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Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at
200C. Find the heat loss from the both surfaces of plate if:
i) Plate is kept vertical, ii) plate is kept horizontal.
Use following properties of air at 600C:- ρ = 1.06 kg/m3
, υ = 18.97x10-6 m2
/s,
Cp= 1008 J/Kg-K, k = 0.028 W/m-K.
Use the following correlations:
i) Nu = 0.13 (Gr.Pr)1/3 for vertical position,
ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface,
iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?
200C. Find the heat loss from the both surfaces of plate if:
i) Plate is kept vertical, ii) plate is kept horizontal.
Use following properties of air at 600C:- ρ = 1.06 kg/m3
, υ = 18.97x10-6 m2
/s,
Cp= 1008 J/Kg-K, k = 0.028 W/m-K.
Use the following correlations:
i) Nu = 0.13 (Gr.Pr)1/3 for vertical position,
ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface,
iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?
Most Upvoted Answer
Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air a...
Introduction
To determine the heat loss from a hot square plate (50 cm x 50 cm) at 1000°C exposed to atmospheric air at 200°C, we will analyze two scenarios: vertical and horizontal orientations. We will use properties of air at 600°C and the provided Nusselt number correlations.
Properties of Air
- Density (ρ) = 1.06 kg/m³
- Kinematic viscosity (ν) = 18.97 x 10⁻⁶ m²/s
- Specific heat (Cp) = 1008 J/kg-K
- Thermal conductivity (k) = 0.028 W/m-K
Calculating Grashof Number (Gr) and Prandtl Number (Pr)
- Grashof Number (Gr) is calculated using:
\[ Gr = \frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot L^3}{\nu^2} \]
where \( g = 9.81 m/s² \), \( \beta \approx \frac{1}{T} \) (in Kelvin), and \( L = 0.5 m \).
- Prandtl Number (Pr) = \( \frac{C_p \cdot \nu}{k} \)
Case i: Plate Kept Vertical
- Using the correlation:
\[ Nu = 0.13 (Gr \cdot Pr)^{1/3} \]
- Heat transfer coefficient (h) = \( \frac{Nu \cdot k}{L} \)
- Heat loss (Q) from both surfaces = \( h \cdot A \cdot (T_s - T_\infty) \)
where \( A = 0.5 m \times 0.5 m = 0.25 m² \).
Case ii: Plate Kept Horizontal
- For upper surface:
\[ Nu_{upper} = 0.71 (Gr \cdot Pr)^{1/4} \]
- For lower surface:
\[ Nu_{lower} = 0.35 (Gr \cdot Pr)^{1/4} \]
- Calculate heat transfer coefficients and total heat loss for both surfaces.
Conclusion
By substituting the calculated values of Gr and Pr into the respective Nusselt number correlations, we can determine the heat loss for both orientations effectively.
To determine the heat loss from a hot square plate (50 cm x 50 cm) at 1000°C exposed to atmospheric air at 200°C, we will analyze two scenarios: vertical and horizontal orientations. We will use properties of air at 600°C and the provided Nusselt number correlations.
Properties of Air
- Density (ρ) = 1.06 kg/m³
- Kinematic viscosity (ν) = 18.97 x 10⁻⁶ m²/s
- Specific heat (Cp) = 1008 J/kg-K
- Thermal conductivity (k) = 0.028 W/m-K
Calculating Grashof Number (Gr) and Prandtl Number (Pr)
- Grashof Number (Gr) is calculated using:
\[ Gr = \frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot L^3}{\nu^2} \]
where \( g = 9.81 m/s² \), \( \beta \approx \frac{1}{T} \) (in Kelvin), and \( L = 0.5 m \).
- Prandtl Number (Pr) = \( \frac{C_p \cdot \nu}{k} \)
Case i: Plate Kept Vertical
- Using the correlation:
\[ Nu = 0.13 (Gr \cdot Pr)^{1/3} \]
- Heat transfer coefficient (h) = \( \frac{Nu \cdot k}{L} \)
- Heat loss (Q) from both surfaces = \( h \cdot A \cdot (T_s - T_\infty) \)
where \( A = 0.5 m \times 0.5 m = 0.25 m² \).
Case ii: Plate Kept Horizontal
- For upper surface:
\[ Nu_{upper} = 0.71 (Gr \cdot Pr)^{1/4} \]
- For lower surface:
\[ Nu_{lower} = 0.35 (Gr \cdot Pr)^{1/4} \]
- Calculate heat transfer coefficients and total heat loss for both surfaces.
Conclusion
By substituting the calculated values of Gr and Pr into the respective Nusselt number correlations, we can determine the heat loss for both orientations effectively.
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Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?
Question Description
Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?.
Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?.
Solutions for Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? in English & in Hindi are available as part of our courses for UPSC.
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Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.?, a detailed solution for Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? has been provided alongside types of Hot square plate 50 cm x50 cm at 1000C is exposed to atmospheric air at 200C. Find the heat loss from the both surfaces of plate if: i) Plate is kept vertical, ii) plate is kept horizontal. Use following properties of air at 600C:- ρ = 1.06 kg/m3 , υ = 18.97x10-6 m2 /s, Cp= 1008 J/Kg-K, k = 0.028 W/m-K. Use the following correlations: i) Nu = 0.13 (Gr.Pr)1/3 for vertical position, ii) Nu = 0.71 (Gr.Pr)1/4 for upper surface, iii) Nu = 0.35 (Gr.Pr)1/4 for lower surface.? theory, EduRev gives you an
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