Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates with one plate moving at a constant velocity of 5 m/s and the other stationary. Consider such a flow with a uniform spacing of 0.5 mm between the plates. The temperatures of the upper and lower plates are 40°C and 15°C, respectively. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Vm/s The properties of oil at the average temperature of (40+15)/2 = 27.5°C are k = 0.145 W/m-K and = 0.605 kg/m-s = 0.605 N-s/m² By simplifying and solving the continuity, momentum, and energy equations, determine the heat flux from the oil to each plate. The heat flux from the oil to the bottom plate is * 104 W/m² The heat flux from the oil to the top plate is [ * 104 W/m².
Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates with one plate moving at a constant velocity of 5 m/s and the other stationary. Consider such a flow with a uniform spacing of 0.5 mm between the plates. The temperatures of the upper and lower plates are 40°C and 15°C, respectively. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Vm/s The properties of oil at the average temperature of (40+15)/2 = 27.5°C are k = 0.145 W/m-K and = 0.605 kg/m-s = 0.605 N-s/m² By simplifying and solving the continuity, momentum, and energy equations, determine the heat flux from the oil to each plate. The heat flux from the oil to the bottom plate is * 104 W/m² The heat flux from the oil to the top plate is [ * 104 W/m².
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
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Chapter5: Analysis Of Convection Heat Transfer
Section: Chapter Questions
Problem 5.57P
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
Transcribed Image Text:Required information
Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates with one plate moving at a
constant velocity of 5 m/s and the other stationary. Consider such a flow with a uniform spacing of 0.5 mm between the
plates. The temperatures of the upper and lower plates are 40°C and 15°C, respectively.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Vm/s
uy)
The properties of oil at the average temperature of (40+15)/2 = 27.5°C are k= 0.145 W/m-K and = 0.605 kg/m-s = 0.605
N-s/m²
By simplifying and solving the continuity, momentum, and energy equations, determine the heat flux from the oil to each plate.
The heat flux from the oil to the bottom plate is
* 104 W/m²
The heat flux from the oil to the top plate is
* 104 W/m².
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Please help answer the question show. The topic is heat transfer. Thank you.
![### Required Information
**Oil Flow in a Journal Bearing:**
Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates, with one plate moving at a constant velocity of 5 m/s and the other remaining stationary. This setup considers a uniform spacing of 0.5 mm between the plates. The upper plate is at a temperature of 40°C and the lower plate is at 15°C.
**Note:**
This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
**Diagram Explanation:**
A diagram shows two parallel plates with the top plate moving at a velocity of \( V \, \text{m/s} \). The velocity profile between the plates is indicated as \( u(y) \).
**Properties of Oil:**
- Average Temperature: \( \frac{40 + 15}{2} = 27.5^\circ\text{C} \)
- Thermal Conductivity (\( k \)): 0.145 W/m·K
- Dynamic Viscosity (\( \mu \)): 0.605 kg/m·s = 0.605 N·s/m\(^2\)
### Calculation
By simplifying and solving the continuity, momentum, and energy equations, determine the heat flux from the oil to each plate.
- **Heat Flux from the Oil to the Bottom Plate:**
\[ -1.5125 \times 10^4 \, \text{W/m}^2 \]
- **Heat Flux from the Oil to the Top Plate:**
\[ 1.5125 \times 10^4 \, \text{W/m}^2 \]](https://content.bartleby.com/qna-images/question/fb9a7c18-65f1-48ed-bb7e-4937a04e4157/7467fd78-69c3-4478-990d-1921ca8ab6b7/c8o3l38_thumbnail.png)
Transcribed Image Text:### Required Information
**Oil Flow in a Journal Bearing:**
Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates, with one plate moving at a constant velocity of 5 m/s and the other remaining stationary. This setup considers a uniform spacing of 0.5 mm between the plates. The upper plate is at a temperature of 40°C and the lower plate is at 15°C.
**Note:**
This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
**Diagram Explanation:**
A diagram shows two parallel plates with the top plate moving at a velocity of \( V \, \text{m/s} \). The velocity profile between the plates is indicated as \( u(y) \).
**Properties of Oil:**
- Average Temperature: \( \frac{40 + 15}{2} = 27.5^\circ\text{C} \)
- Thermal Conductivity (\( k \)): 0.145 W/m·K
- Dynamic Viscosity (\( \mu \)): 0.605 kg/m·s = 0.605 N·s/m\(^2\)
### Calculation
By simplifying and solving the continuity, momentum, and energy equations, determine the heat flux from the oil to each plate.
- **Heat Flux from the Oil to the Bottom Plate:**
\[ -1.5125 \times 10^4 \, \text{W/m}^2 \]
- **Heat Flux from the Oil to the Top Plate:**
\[ 1.5125 \times 10^4 \, \text{W/m}^2 \]
Solution
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