location 2 and Location 1 (including its sign)? Ignore viscosity.
The Correct Answer and Explanation is:
To calculate the pressure difference ΔP\Delta PΔP between Locations 2 and 1 in the pipe, we can use the Bernoulli equation and the continuity equation.
Step 1: Apply the Continuity Equation
The continuity equation states that for an incompressible fluid (like water or most liquids), the flow rate must remain constant across different sections of the pipe. This is expressed as:A1v1=A2v2A_1 v_1 = A_2 v_2A1v1=A2v2
Where:
- A1A_1A1 and A2A_2A2 are the cross-sectional areas at Locations 1 and 2, respectively,
- v1v_1v1 and v2v_2v2 are the flow speeds at Locations 1 and 2, respectively.
The cross-sectional area of a pipe is given by the formula A=πd2/4A = \pi d^2 / 4A=πd2/4, where ddd is the diameter of the pipe.
At Location 1:A1=πd124A_1 = \frac{\pi d_1^2}{4}A1=4πd12
At Location 2:A2=πd224A_2 = \frac{\pi d_2^2}{4}A2=4πd22
Now, we can use the continuity equation to find the flow speed v2v_2v2 at Location 2.πd124v1=πd224v2\frac{\pi d_1^2}{4} v_1 = \frac{\pi d_2^2}{4} v_24πd12v1=4πd22v2
Simplifying:d12v1=d22v2d_1^2 v_1 = d_2^2 v_2d12v1=d22v2
Solving for v2v_2v2:v2=d12v1d22v_2 = \frac{d_1^2 v_1}{d_2^2}v2=d22d12v1
Substituting the values:
- d1=5.45 cm=0.0545 md_1 = 5.45 \, \text{cm} = 0.0545 \, \text{m}d1=5.45cm=0.0545m,
- d2=2.83 cm=0.0283 md_2 = 2.83 \, \text{cm} = 0.0283 \, \text{m}d2=2.83cm=0.0283m,
- v1=2.83 m/sv_1 = 2.83 \, \text{m/s}v1=2.83m/s.
v2=(0.0545)2×2.83(0.0283)2v_2 = \frac{(0.0545)^2 \times 2.83}{(0.0283)^2}v2=(0.0283)2(0.0545)2×2.83v2≈11.38 m/sv_2 \approx 11.38 \, \text{m/s}v2≈11.38m/s
Step 2: Apply the Bernoulli Equation
The Bernoulli equation for an incompressible fluid is:P1+12ρv12=P2+12ρv22P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2P1+21ρv12=P2+21ρv22
Where:
- P1P_1P1 and P2P_2P2 are the pressures at Locations 1 and 2,
- ρ\rhoρ is the fluid density,
- v1v_1v1 and v2v_2v2 are the velocities at Locations 1 and 2.
Rearranging to solve for the pressure difference ΔP=P2−P1\Delta P = P_2 – P_1ΔP=P2−P1:P2−P1=12ρ(v12−v22)P_2 – P_1 = \frac{1}{2} \rho (v_1^2 – v_2^2)P2−P1=21ρ(v12−v22)
Substitute the values:
- ρ=1110 kg/m3\rho = 1110 \, \text{kg/m}^3ρ=1110kg/m3,
- v1=2.83 m/sv_1 = 2.83 \, \text{m/s}v1=2.83m/s,
- v2=11.38 m/sv_2 = 11.38 \, \text{m/s}v2=11.38m/s.
ΔP=12×1110×(2.832−11.382)\Delta P = \frac{1}{2} \times 1110 \times \left( 2.83^2 – 11.38^2 \right)ΔP=21×1110×(2.832−11.382)ΔP=12×1110×(8.0089−129.6336)\Delta P = \frac{1}{2} \times 1110 \times (8.0089 – 129.6336)ΔP=21×1110×(8.0089−129.6336)ΔP≈12×1110×(−121.6247)\Delta P \approx \frac{1}{2} \times 1110 \times (-121.6247)ΔP≈21×1110×(−121.6247)ΔP≈−67460.1 Pa\Delta P \approx -67460.1 \, \text{Pa}ΔP≈−67460.1Pa
Final Answer:
The pressure difference ΔP\Delta PΔP between Locations 2 and 1 is approximately -67.46 kPa, meaning the pressure at Location 2 is lower than the pressure at Location 1.
