/*! 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

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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.

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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.

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Most popular questions from this chapter

The approximate diameter of the aorta is \(0.50 \mathrm{~cm}\); that of a capillary is \(10 \mu \mathrm{m}\). The approximate average blood flow speed is \(1.0 \mathrm{~m} / \mathrm{s}\) in the aorta and \(1.0 \mathrm{~cm} / \mathrm{s}\) in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the circulatory system.

The average human has a density of \(945 \mathrm{~kg} / \mathrm{m}^{3}\) after inhaling and \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) after exhaling. (a) Without making any swimming movements, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about \(1230 \mathrm{~kg} / \mathrm{m}^{3}\) ) in each of these cases? (b) Given that bone and muscle are denser than fat, what physical characteristics differentiate "sinkers" (those who tend to sink in water) from "floaters" (those who readily float)?

Old Faithful geyser in Yellowstone Park erupts at approximately 1-hour intervals, and the height of the fountain reaches \(40.0 \mathrm{~m}\) (Fig. P9.57). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c) What is the pressure (above atmospheric pressure) in the heated underground chamber \(175 \mathrm{~m}\) below the vent? You may assume the chamber is large compared with the geyser vent.

Figure P9.85 shows a water tank with a valve. If the valve is opened, what is the maximum height attained by the stream of water coming out of the right side of the tank? Assume \(h=10.0 \mathrm{~m}, L=2.00 \mathrm{~m}\), and \(\theta=30.0^{\circ}\), and that the cross-sectional area at \(A\) is very large compared with that at \(B\).

A spherical weather balloon is filled with hydrogen until its radius is \(3.00 \mathrm{~m}\). Its total mass including the instruments it carries is \(15.0 \mathrm{~kg}\). (a) Find the buoyant force acting on the balloon, assuming the density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?

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