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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.
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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.
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