water at 20degC is flowing in a horizontal pipe that is 20m long. the flow

is laminar and the water completely fills the pipe.A pump maintains a gause pressure of 1400pa,at a large tank at one end of the pipe.The other end of the pipe is open to the air,The viscosity of water at 20degC is 1.005poise.

(a)If the pipe has diameter 8cm ,what is the volume flow rate?

(b)What gauge pressure must the pump provide to achive the same volume flow rate for a pipe with a diameter of 4cm?

(c)For a pipe in part (a) and the same gauge pressure maintened by a pump,what does the volume flow rate become if the water is at a temperature of 60degC (the viscosity of water at 60degC is 0.469poise) ans-(a)7*10^-4m^3/s (b)2.24*10^4Pa (c)0.5*10^-3 m^3/s how???????????

To tackle this problem, we need to apply some principles from fluid mechanics, particularly focusing on laminar flow and the relationship between pressure, viscosity, and flow rate. Let's break down each part of your question step by step.

Understanding Volume Flow Rate in Laminar Flow

In laminar flow, the volume flow rate (Q) can be calculated using the Hagen-Poiseuille equation, which is given by:

Q = (π * r^4 * ΔP) / (8 * μ * L)

Where:

• Q = volume flow rate (m³/s)
• r = radius of the pipe (m)
• ΔP = pressure difference (Pa)
• μ = dynamic viscosity (Pa·s)
• L = length of the pipe (m)

Part (a): Volume Flow Rate for an 8 cm Diameter Pipe

First, we need to convert the diameter of the pipe into radius:

Diameter = 8 cm = 0.08 m, thus radius (r) = 0.08 m / 2 = 0.04 m.

Given:

• Gauge pressure (ΔP) = 1400 Pa
• Length of the pipe (L) = 20 m
• Viscosity (μ) = 1.005 poise = 1.005 * 0.1 Pa·s = 0.1005 Pa·s (since 1 poise = 0.1 Pa·s)

Now, substituting these values into the Hagen-Poiseuille equation:

Q = (π * (0.04 m)^4 * 1400 Pa) / (8 * 0.1005 Pa·s * 20 m)

Calculating this gives:

Q ≈ 7 * 10^-4 m³/s.

Part (b): Volume Flow Rate for a 4 cm Diameter Pipe

Next, we need to find the gauge pressure required to achieve the same volume flow rate (Q) for a pipe with a diameter of 4 cm.

For the new diameter:

Diameter = 4 cm = 0.04 m, thus radius (r) = 0.04 m / 2 = 0.02 m.

We will use the same volume flow rate (Q = 7 * 10^-4 m³/s) and the same length (L = 20 m) and viscosity (μ = 0.1005 Pa·s).

Rearranging the Hagen-Poiseuille equation to solve for ΔP:

ΔP = (8 * μ * L * Q) / (π * r^4)

Substituting the values:

ΔP = (8 * 0.1005 Pa·s * 20 m * 7 * 10^-4 m³/s) / (π * (0.02 m)^4)

This calculation results in:

ΔP ≈ 2.24 * 10^4 Pa.

Part (c): Volume Flow Rate at 60 degC

Finally, we need to determine the volume flow rate when the water temperature is increased to 60 degC, where the viscosity is 0.469 poise (or 0.0469 Pa·s).

Using the same diameter (8 cm) and the same gauge pressure (1400 Pa), we apply the Hagen-Poiseuille equation again:

Q = (π * (0.04 m)^4 * 1400 Pa) / (8 * 0.0469 Pa·s * 20 m)

Calculating this gives:

Q ≈ 0.5 * 10^-3 m³/s.

Summary

To summarize:

• For the 8 cm diameter pipe, the volume flow rate is approximately 7 * 10^-4 m³/s.
• For the 4 cm diameter pipe, the required gauge pressure is about 2.24 * 10^4 Pa.
• At 60 degC, the volume flow rate for the 8 cm diameter pipe becomes approximately 0.5 * 10^-3 m³/s.

This analysis shows how viscosity and pipe diameter significantly influence flow rates in laminar flow conditions. Understanding these relationships is crucial for applications in fluid dynamics and engineering.