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An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.?
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An imaginary engine receives heat and performs work on a slowly moving...
Given:
- Cycle of operation of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram
- Scale: 1 cm = 300 kPa on the Y-axis, 1 cm = 0.1 m^3 on the X-axis
To Find:
- Net work done
Assumptions:
- The fluid undergoes a reversible process.
Formula Used:
- The work done by the fluid during a cycle is given by the area enclosed by the cycle on the P-V diagram.
- Net work done = Area enclosed by the cycle
Calculations:
Step 1: Calculate the radius of the circle represented by the fluid on the P-V diagram.
Given that the diameter of the circle is 10 cm, the radius (r) can be calculated as:
r = diameter/2 = 10 cm / 2 = 5 cm = 0.05 m
Step 2: Calculate the area of the circle represented by the fluid on the P-V diagram.
The area of a circle is given by the formula:
Area = π * r^2
Substituting the values, we get:
Area = π * (0.05 m)^2
Step 3: Convert the area from square meters to square centimeters.
Since the scale on the X-axis is 1 cm = 0.1 m^3, we need to convert square meters to square centimeters.
1 m^2 = (1/0.1)^2 cm^2 = 100 cm^2
Therefore, the area in square centimeters is:
Area = π * (0.05 m)^2 * 100 cm^2/m^2
Step 4: Calculate the net work done by multiplying the area by the scale on the Y-axis.
Since the scale on the Y-axis is 1 cm = 300 kPa, we need to multiply the area by 300 kPa to get the net work done.
Net work done = Area * 300 kPa
Step 5: Convert the net work done from kPa*cm^2 to Joules (J).
1 J = 1 kPa * m^3
1 cm^2 = (1/100)^2 m^2 = 0.0001 m^2
Therefore, the net work done in Joules is:
Net work done = Area * 300 kPa * 0.0001 m^2/cm^2
Final Answer:
The net work done by the imaginary engine is the calculated value of the net work done in Joules.
- Cycle of operation of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram
- Scale: 1 cm = 300 kPa on the Y-axis, 1 cm = 0.1 m^3 on the X-axis
To Find:
- Net work done
Assumptions:
- The fluid undergoes a reversible process.
Formula Used:
- The work done by the fluid during a cycle is given by the area enclosed by the cycle on the P-V diagram.
- Net work done = Area enclosed by the cycle
Calculations:
Step 1: Calculate the radius of the circle represented by the fluid on the P-V diagram.
Given that the diameter of the circle is 10 cm, the radius (r) can be calculated as:
r = diameter/2 = 10 cm / 2 = 5 cm = 0.05 m
Step 2: Calculate the area of the circle represented by the fluid on the P-V diagram.
The area of a circle is given by the formula:
Area = π * r^2
Substituting the values, we get:
Area = π * (0.05 m)^2
Step 3: Convert the area from square meters to square centimeters.
Since the scale on the X-axis is 1 cm = 0.1 m^3, we need to convert square meters to square centimeters.
1 m^2 = (1/0.1)^2 cm^2 = 100 cm^2
Therefore, the area in square centimeters is:
Area = π * (0.05 m)^2 * 100 cm^2/m^2
Step 4: Calculate the net work done by multiplying the area by the scale on the Y-axis.
Since the scale on the Y-axis is 1 cm = 300 kPa, we need to multiply the area by 300 kPa to get the net work done.
Net work done = Area * 300 kPa
Step 5: Convert the net work done from kPa*cm^2 to Joules (J).
1 J = 1 kPa * m^3
1 cm^2 = (1/100)^2 m^2 = 0.0001 m^2
Therefore, the net work done in Joules is:
Net work done = Area * 300 kPa * 0.0001 m^2/cm^2
Final Answer:
The net work done by the imaginary engine is the calculated value of the net work done in Joules.
Community Answer
An imaginary engine receives heat and performs work on a slowly moving...
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An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.?
Question Description
An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.?.
An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.?.
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An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.?, a detailed solution for An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? has been provided alongside types of An imaginary engine receives heat and performs work on a slowly moving piston at such a rate that the cycle moving of operation of 1 kg of Fluid can be represented as a circle of 1 kg of fluid can be represented as a circle of 10 cm diameter on P-V diagram. The scale is 1 cm =300 kpa on Y-axis and 1 cm = 0.1 m^3 on X-axis. Find the net work done.? theory, EduRev gives you an
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