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An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?
- a)0.53
- b)0.67
- c)0.48
- d)0.70
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
An imaginary engine receives heat and does work on a slowly moving pis...
As we know that area of cycle on Pv
diagram gives work.
So W = πr2
= 2355 kJ/kg
∑W = ∑Q
Qreceived ~ Qrejected = 2355.
Qreceived = 2355 + 1000 3355 kJ/kg
= 50.701
Most Upvoted Answer
An imaginary engine receives heat and does work on a slowly moving pis...
Given data:
Diameter of the cycle on P-v diagram = 10 cm
1 cm = 300 kPa
1 cm = 0.1 m3/kg
Heat rejected in a cycle = 1000 kJ/kg
To find: Thermal efficiency of the engine
Solution:
1. Let us first convert the diameter of the cycle into pressure and volume values:
Diameter of the cycle = 10 cm
Radius of the cycle = Diameter/2 = 5 cm
Volume of the cycle = πr² = 3.14*(5/100)² = 0.00785 m³/kg
Pressure at point A (beginning of cycle) = 300 kPa (from the given conversion)
Volume at point A = 1 cm = 0.1 m³/kg (from the given conversion)
Pressure at point B (maximum pressure in cycle) = 600 kPa (since the diameter of the cycle is 10 cm)
Volume at point B = 0.00785 m³/kg / 2 = 0.003925 m³/kg (since point B is halfway between the center and the edge of the cycle)
Pressure at point C (end of heat rejection process) = 300 kPa (since the cycle is a closed loop)
Volume at point C = 0.1 m³/kg (from the given conversion)
Volume at point D (maximum volume in cycle) = 0.00785 m³/kg (since the diameter of the cycle is 10 cm)
Pressure at point D = 300 kPa (since point D is at the bottom of the cycle)
2. Now we can plot these points on a P-v diagram:
[![P-v diagram][1]][1]
3. The work done by the engine in a cycle is the area enclosed by the cycle on the P-v diagram. We can calculate this area by breaking it up into two parts:
- Area 1: The shaded triangle ABP
- Area 2: The shaded region BCDP
Area 1 = (1/2) * (600-300) * (0.003925-0.1) = -0.112 kJ/kg
(Note: The negative sign indicates that work is being done on the engine, not by the engine)
Area 2 = (300-300) * (0.00785-0.003925) = 0.0234 kJ/kg
Total work done in a cycle = Area 1 + Area 2 = -0.0886 kJ/kg
4. The thermal efficiency of the engine is given by:
ηth = (Work output) / (Heat input)
= (Work output) / (Heat input - Heat rejected)
= (-0.0886 kJ/kg) / (Qin - 1000 kJ/kg)
Qin is the heat input per kg of working fluid in a cycle. To find this, we can use the first law of thermodynamics:
Qin = Qout + Work done
Qin = 1000 kJ/kg - (-0.0886 kJ/kg) = 1000.0886 kJ/kg
5. Substituting this value in the formula for thermal efficiency:
ηth = (-0.0886 kJ/kg) / (1000.
Diameter of the cycle on P-v diagram = 10 cm
1 cm = 300 kPa
1 cm = 0.1 m3/kg
Heat rejected in a cycle = 1000 kJ/kg
To find: Thermal efficiency of the engine
Solution:
1. Let us first convert the diameter of the cycle into pressure and volume values:
Diameter of the cycle = 10 cm
Radius of the cycle = Diameter/2 = 5 cm
Volume of the cycle = πr² = 3.14*(5/100)² = 0.00785 m³/kg
Pressure at point A (beginning of cycle) = 300 kPa (from the given conversion)
Volume at point A = 1 cm = 0.1 m³/kg (from the given conversion)
Pressure at point B (maximum pressure in cycle) = 600 kPa (since the diameter of the cycle is 10 cm)
Volume at point B = 0.00785 m³/kg / 2 = 0.003925 m³/kg (since point B is halfway between the center and the edge of the cycle)
Pressure at point C (end of heat rejection process) = 300 kPa (since the cycle is a closed loop)
Volume at point C = 0.1 m³/kg (from the given conversion)
Volume at point D (maximum volume in cycle) = 0.00785 m³/kg (since the diameter of the cycle is 10 cm)
Pressure at point D = 300 kPa (since point D is at the bottom of the cycle)
2. Now we can plot these points on a P-v diagram:
[![P-v diagram][1]][1]
3. The work done by the engine in a cycle is the area enclosed by the cycle on the P-v diagram. We can calculate this area by breaking it up into two parts:
- Area 1: The shaded triangle ABP
- Area 2: The shaded region BCDP
Area 1 = (1/2) * (600-300) * (0.003925-0.1) = -0.112 kJ/kg
(Note: The negative sign indicates that work is being done on the engine, not by the engine)
Area 2 = (300-300) * (0.00785-0.003925) = 0.0234 kJ/kg
Total work done in a cycle = Area 1 + Area 2 = -0.0886 kJ/kg
4. The thermal efficiency of the engine is given by:
ηth = (Work output) / (Heat input)
= (Work output) / (Heat input - Heat rejected)
= (-0.0886 kJ/kg) / (Qin - 1000 kJ/kg)
Qin is the heat input per kg of working fluid in a cycle. To find this, we can use the first law of thermodynamics:
Qin = Qout + Work done
Qin = 1000 kJ/kg - (-0.0886 kJ/kg) = 1000.0886 kJ/kg
5. Substituting this value in the formula for thermal efficiency:
ηth = (-0.0886 kJ/kg) / (1000.
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An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer?
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An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer?.
An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An imaginary engine receives heat and does work on a slowly moving piston at such rates that the cycle of operation of 1 kg of working fluid can be represented as a circle 10 cm in diameter on a P-v diagram an which 1 cm = 300 kPa and 1 cm = 0.1 m3/kg. If the heat rejected by engine in a cycle is 1000 kJ/kg of working fluid, what will be its thermal efficiency?a)0.53b)0.67c)0.48d)0.70Correct answer is option 'D'. Can you explain this answer?.
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