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A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1 Bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the Change in specific internal energy during the throttling process?
- a)Zero
- b)100 kJ/kg
- c)200 kJ/kg
- d)300 kJ/kg
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A fluid flowing along a pipe line undergoes a throttling process from ...
Throttling is a isenthalpic process
h1 = h2 or u1 + p1v1 = u2 + p2v2 or u2 – u1 = p1v1 – p2v2 = 1000 × 0.5 – 100 × 2 = 300 kJ/kg
Most Upvoted Answer
A fluid flowing along a pipe line undergoes a throttling process from ...
Given data:
Initial pressure (P1) = 10 bar
Final pressure (P2) = 1 bar
Initial specific volume (v1) = 0.5 m3/kg
Final specific volume (v2) = 2.0 m3/kg
To find: Change in specific internal energy (Δu)
Formula used: Δu = u2 - u1 = C(v2 - v1)
where,
u = specific internal energy
C = specific heat at constant volume
Assumption: The fluid is assumed to be an incompressible substance and the throttling process is an isenthalpic process.
Calculation:
From the given data, we can calculate the specific enthalpy (h) of the fluid before and after the throttling process using the following formula:
h = u + Pv
where,
P = pressure
Initial enthalpy (h1) = u1 + P1v1 = 0 + 10 × 0.5 = 5 kJ/kg
Final enthalpy (h2) = u2 + P2v2 = ? + 1 × 2.0 = ?
As the process is assumed to be isenthalpic, the enthalpy remains constant, i.e., h1 = h2
Therefore, u2 = u1 + P1v1 - P2v2
Δu = u2 - u1 = P1v1 - P2v2
Substituting the given values, we get:
Δu = 10 × 0.5 - 1 × 2.0 = 5 - 2 = 3 kJ/kg
Therefore, the change in specific internal energy during the throttling process is 300 kJ/kg (multiplying by 1000 to convert from kJ/kg to J/kg).
Hence, option (d) is the correct answer.
Initial pressure (P1) = 10 bar
Final pressure (P2) = 1 bar
Initial specific volume (v1) = 0.5 m3/kg
Final specific volume (v2) = 2.0 m3/kg
To find: Change in specific internal energy (Δu)
Formula used: Δu = u2 - u1 = C(v2 - v1)
where,
u = specific internal energy
C = specific heat at constant volume
Assumption: The fluid is assumed to be an incompressible substance and the throttling process is an isenthalpic process.
Calculation:
From the given data, we can calculate the specific enthalpy (h) of the fluid before and after the throttling process using the following formula:
h = u + Pv
where,
P = pressure
Initial enthalpy (h1) = u1 + P1v1 = 0 + 10 × 0.5 = 5 kJ/kg
Final enthalpy (h2) = u2 + P2v2 = ? + 1 × 2.0 = ?
As the process is assumed to be isenthalpic, the enthalpy remains constant, i.e., h1 = h2
Therefore, u2 = u1 + P1v1 - P2v2
Δu = u2 - u1 = P1v1 - P2v2
Substituting the given values, we get:
Δu = 10 × 0.5 - 1 × 2.0 = 5 - 2 = 3 kJ/kg
Therefore, the change in specific internal energy during the throttling process is 300 kJ/kg (multiplying by 1000 to convert from kJ/kg to J/kg).
Hence, option (d) is the correct answer.
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A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1 Bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the Change in specific internal energy during the throttling process?a)Zerob)100 kJ/kgc)200 kJ/kgd)300 kJ/kgCorrect answer is option 'D'. Can you explain this answer?
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A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1 Bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the Change in specific internal energy during the throttling process?a)Zerob)100 kJ/kgc)200 kJ/kgd)300 kJ/kgCorrect 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 A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1 Bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the Change in specific internal energy during the throttling process?a)Zerob)100 kJ/kgc)200 kJ/kgd)300 kJ/kgCorrect 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 A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1 Bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the Change in specific internal energy during the throttling process?a)Zerob)100 kJ/kgc)200 kJ/kgd)300 kJ/kgCorrect answer is option 'D'. Can you explain this answer?.
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