Question

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 kg/m\(^3\)) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 Pa, what is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 12.6) of the fluid.

Short answer

The minimum height \( h \) is approximately 0.580 meters.

Step-by-step solution

Understanding the Problem

We need to find the minimum height \( h \) such that the fluid, having a density of 1050 kg/m^3, moves from the reservoir into the vein, where the gauge pressure is 5980 Pa.

Applying the Hydrostatic Pressure Equation

Since the top of the reservoir is open to the air, the pressure at the top of the reservoir is equal to the atmospheric pressure. To move the fluid into the vein, the pressure due to the fluid column must at least equal the gauge pressure in the vein. The relationship is given by the hydrostatic pressure equation: \[ P = \rho g h \]Where \( P \) is the pressure due to the fluid column, \( \rho \) is the fluid density \( (1050 \text{ kg/m}^3) \), \( g \) is the acceleration due to gravity \( (9.81 \text{ m/s}^2) \), and \( h \) is the height.

Setting Up the Equation for Minimum Height

We set the hydrostatic pressure equal to the vein's gauge pressure to find the minimum height required for the fluid to enter the vein:\[ \rho g h = 5980 \text{ Pa} \]

Solving for Height

Rearrange the equation to solve for \( h \):\[ h = \frac{5980}{1050 \times 9.81} \]Calculate this value to determine \( h \).

Calculating the Result

Substitute the known values:\[ h = \frac{5980}{1050 \times 9.81} \approx 0.580 \text{ meters} \]Therefore, the minimum height \( h \) is approximately 0.580 meters.

Key concepts

Pressure calculation

Pressure calculation is a critical concept in hydrostatics. It involves understanding the force applied per unit area within a fluid. For fluids, the pressure often varies with depth because of the weight of the fluid above.

Some key points in pressure calculation include:

• The formula to calculate pressure in a fluid is given by: \[ P = \rho g h \] Here, \( P \) denotes the pressure, \( \rho \) is the fluid density, \( g \) is the acceleration due to gravity, and \( h \) is the height or depth of the fluid column.
• Pressure is fundamentally a reflection of how much force the fluid exerts on a surface due to gravity replacing the atmosphere at that point.
• Since we're dealing with gauge pressure (pressure above atmospheric pressure), it's essential to recognize the interplay between atmospheric pressure and fluid pressure.

Understanding pressure calculation helps determine how deep or high a fluid needs to be to exert enough pressure to achieve a particular task, such as flowing into a vein as in the exercise example.

Fluid density

Fluid density is a measure of how much mass is contained within a given volume of fluid. It's a fundamental property that impacts how fluids behave and interact with other substances. For the exercise at hand:

• Density is denoted by \( \rho \) and measured in \( \text{kg/m}^3 \). For the intravenous fluid, the density is given as 1050 \( \text{kg/m}^3 \).
• Density directly influences pressure calculations since denser fluids exert more pressure at a given height compared to less dense fluids.
• In practical applications, fluid density can influence the delivery and absorption rates in intravenous treatments or engineering designs in hydrostatic systems.

Therefore, understanding fluid density is crucial for predicting the pressure fluid can exert and ensuring it functions correctly in its application.

Gauge pressure

Gauge pressure is the pressure of a fluid relative to the ambient atmospheric pressure. It measures how much force the fluid in a system is exerting over what is naturally present in the environment.

In solving the exercise:

• The gauge pressure inside the vein is given as 5980 Pa. This indicates the additional pressure the fluid needs to overcome to enter the vein.
• The equation used in hydrostatics problems, like this one, often needs the gauge pressure because it directly affects how much pressure is needed beyond atmospheric pressure to achieve fluid flow.
• Since the reservoir is open to air, the additional height of fluid needed is entirely to overcome this gauge pressure, confirming the calculations for \( h \).

By focusing on gauge pressure exclusively, we can tailor solutions that ensure safe and effective fluid delivery without unnecessary complications due to atmospheric variations.