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A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill \(2200.355-\) L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 \(\mathrm{cm}^{2} .\) At point \(1,1.35\) above point \(2,\) the cross-sectional area is 2.00 \(\mathrm{cm}^{2} .\) Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point \(1 .\)

Short Answer

Expert verified
Mass flow: 36.67 kg/s, Volume flow: 0.03667 m³/s, Speeds: 44.84 m/s at point 2, 179.36 m/s at point 1, Pressure: Negative, suggesting cavitation.

Step by step solution

01

Mass Flow Rate

To find the mass flow rate \( \dot{m} \), we start with the given information that 2200.355 liters (L) of the soft drink fills per minute. Note that 1 L of water has a mass of 1 kg (since it is mostly water). Thus, the mass flow rate \( \dot{m} \) is 2200.355 kg/min. To convert this to kg/s, divide by 60 s/min: \[ \dot{m} = \frac{2200.355 \ \text{kg/min}}{60} = 36.67258 \ \text{kg/s} \].
02

Volume Flow Rate

The volume flow rate \( \dot{Q} \) is found by the volume of liquid being moved per unit time. Since we have 2200.355 L/min, the volume flow rate is \( 2200.355 \ \text{L/min} \). Converting to \( \text{m}^3/\text{s} \) (as 1 L = 0.001 \( \text{m}^3 \)): \[ \dot{Q} = \frac{2200.355 \cdot 0.001}{60} \approx 0.03667258 \ \text{m}^3/s \].
03

Flow Speed at Point 2

To find the flow speed \( v_2 \) at point 2, use the formula \( \dot{Q} = A_2 v_2 \). Rearrange to solve for \( v_2 \): \( v_2 = \frac{\dot{Q}}{A_2} \). \( A_2 = 8.00 \ \text{cm}^2 = 8.00 \times 10^{-4} \ \text{m}^2 \). \[ v_2 = \frac{0.03667258}{8.00 \times 10^{-4}} \approx 45.84 \ \text{m/s} \].
04

Flow Speed at Point 1

Use the continuity equation \( A_1 v_1 = A_2 v_2 \) to calculate \( v_1 \). Rearranging gives \( v_1 = \frac{A_2 v_2}{A_1} \). With \( A_1 = 2.00 \times 10^{-4} \ \text{m}^2 \): \[ v_1 = \frac{8.00 \times 10^{-4} \times 45.84}{2.00 \times 10^{-4}} = 183.36 \ \text{m/s} \].
05

Gauge Pressure at Point 1

Using Bernoulli’s equation, \( P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \) simplifying and rearranging for \( P_1 \): Since \( h_2 = 0 \) and \( h_1 = 1.35 \): \[ P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g (-1.35) \]. With \( \rho = 1000 \ \text{kg/m}^3 \), \( g = 9.81 \ \text{m/s}^2 \), \( P_2 = 152000 \ \text{Pa} \): \[ P_1 = 152000 + \frac{1}{2} \times 1000 \times (45.84^2 - 183.36^2) - 1000 \times 9.81 \times 1.35 \approx -152938 \ \text{Pa} \]. Negative gauge pressure suggests cavitation at this location.

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Key Concepts

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

Mass Flow Rate
Mass flow rate, often denoted as \( \dot{m} \), is a measure of the mass of a fluid that passes through a cross-section per unit of time. It is typically expressed in units such as kilograms per second (kg/s). Think of it as the weight of the fluid moving through the pipe every second.
In the context of our problem, the soft drink, which is mostly water, has a mass flow rate based on the volume that can fill 2200.355-liter cans per minute. Since water has a density of about 1 kg/L, this volume translates directly to mass in kilograms. To find \( \dot{m} \) in kg/s, you need to convert from minutes to seconds by dividing by 60. Thus, the mass flow rate becomes 36.67258 kg/s.
This calculation is essential because it sets the stage for understanding how much substance is being moved through different sections of the pipeline. It also helps in determining other variables like volume flow rate and the changes in flow speed at different points.
Volume Flow Rate
Volume flow rate, represented as \( \dot{Q} \), measures how much volume of fluid passes through a given surface per unit time. It typically uses units such as cubic meters per second (m\(^3\)/s).
For the problem at hand, the given volume flow rate is 2200.355 liters per minute, which can be converted to m\(^3\) by recognizing that 1 liter equals 0.001 m\(^3\). By doing the conversion and adjusting for time (i.e., dividing by 60 to switch from minutes to seconds), we find \( \dot{Q} = 0.03667258 \, \text{m}^3/s \).
Understanding the volume flow rate aids in deriving critical insights about how fast the fluid is moving in terms of volume. It is a key link between mass flow rate and velocity, as seen in various applications such as calculating the velocity of fluids through different cross-sectional areas.
Bernoulli's Equation
Bernoulli's equation is a cornerstone in fluid dynamics, providing a relationship between pressure, velocity, and elevation in a flowing fluid. It essentially states that an increase in the fluid's speed occurs simultaneously with a decrease in pressure or potential energy of the fluid.
For the purpose of our exercise, Bernoulli's equation is used to find the gauge pressure at point 1 of the pipe. It is vital to note the equation simplifies the concept by assuming steady, incompressible flow with no friction losses, describing the fluid energy conservation along a streamline.
The equation looks like this: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \]By rearranging, we can solve for \( P_1 \), which gives the gauge pressure at the higher level (point 1). In the given context, after inputting known values and solving the equation, a negative result indicates possible cavitation, a phenomenon where vapor bubbles form due to low pressure.
Continuity Equation
The continuity equation is a fundamental concept ensuring that mass is conserved in a fluid flow. Simply put, it states that for any incompressible, steady flow through a pipe, the mass flow rate must remain constant across any cross-section of the pipe.
In mathematical terms, this is expressed as:\[ A_1v_1 = A_2v_2 \]where \( A \) represents the cross-sectional area of the pipe and \( v \) is the flow speed. This equation helps us find the flow speed at different points along the pipeline.
For example, given the flow speed at point 2, we can find the flow speed at point 1 by rearranging the continuity equation. It shows that as the pipe narrows (decreased cross-sectional area), the flow speed must increase to maintain the mass flow rate. This principle is crucial for designing systems where fluid speeds must be controlled or adjusted.

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Most popular questions from this chapter

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