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Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.?
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Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . ...
Introduction
In this analysis, we will calculate the exit velocity of steam and the exit area of a nozzle expanding isentropically from given inlet conditions.
Given Data
- Inlet Pressure (P1): 1 MPa
- Inlet Temperature (T1): 250 °C
- Exit Pressure (P2): 10 kPa
- Mass Flow Rate (ṁ): 1 kg/s
Step 1: Determine Enthalpy at Inlet
Using steam tables, find the enthalpy (h1) at 1 MPa and 250 °C. For superheated steam, h1 is approximately 2800 kJ/kg.
Step 2: Determine Enthalpy at Exit
For the exit pressure of 10 kPa, we find the enthalpy (h2) at this pressure. At this state, h2 is approximately 1900 kJ/kg.
Step 3: Apply Energy Equation
Using the energy equation for steady flow:
- h1 + (KE1/2) = h2 + (KE2/2)
Neglecting KE1, we have:
- h1 = h2 + (KE2/2)
Rearranging gives:
- (KE2/2) = h1 - h2
Substituting values:
- (KE2/2) = 2800 - 1900 = 900 kJ/kg
Thus, KE2 = 1800 kJ/kg.
Step 4: Calculate Exit Velocity
The kinetic energy is related to velocity by:
- KE = (1/2) * ṁ * v²
Substituting values:
- 1800 = (1/2) * 1 * v²
Solving for v:
- v = √(3600) = 60 m/s
Step 5: Calculate Exit Area
Using the continuity equation:
- ṁ = ρ * A * v
Where ρ (density at 10 kPa) is approximately 0.6 kg/m³. Rearranging gives:
- A = ṁ / (ρ * v)
Substituting values:
- A = 1 / (0.6 * 60) = 0.0278 m² or 278 cm².
Conclusion
- Exit Velocity: 60 m/s
- Exit Area: 278 cm²
In this analysis, we will calculate the exit velocity of steam and the exit area of a nozzle expanding isentropically from given inlet conditions.
Given Data
- Inlet Pressure (P1): 1 MPa
- Inlet Temperature (T1): 250 °C
- Exit Pressure (P2): 10 kPa
- Mass Flow Rate (ṁ): 1 kg/s
Step 1: Determine Enthalpy at Inlet
Using steam tables, find the enthalpy (h1) at 1 MPa and 250 °C. For superheated steam, h1 is approximately 2800 kJ/kg.
Step 2: Determine Enthalpy at Exit
For the exit pressure of 10 kPa, we find the enthalpy (h2) at this pressure. At this state, h2 is approximately 1900 kJ/kg.
Step 3: Apply Energy Equation
Using the energy equation for steady flow:
- h1 + (KE1/2) = h2 + (KE2/2)
Neglecting KE1, we have:
- h1 = h2 + (KE2/2)
Rearranging gives:
- (KE2/2) = h1 - h2
Substituting values:
- (KE2/2) = 2800 - 1900 = 900 kJ/kg
Thus, KE2 = 1800 kJ/kg.
Step 4: Calculate Exit Velocity
The kinetic energy is related to velocity by:
- KE = (1/2) * ṁ * v²
Substituting values:
- 1800 = (1/2) * 1 * v²
Solving for v:
- v = √(3600) = 60 m/s
Step 5: Calculate Exit Area
Using the continuity equation:
- ṁ = ρ * A * v
Where ρ (density at 10 kPa) is approximately 0.6 kg/m³. Rearranging gives:
- A = ṁ / (ρ * v)
Substituting values:
- A = 1 / (0.6 * 60) = 0.0278 m² or 278 cm².
Conclusion
- Exit Velocity: 60 m/s
- Exit Area: 278 cm²
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Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.?
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Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.? 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 Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.?.
Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.? 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 Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Steam expands isentropically in a nozzle from 1Mpa ,250°c to 10 kpa . the steam flow rate is 1kg/s . neglecting the KE of the at inlet to the nozzle, find the velocity of steam at exit from the nozzle and the exit area of the nozzle.?.
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