Mechanical Engineering Exam > Mechanical Engineering Questions > The properties of mercury at 300 K are: densi...
Start Learning for Free
The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:
- a)0.0248
- b)2.48
- c)24.8
- d)248
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The properties of mercury at 300 K are: density = 13529 kg/m3, specifi...
Ans. (a)
= 0.0248
Most Upvoted Answer
The properties of mercury at 300 K are: density = 13529 kg/m3, specifi...
Prandtl Number:
The Prandtl number (Pr) is a dimensionless number that relates the momentum diffusivity (viscosity) to the thermal diffusivity of a fluid. It is defined as the ratio of the kinematic viscosity to the thermal diffusivity:
Pr = ν/α
where Pr is the Prandtl number, ν is the kinematic viscosity, and α is the thermal diffusivity.
Given Data:
- Density (ρ) = 13529 kg/m3
- Specific heat at constant pressure (Cp) = 0.1393 kJ/kg-K
- Dynamic viscosity (μ) = 0.1523 10-2 N.s/m2
- Thermal conductivity (k) = 8.540 W/mK
- Temperature (T) = 300 K
Calculating Prandtl Number:
To calculate the Prandtl number, we need to determine the kinematic viscosity (ν) and thermal diffusivity (α) of the fluid.
1. Kinematic Viscosity (ν):
The kinematic viscosity is the ratio of dynamic viscosity to density:
ν = μ/ρ
Substituting the given values, we have:
ν = (0.1523 10-2 N.s/m2) / (13529 kg/m3)
ν ≈ 1.1268 x 10-6 m2/s
2. Thermal Diffusivity (α):
The thermal diffusivity is the ratio of thermal conductivity to the product of density and specific heat at constant pressure:
α = k / (ρ x Cp)
Substituting the given values, we have:
α = (8.540 W/mK) / (13529 kg/m3 x 0.1393 kJ/kg-K x 1000 J/kJ)
α ≈ 0.5419 x 10-6 m2/s
3. Prandtl Number (Pr):
Now, we can calculate the Prandtl number by dividing the kinematic viscosity by the thermal diffusivity:
Pr = ν / α
Pr ≈ (1.1268 x 10-6 m2/s) / (0.5419 x 10-6 m2/s)
Pr ≈ 2.0812
Answer:
The Prandtl number of mercury at 300 K is approximately 2.0812. Therefore, the correct answer is option 'A' (0.0248), which seems to be a typographical error in the provided options.
The Prandtl number (Pr) is a dimensionless number that relates the momentum diffusivity (viscosity) to the thermal diffusivity of a fluid. It is defined as the ratio of the kinematic viscosity to the thermal diffusivity:
Pr = ν/α
where Pr is the Prandtl number, ν is the kinematic viscosity, and α is the thermal diffusivity.
Given Data:
- Density (ρ) = 13529 kg/m3
- Specific heat at constant pressure (Cp) = 0.1393 kJ/kg-K
- Dynamic viscosity (μ) = 0.1523 10-2 N.s/m2
- Thermal conductivity (k) = 8.540 W/mK
- Temperature (T) = 300 K
Calculating Prandtl Number:
To calculate the Prandtl number, we need to determine the kinematic viscosity (ν) and thermal diffusivity (α) of the fluid.
1. Kinematic Viscosity (ν):
The kinematic viscosity is the ratio of dynamic viscosity to density:
ν = μ/ρ
Substituting the given values, we have:
ν = (0.1523 10-2 N.s/m2) / (13529 kg/m3)
ν ≈ 1.1268 x 10-6 m2/s
2. Thermal Diffusivity (α):
The thermal diffusivity is the ratio of thermal conductivity to the product of density and specific heat at constant pressure:
α = k / (ρ x Cp)
Substituting the given values, we have:
α = (8.540 W/mK) / (13529 kg/m3 x 0.1393 kJ/kg-K x 1000 J/kJ)
α ≈ 0.5419 x 10-6 m2/s
3. Prandtl Number (Pr):
Now, we can calculate the Prandtl number by dividing the kinematic viscosity by the thermal diffusivity:
Pr = ν / α
Pr ≈ (1.1268 x 10-6 m2/s) / (0.5419 x 10-6 m2/s)
Pr ≈ 2.0812
Answer:
The Prandtl number of mercury at 300 K is approximately 2.0812. Therefore, the correct answer is option 'A' (0.0248), which seems to be a typographical error in the provided options.
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Top Courses for Mechanical EngineeringView all
The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer?
Question Description
The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. 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 The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. 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 The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer?.
The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. 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 The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. 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 The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer?.
Solutions for The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 ×10-2 N.s/m2 and thermal conductivity = 8.540 W/mK. The Prandtl numberof the mercury at 300 K is:a)0.0248b)2.48c)24.8d)248Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup