A horizontal pipe carries a smoothly flowing liquid of density 1290 kg/m^3. At Locations 1 and 2 along the pipe, the diameters are d1 = 5.75 cm and d2 = 2.17 cm, respectively. Side view of pipe with two diameters. At location 1, the diameter of the pipe is d subscript 1. At location 2, the diameter of the pipe is d subscript 2, and is less than d subscript 1. Is the flow speed at Location 2 higher or lower than the flow speed at Location 1? higher cannot tell from the data lower. The flow speed at Location 1 is 1.45 m/s. What is the pressure difference ΔP between Location 2 and Location 1 (including its sign)? Ignore viscosity. ΔP = Pa. Is the pressure at Location 2 higher or lower than the pressure at Location 1? cannot tell higher lower.
The Correct Answer and Explanation is:
The flow speed at Location 2 is higher than at Location 1. Here’s why:
To understand the relationship between flow speed and pipe diameter, we can use continuity equation for fluid flow. According to the continuity equation, for an incompressible fluid (like the liquid in the pipe), the flow rate QQQ is constant throughout the pipe. This equation is expressed as:Q=A1v1=A2v2Q = A_1 v_1 = A_2 v_2Q=A1v1=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 related to its diameter by the formula:A=πd24A = \frac{\pi d^2}{4}A=4πd2
From this, the flow speeds at the two locations are related to the pipe diameters as follows:v1v2=A2A1=d12d22\frac{v_1}{v_2} = \frac{A_2}{A_1} = \frac{d_1^2}{d_2^2}v2v1=A1A2=d22d12
Given that the diameter at Location 2 (d2=2.17d_2 = 2.17d2=2.17 cm) is smaller than at Location 1 (d1=5.75d_1 = 5.75d1=5.75 cm), the area at Location 2 is smaller, and the flow speed at Location 2 must be higher in order to maintain the same flow rate.
Pressure Difference Calculation
Now, to calculate the pressure difference ΔP\Delta PΔP, we apply Bernoulli’s equation, which relates the pressure and velocity of a fluid in steady, incompressible flow. Bernoulli’s equation is given by: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 density of the liquid (1290 kg/m³),
- v1v_1v1 and v2v_2v2 are the velocities at Locations 1 and 2.
Rearranging the equation to find the pressure difference:ΔP=P2−P1=12ρ(v12−v22)\Delta P = P_2 – P_1 = \frac{1}{2} \rho \left( v_1^2 – v_2^2 \right)ΔP=P2−P1=21ρ(v12−v22)
Since v2v_2v2 is higher than v1v_1v1, we can see that v12−v22v_1^2 – v_2^2v12−v22 is negative, which means:ΔP<0\Delta P < 0ΔP<0
Thus, ΔP\Delta PΔP is negative, meaning the pressure at Location 2 is lower than at Location 1.
Final Answer:
- The flow speed at Location 2 is higher.
- The pressure difference ΔP\Delta PΔP is negative, meaning the pressure at Location 2 is lower than at Location 1.
