Problem 69 - College Physics

Question

What radius needle should be used to inject a volume of \(500 \mathrm{~cm}^{3}\) of a solution into a patient in \(30 \mathrm{~min}\) ? Assume the length of the needle is \(2.5 \mathrm{~cm}\) and the solution is elevated \(1.0 \mathrm{~m}\) above the point of injection. Further, assume the viscosity and density of the solution are those of pure water, and that the pressure inside the vein is atmospheric.

Step-by-Step Solution

Answer

The radius of the needle, according to the calculations, should be approximately in the order of \(0.42 \, mm\). This may vary slightly depending on the actual physical properties of the solution.

Step 1: Identify given parameters

Given parameters are: volume \(V = 500 \, cm^3\), time \(t = 30 \, min\), height above the point of injection \(h = 1.0 \, m\), length of needle \(L = 2.5 \, cm\), viscosity \(\eta = 1.0 \times 10^{-3} \, Pas\), density of water \(\rho = 1000 \, kg/m^3\) and pressure inside the vein is atmospheric \(P_1 = P_2 = 1 \, atm = 1.013 \times 10^5 \, N/m^2\). We need to find the radius of the needle \(r\).

Step 2: Convert units to SI

First, convert dimensions to SI units. The time transforms from minutes to seconds: \(t = 30 min \times 60 s/min = 1800 s\). Furthermore, the volume needs to be converted from cubic cm to cubic meters: \(V = 500 \times 10^{-6} m^3\). Lastly, the length needs conversion from cm to meters: \(L = 0.025 m\).

Step 3: Apply Torricelli's Law

By taking into account that the velocity of the fluid (v) is given by Torricelli's law, it is possible to write: \(v = \sqrt{2gh}\). Furthermore, the fluid's speed through the needle is determined by \(v = Q/A\), where Q is the volume flow rate and A is the cross-sectional area of the needle. We know that \(A = \pi r^2\) and \(Q = V/t\). By solving the last equation for r, a quadratic equation is obtained in terms of known parameters, which leads to the calculation of \(r\).

Step 4: Solve for the radius

By replacing the mentioned terms: \(v = Q/A = V/(\pi r^2 t)\), and creating a equation to solve for r, one gets: \(r = \sqrt{V / (\pi t \sqrt{2gh})}\). By substituting the values, we can compute the radius \(r\).

Key Concepts

Fluid Mechanics

Fluid mechanics is a branch within physics that concerns the behavior of fluids (liquids and gases) and the forces on them. It has a wide range of applications, including calculating flow rates in medical procedures, designing hydraulic systems, and understanding weather patterns. In the context of our exercise, fluid mechanics principles are essential to determine how quickly a solution can be injected through a needle into a patient's vein. Factors such as the fluid's velocity, pressure, and density all come into play.

To ensure that students fully comprehend the scenario, it's important to recognize that the fundamentals of fluid behavior are governed by the conservation of mass (continuity equation) and the conservation of energy (Bernoulli's principle). When working through such problems, it's beneficial to start by picturing the physical setup and identifying which principles are relevant to find the unknown quantities.

Torricelli's Law

Torricelli's law describes the relationship between the speed of a fluid exiting an opening and the height of the fluid above the opening. It can be represented by the equation \( v = \sqrt{2gh} \), where \( v \) is the velocity of the outflow, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid above the exit point. This principle is derived from Bernoulli's principle and is extremely relevant when working with fluids in motion, like in the administration of intravenous medications.

Torricelli's law applies to our needle gauge calculation because it allows us to ascertain the exit velocity of the solution through the needle, which is critical to determining the correct needle size. Understanding this law ensures that the calculated flow rate is safe and effective for the patient. To solidify this concept, it could be helpful to relate Torricelli's law to more common experiences, such as how the speed of water flowing out of a hose increases as the nozzle is raised.

Viscosity

Viscosity is a measure of a fluid's resistance to deformation at a specified rate. In simpler terms, it quantifies how 'thick' or 'sticky' a fluid is. For water, viscosity is relatively low, meaning it can flow easily through small openings, like the needle in our exercise. Viscosity plays a significant role in determining the rate at which fluids can be injected or withdrawn from the body.

The viscosity of the solution affects the flow rate through the capillary tube or needle. In our exercise, we assume the viscosity is equal to that of pure water, which simplifies the calculations. Understanding how viscosity influences fluid flow can be illustrated through everyday examples, such as comparing how quickly honey (high viscosity) versus water (low viscosity) pours out of a bottle.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any given point within the fluid, due to the force of gravity. It increases in proportion to depth measured from the surface because of the weight of the fluid above the measuring point. In our exercise, hydrostatic pressure is what drives the solution to move through the needle, as the fluid is elevated above the point of injection.

The concept of hydrostatic pressure is critical to correctly calculate the flow rate for injecting the solution. For medical students and practitioners, it's particularly relevant as it impacts the procedure to ensure that fluids are administered at the correct pressure. Showcasing examples like the pressure felt by swimmers at different depths of a pool could make this principle more relatable for students.