To calculate the value of $P_2$, we will use the given formula for the flow rate:

$V = \frac{\pi p r^4}{8 \eta l}$

Where:

• $V$ is the flow rate of the fluid (1.25 × 10$^{-3}$ m$^3$/sec)
• $p$ is the pressure gradient between the two ends of the pipe
• $r$ is the radius of the needle (0.125 mm = 0.125 × 10$^{-3}$ m)
• $\eta$ is the viscosity of the fluid (1 × 10$^{-3}$ Ns/m²)
• $l$ is the length of the needle (2.50 cm = 2.50 × 10$^{-2}$ m)

We are given the pressure difference $p = P_1 - P_2$, and $P_1$ is provided as 1.066 × 10² N/m². We need to solve for $P_2$.

Step 1: Rearrange the equation to solve for $p$:

We can express the pressure difference $p$ as:

$p = \frac{8 \eta l V}{\pi r^4}$

Step 2: Plug in the known values:

$p = \frac{8 \times (1 \times 10^{-3}) \times (2.50 \times 10^{-2}) \times (1.25 \times 10^{-3})}{\pi \times (0.125 \times 10^{-3})^4}$

Step 3: Simplify and compute:

Let's break it down to get the value of $p$.

Let me calculate that for you. math

Given values

V = 1.25 * 10**-3 # m^3/sec eta = 1 * 10**-3 # Ns/m^2 l = 2.50 * 10**-2 # m r = 0.125 * 10**-3 # m

Calculate pressure gradient p

p = (8 * eta * l * V) / (math.pi * r**4) pThe calculated pressure gradient $p$ is approximately $3.26 \times 10^8 \, \text{N/m}^2$.

Step 4: Solve for $P_2$:

We know that:

$p = P_1 - P_2$

Rearranging this equation to find $P_2$:

$P_2 = P_1 - p$

Substitute the given value of $P_1 = 1.066 \times 10^2 \, \text{N/m}^2$ and the calculated value of $p$:

$P_2 = 1.066 \times 10^2 - 3.26 \times 10^8$

Let me calculate that for you. Given P1 P1 = 1.066 * 10**2 # N/m^2

Calculate P2

P2 = P1 - p P2The calculated value of $P_2$ is approximately $-3.26 \times 10^8 \, \text{N/m}^2$.

This negative value indicates that the pressure at the second end of the needle is significantly lower than the pressure at the first end, which may correspond to a suction or draw effect in the intravenous fluid system.

If you need further clarification or have more questions, feel free to ask!

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Related Questions:

1. What happens if the pressure gradient in the intravenous system increases?
2. How does the radius of the needle affect the flow rate of the saline solution?
3. Why is the viscosity of saline relevant in calculating the flow rate through a needle?
4. How would the flow rate change if the viscosity of the fluid were doubled?
5. What other factors influence the efficiency of an intravenous fluid delivery system?

Tip: The flow rate in a pipe or needle depends heavily on the fourth power of the radius ($r^4$), so even a small change in radius can have a large impact on the flow rate.