/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A collapsible plastic bag (Fig. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A collapsible plastic bag (Fig. P9.23) contains a glucose solution. If the average gauge pressure in the vein is \(1.33 \times 10^{3} \mathrm{~Pa}\), what must be the minimum height \(h\) of the bag in order to infuse glucose into the vein? Assume the specific gravity of the solution is \(1.02\).

Short Answer

Expert verified
The minimum height \( h \) of the glucose bag must be approximately \(0.13 m\) or \(13 cm\) to infuse glucose into the vein.

Step by step solution

01

- Compute Pressure of Glucose Solution

To start with this problem, we have to first compute the pressure exerted by the glucose solution in the bag at depth \(h\). According to Pascal’s principle, the pressure at a point in a fluid is determined by the weight of the fluid above it. In mathematical terms, it's given by \[ P = \rho gh \] where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity and \( h \) is the height or depth of the fluid column above the point. The density \(\rho\) can be calculated as the specific gravity times the density of water, which is \( 1.02 \times 10^{3} \, \mathrm{kg/m^3} \)
02

- Solve Pressure Equation

Given the pressure of the vein \(1.33 \times 10^{3} Pa\), we equalize this to the weight of the glucose solution to find the minimum \( h \). So we solve the following equation for h: \( h= \frac{P_{\mathrm{vein}}}{\rho g} \). We substitute the known values: \(1.33 \times 10^{3} \, \mathrm{Pa}\) for \(P_{\mathrm{vein}}\), \(1.02 \times 10^{3} \, kg/m^3\) for \(rho\) and \(9.81 \, m/s^2\) for \(g\).
03

- Compute Minimum Height

After solving the equation, we find the minimum height required to infuse glucose into the vein. This step is mostly arithmetical and requires careful calculation to get an accurate result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Calculating pressure in fluid mechanics is essential, especially when dealing with a fluid column. Pressure (P) in a fluid is the force exerted by the fluid per unit area on the walls of its container. In general, it is defined as P = \frac{F}{A}, where F is the force applied and A is the area over which the force is distributed.
For fluids specifically, we use the formula P = \rho gh to calculate pressure due to a fluid column, where:
  • \( \rho \) is the density of the fluid
  • \( g \) is the acceleration due to gravity
  • \( h \) is the height of the fluid column
The weight of the fluid column is responsible for exerting pressure beneath it. Understanding this relationship is crucial for determining how fluids will behave under different conditions.
Specific Gravity
Specific gravity is a dimensionless number that is used to compare the density of a substance to the density of a reference substance, generally water at 4 °C. The formula to find specific gravity is:\[\text{Specific Gravity} = \frac{\text{Density of substance}}{\text{Density of reference}}\]In the context of fluid mechanics, specific gravity is handy because it lets us understand how a liquid is "heavier" or "lighter" than water. For example, a specific gravity of 1.02 signifies that the solution is 1.02 times heavier than water.
To find the actual density of a glucose solution with a specific gravity of 1.02, multiply by the density of water (1000 kg/m³). This gives you a solution density of 1020 kg/m³. Specific gravity helps in calculating other properties of fluids, such as pressure, necessary for understanding fluid dynamics in various systems.
Pascal’s Principle
Pascal's Principle, also known as the principle of transmission of fluid-pressure, states that changes in pressure applied to a contained fluid are transmitted undiminished to every point of the fluid and to the walls of its container. Essentially, this means when you apply pressure at one point in a fluid, that pressure is maintained throughout the fluid.
This principle is vital in explaining how fluid systems work, for instance in hydraulic lifts or in our case, intravenous infusion. If the bag containing the glucose solution is elevated to a sufficient height, the pressure exerted by the fluid overcomes the pressure in the vein allowing infusion. In simple terms, the pressure differential due to gravity ensures the flow from higher to lower pressure areas, ensuring nutrients reach the bloodstream effectively.
Fluid Column Height
The concept of fluid column height is crucial in understanding how fluids create pressure in contained areas. For the infusion of glucose into a vein, the height of the glucose solution bag actually changes the pressure exerted by the fluid.
In the exercise, we calculate the minimum height using the relationship \[ h = \frac{P_{\text{vein}}}{\rho g} \]where \( P_{\text{vein}} \)is the pressure of the vein, \( \rho \)is the density of the fluid, and \( g \)is the gravitational acceleration (9.81 m/s²). The height \( h \)is critical to ensure the fluid overcomes the venous pressure and enters the bloodstream. Elevating the bag to this height ensures gravitational potential energy is converted to pressure, facilitating flow due to not only the density of the fluid but also Pascal’s principle that makes sure pressure is sustained throughout the fluid.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hypodermic syringe contains a medicine with the density of water (Fig. P9.47). The barrel of the syringe has a cross-sectional area of \(2.50 \times 10^{-5} \mathrm{~m}^{2}\). In the absence of a force on the plunger, the pressure everywhere is \(1.00 \mathrm{~atm}\). A force \(\overrightarrow{\mathbf{F}}\) of magnitude \(2.00 \mathrm{~N}\) is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle. Assume the pressure in the needle remains equal to \(1.00 \mathrm{~atm}\) and that the syringe is horizontal.

A sample of an unknown material appears to weigh \(300 \mathrm{~N}\) in air and \(200 \mathrm{~N}\) when immersed in alcohol of specific gravity \(0.700\). What are (a) the volume and (b) the density of the material?

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0\) - L bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately \(2.4 \mathrm{~cm}^{2}\). During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of \(80 \mathrm{~km} / \mathrm{h}\) in \(5.0 \mathrm{~ms}\). If the arm has an effective mass of \(3.0 \mathrm{~kg}\) and bone material can withstand a maximum compressional stress of \(16 \times 10^{7} \mathrm{~Pa}\), is the arm likely to withstand the crash?

A \(200-\mathrm{kg}\) load is hung on a wire of length \(4.00 \mathrm{~m}\), cross-sectional area \(0.200 \times 10^{-4} \mathrm{~m}^{2}\), and Young's modulus \(8.00 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). What is its increase in length?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.