Question

onsider Couette flow of an oil of density 1500 kg m−³, thermal conductivity 0.2 W m-¹K-¹ and viscosity 0.2 Pas between two horizontal plates separated by a distance of 0.1 mm. The upper plate is moving at a velocity of 5 ms and the lower plate is stationary. Both the plates are at 30°C. (a) Estimate maximum temperature of the oil due to viscous dissipation at steady state

Answer 1

So, the maximum temperature rise
`(\(ΔT_{\text{max}}\))` due to viscous dissipation is 0.25 degrees Celsius.

Step-by-step explanation:

To estimate the maximum temperature of the oil due to viscous dissipation in Couette flow, we can use the energy equation for steady-state flow, which takes into account both conduction and viscous dissipation. The energy equation for this situation is:

`\[q = (kA)/(d) \Delta T + \mu \cdot \frac{{du^2}}{dy^2}\]`

Where:

- \(q\) is the heat transfer rate per unit area.

- \(k\) is the thermal conductivity of the fluid (0.2 W/m·K).

- \(A\) is the cross-sectional area of the plates.

- \(d\) is the distance between the plates (0.1 mm = 0.0001 m).

- \(\Delta T\) is the temperature difference between the plates.

- \(\mu\) is the dynamic viscosity of the fluid (0.2 Pa·s).

- \(u\) is the velocity of the upper plate (5 m/s).

- \(y\) is the distance from the lower plate.

First, let's calculate the temperature difference
`\(\Delta T\)`:

Now, let's calculate the heat transfer rate per unit area
`(\(q\))` due to viscous dissipation. We'll consider the upper plate to be at a higher temperature due to viscous dissipation:

`\[q = \mu \cdot \frac{{du^2}}{dy^2}\]`

Since we want to estimate the maximum temperature,
`\[\Delta T = T_{\text{upper}} - T_{\text{lower}} = 30°C - 30°C = 0K\]`

we need to find the maximum value of
`\(q\)`, which occurs at
`\(y = 0\)`. The derivative
`\((du^2)/(dy^2)\)` is maximized at
`\(y = 0\)` for Couette flow.

`\[(du^2)/(dy^2) = (d)/(dy)(du^2/dy) = 0\]`

So,
`\(q\)` is maximized at
`\(y = 0\)`, and we can calculate it there:

`\[q_{\text{max}} = \mu \cdot \frac{{du^2}}{dy^2} \bigg|_(y=0)\]`

Now, plug in the known values:

`\[q_{\text{max}} = (0.2 Pa·s) \cdot \frac{{(5 m/s)^2}}{0.0001 m^2} = 500,000 W/m^2 = 500 kW/m^2\]`

Now, let's calculate the maximum temperature rise
`(\(ΔT_{\text{max}}\))` due to this heat transfer using the formula:

`\[q_{\text{max}} = (kA)/(d) \cdot ΔT_{\text{max}}\]`

Rearrange the formula to solve for
`\(ΔT_{\text{max}}\)`:

`\[ΔT_{\text{max}} = \frac{q_{\text{max}} \cdot d}{kA}\]`

Now, plug in the values:

`\[ΔT_{\text{max}} = ((500 kW/m^2) \cdot 0.0001 m)/(0.2 W/m·K \cdot A)\]`

The value of
`\(A\)` depends on the geometry of your plates. Assuming a unit width (1 m) and considering only one plate (as the other one is stationary), we have
`\(A = 1 m^2\)`.

`\[ΔT_{\text{max}} = ((500 kW/m^2) \cdot 0.0001 m)/(0.2 W/m·K \cdot 1 m^2) = 0.25 K\]`

So, the maximum temperature rise
`(\(ΔT_{\text{max}}\))` due to viscous dissipation is 0.25 degrees Celsius.