he working fluid is water. There is no heat transfer to or from the water, and the f the water per unit mass is constant. The water may be considered incompressibl a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system he closed-system approach), calculate the work necessary to raise the rack and ca he change in potential energy of the water in the system). b) Repeat part (a), taking all the water plus the car and the rack as the system. c) Repeat part (a), taking an open-system approach; choose as your system the vo
he working fluid is water. There is no heat transfer to or from the water, and the f the water per unit mass is constant. The water may be considered incompressibl a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system he closed-system approach), calculate the work necessary to raise the rack and ca he change in potential energy of the water in the system). b) Repeat part (a), taking all the water plus the car and the rack as the system. c) Repeat part (a), taking an open-system approach; choose as your system the vo
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:3. A hydraulic lift is shown below. The combined mass of the piston, rack, and car is 4000 lbm.
The working fluid is water. There is no heat transfer to or from the water, and the internal energy
of the water per unit mass is constant. The water may be considered incompressible.
(a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system (i.e., taking
the closed-system approach), calculate the work necessary to raise the rack and car 1 ft (neglect
the change in potential energy of the water in the system).
(b) Repeat part (a), taking all the water plus the car and the rack as the system.
(c) Repeat part (a), taking an open-system approach; choose as your system the volume of the
hydraulic cylinder, excluding the piston, rack, and car. If the absolute pressure in the system is
1000 lbf/in², calculate the volume that must flow in to raise the car 1 ft.
Reservoir
Pump
Hydraulic
cylinder
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i think your part B in this solution is for part A, how do you solve for part B now that we have to take water into account
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I am confused on the (-) & how u went from ft/in^2 to m^3
![### Mathematical Transcription
The image contains a series of equations related to energy flow and mathematical substitutions. Here's a step-by-step description:
1. **Energy Flow Equation:**
\[
dW_{energyflow} = - \left( Pv_{in} \right) dm_{in}
\]
2. **Velocity and Mass Flow Rate Relationship:**
\[
v_{in} dm_{in} = \frac{dW_{energyflow}}{-P}
\]
3. **Refined Equation Considering Absolute Pressure:**
\[
v_{in} dm_{in} = \frac{dW_{energyflow}}{\left( P_{abs} - P_{am} \right)}
\]
4. **Substitution with Given Values:**
\[
v_{in} dm_{in} = \frac{-4000}{(1000 - 14.7)}
\]
5. **Result in Square Inches:**
\[
v_{in} dm_{in} = -4.06 \, \text{ft} - \text{in}^2
\]
6. **Result in Cubic Meters:**
\[
v_{in} dm_{in} = 7.93 \times 10^{-4} \, \text{m}^3
\]
### Explanation
- The equations involve energy flow (\(dW_{energyflow}\)) and relate it to velocity (\(v_{in}\)) and mass flow rate (\(dm_{in}\)).
- The adjustments account for pressure differences using absolute and ambient pressure terms (\(P_{abs}\) and \(P_{am}\)).
- Two units are provided for the result: square feet-inches and cubic meters, showing both imperial and metric measurements.](https://content.bartleby.com/qna-images/question/f4dfc165-1571-419f-99fc-a4587d76c340/f33b6735-a447-4a54-8591-4139a3ed8be4/jfyw3db_processed.jpeg)
Transcribed Image Text:### Mathematical Transcription
The image contains a series of equations related to energy flow and mathematical substitutions. Here's a step-by-step description:
1. **Energy Flow Equation:**
\[
dW_{energyflow} = - \left( Pv_{in} \right) dm_{in}
\]
2. **Velocity and Mass Flow Rate Relationship:**
\[
v_{in} dm_{in} = \frac{dW_{energyflow}}{-P}
\]
3. **Refined Equation Considering Absolute Pressure:**
\[
v_{in} dm_{in} = \frac{dW_{energyflow}}{\left( P_{abs} - P_{am} \right)}
\]
4. **Substitution with Given Values:**
\[
v_{in} dm_{in} = \frac{-4000}{(1000 - 14.7)}
\]
5. **Result in Square Inches:**
\[
v_{in} dm_{in} = -4.06 \, \text{ft} - \text{in}^2
\]
6. **Result in Cubic Meters:**
\[
v_{in} dm_{in} = 7.93 \times 10^{-4} \, \text{m}^3
\]
### Explanation
- The equations involve energy flow (\(dW_{energyflow}\)) and relate it to velocity (\(v_{in}\)) and mass flow rate (\(dm_{in}\)).
- The adjustments account for pressure differences using absolute and ambient pressure terms (\(P_{abs}\) and \(P_{am}\)).
- Two units are provided for the result: square feet-inches and cubic meters, showing both imperial and metric measurements.
Solution
by Bartleby ExpertKnowledge Booster
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