UPSC Exam > UPSC Questions > A vertical pipe 5 cm diameter carrying hot wa...
Start Learning for Free
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient
air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from
1 m length of the pipe.
Use following properties of air at 40C:- ρ = 1.12 kg/m3
, µ = 19.1x10-6 kg/ms,
υ= 16.96x10-6 m2
/s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K.
Use the correlation: Nu = 0.1 (Gr.Pr)0.33?
air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from
1 m length of the pipe.
Use following properties of air at 40C:- ρ = 1.12 kg/m3
, µ = 19.1x10-6 kg/ms,
υ= 16.96x10-6 m2
/s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K.
Use the correlation: Nu = 0.1 (Gr.Pr)0.33?
Most Upvoted Answer
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient...
Introduction
To calculate the heat loss from a vertical pipe carrying hot water, we can use the principles of convection heat transfer. Given the parameters, we can apply the Nusselt number correlation for natural convection.
Given Data
- Diameter of the pipe (D) = 5 cm = 0.05 m
- Length of the pipe (L) = 1 m
- Surface temperature (Ts) = 65°C
- Ambient temperature (Ta) = 15°C
- Properties of air at 40°C:
- Density (ρ) = 1.12 kg/m³
- Dynamic viscosity (µ) = 19.1 × 10⁻⁶ kg/ms
- Kinematic viscosity (ν) = 16.96 × 10⁻⁶ m²/s
- Specific heat (Cp) = 1007 J/kg-K
- Thermal conductivity (k) = 0.027 W/m-K
Calculating Grashof Number (Gr)
The Grashof number (Gr) is defined as:
Where:
- g = acceleration due to gravity ≈ 9.81 m/s²
- β = 1/Ta (in Kelvin) = 1/(15 + 273) = 0.00337 K⁻¹
Calculating Gr:
Calculating Prandtl Number (Pr)
The Prandtl number (Pr) is defined as:
Calculating Pr:
Calculating Nusselt Number (Nu)
Using the correlation provided:
Heat Transfer Coefficient (h)
The heat transfer coefficient (h) can be calculated as:
Final Heat Loss Calculation
The heat loss (Q) from the pipe can be calculated using:
Where the area (A) for one meter length of the pipe is:
Conclusion
By substituting all the calculated values into the equations, the final heat loss can be determined. Ensure to perform the calculations step-by-step for accuracy.
To calculate the heat loss from a vertical pipe carrying hot water, we can use the principles of convection heat transfer. Given the parameters, we can apply the Nusselt number correlation for natural convection.
Given Data
- Diameter of the pipe (D) = 5 cm = 0.05 m
- Length of the pipe (L) = 1 m
- Surface temperature (Ts) = 65°C
- Ambient temperature (Ta) = 15°C
- Properties of air at 40°C:
- Density (ρ) = 1.12 kg/m³
- Dynamic viscosity (µ) = 19.1 × 10⁻⁶ kg/ms
- Kinematic viscosity (ν) = 16.96 × 10⁻⁶ m²/s
- Specific heat (Cp) = 1007 J/kg-K
- Thermal conductivity (k) = 0.027 W/m-K
Calculating Grashof Number (Gr)
The Grashof number (Gr) is defined as:
- Gr = (g × β × (Ts - Ta) × L³) / (ν²)
Where:
- g = acceleration due to gravity ≈ 9.81 m/s²
- β = 1/Ta (in Kelvin) = 1/(15 + 273) = 0.00337 K⁻¹
Calculating Gr:
- Gr = (9.81 × 0.00337 × (65 - 15) × (1)³) / (16.96 × 10⁻⁶)²
Calculating Prandtl Number (Pr)
The Prandtl number (Pr) is defined as:
- Pr = (µ × Cp) / k
Calculating Pr:
- Pr = (19.1 × 10⁻⁶ × 1007) / 0.027
Calculating Nusselt Number (Nu)
Using the correlation provided:
- Nu = 0.1 (Gr × Pr)⁰ᴾˢ³³
Heat Transfer Coefficient (h)
The heat transfer coefficient (h) can be calculated as:
- h = (Nu × k) / D
Final Heat Loss Calculation
The heat loss (Q) from the pipe can be calculated using:
- Q = h × A × (Ts - Ta)
Where the area (A) for one meter length of the pipe is:
- A = π × D × L
Conclusion
By substituting all the calculated values into the equations, the final heat loss can be determined. Ensure to perform the calculations step-by-step for accuracy.
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Similar UPSC Doubts
Top Courses for UPSCView all
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33?
Question Description
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33?.
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33?.
Solutions for A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? defined & explained in the simplest way possible. Besides giving the explanation of
A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33?, a detailed solution for A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? has been provided alongside types of A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? theory, EduRev gives you an
ample number of questions to practice A vertical pipe 5 cm diameter carrying hot water is exposed to ambient air at 15C. If the outer surface of the pipe is at 65C, find the heat loss from 1 m length of the pipe. Use following properties of air at 40C:- ρ = 1.12 kg/m3 , µ = 19.1x10-6 kg/ms, υ= 16.96x10-6 m2 /s, Cp= 1007 J/Kg-K, k = 0.027 W/m-K. Use the correlation: Nu = 0.1 (Gr.Pr)0.33? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC 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