Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is 5 ft wide and 15 ft long, and the average temperature of the exposed surface of the collector is 94°F. The emissivity of the exposed surface of the collector is 0.9. Determine the rate of heat loss from the collector by convection and radiation on a calm day when the ambient air temperature is 70 °F and the effective sky temperature for radiation exchange is 50°F. Take the convection heat transfer coefficient on the exposed surface as 2.5 Btu/h-ft2. F.
Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is 5 ft wide and 15 ft long, and the average temperature of the exposed surface of the collector is 94°F. The emissivity of the exposed surface of the collector is 0.9. Determine the rate of heat loss from the collector by convection and radiation on a calm day when the ambient air temperature is 70 °F and the effective sky temperature for radiation exchange is 50°F. Take the convection heat transfer coefficient on the exposed surface as 2.5 Btu/h-ft2. F.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Heat Loss Calculation for a Flat-Plate Solar Collector
Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector dimensions are as follows:
- **Width**: 5 ft
- **Length**: 15 ft
The average temperature of the exposed surface of the collector is **94°F**. The emissivity of the exposed surface of the collector is **0.9**.
#### Objective:
Determine the rate of heat loss from the collector by convection and radiation on a calm day when the ambient air temperature is **70°F** and the effective sky temperature for radiation exchange is **50°F**.
Take the convection heat transfer coefficient on the exposed surface as **2.5 Btu/h·ft²·°F**.
#### Given:
- \(\sigma = 0.1714 \times 10^{-8} \text{ Btu/h·ft²·R}^4\)
- Solar collector surface temperature, \(T_s = 94°F\)
- Ambient air temperature, \(T_{air} = 70°F\)
- Effective sky temperature, \(T_{sky} = 50°F\)
- Convection heat transfer coefficient, \(h = 2.5 \text{ Btu/h·ft²·°F}\)
- Emissivity, \(\varepsilon = 0.9\)
#### Calculation of Heat Loss:
Using the given data, the total heat loss from the collector can be broken down into:
1. **Convective Heat Loss**:
\[ Q_{conv} = h \cdot A \cdot (T_s - T_{air}) \]
2. **Radiative Heat Loss**:
\[ Q_{rad} = \varepsilon \cdot \sigma \cdot A \cdot (T_s^4 - T_{sky}^4) \]
where:
- \(A\) is the area of the solar collector.
- \(T_s, T_{air}, T_{sky}\) are the temperatures in Rankine (°R), where \(°R = °F + 459.67\).
Given the area \(A = 5 \times 15 = 75 \text{ ft}^2\):
- \(T_s = 94 + 459.67\)
- \(T_{air} = 70 + 459.67\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a663478-8026-4bb7-815e-65d37446f03b%2Fcef47a66-422f-4766-a4bc-3d49ffc8280f%2Ffn8sigf_processed.png&w=3840&q=75)
Transcribed Image Text:### Heat Loss Calculation for a Flat-Plate Solar Collector
Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector dimensions are as follows:
- **Width**: 5 ft
- **Length**: 15 ft
The average temperature of the exposed surface of the collector is **94°F**. The emissivity of the exposed surface of the collector is **0.9**.
#### Objective:
Determine the rate of heat loss from the collector by convection and radiation on a calm day when the ambient air temperature is **70°F** and the effective sky temperature for radiation exchange is **50°F**.
Take the convection heat transfer coefficient on the exposed surface as **2.5 Btu/h·ft²·°F**.
#### Given:
- \(\sigma = 0.1714 \times 10^{-8} \text{ Btu/h·ft²·R}^4\)
- Solar collector surface temperature, \(T_s = 94°F\)
- Ambient air temperature, \(T_{air} = 70°F\)
- Effective sky temperature, \(T_{sky} = 50°F\)
- Convection heat transfer coefficient, \(h = 2.5 \text{ Btu/h·ft²·°F}\)
- Emissivity, \(\varepsilon = 0.9\)
#### Calculation of Heat Loss:
Using the given data, the total heat loss from the collector can be broken down into:
1. **Convective Heat Loss**:
\[ Q_{conv} = h \cdot A \cdot (T_s - T_{air}) \]
2. **Radiative Heat Loss**:
\[ Q_{rad} = \varepsilon \cdot \sigma \cdot A \cdot (T_s^4 - T_{sky}^4) \]
where:
- \(A\) is the area of the solar collector.
- \(T_s, T_{air}, T_{sky}\) are the temperatures in Rankine (°R), where \(°R = °F + 459.67\).
Given the area \(A = 5 \times 15 = 75 \text{ ft}^2\):
- \(T_s = 94 + 459.67\)
- \(T_{air} = 70 + 459.67\
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