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.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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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.
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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### 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:** 
  \[ 1.5125 \times 10^4 \, \text{W/m}^2 \]
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:** \[ 1.5125 \times 10^4 \, \text{W/m}^2 \]
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