The wall shown in the attached figure moves vertically with velocity U, dragging along with it a
layer of liquid of uniform thickness /. Assuming the liquids density o. viscosity 4, thermal
conducti

Question

The wall shown in the attached figure moves vertically with velocity U, dragging along with it a
layer of liquid of uniform thickness /. Assuming the liquids density o. viscosity 4, thermal
conductivity k, and specific heat ¢ are known, the following is requested:

3.1 Determine the velocity distribution in the liquid. In the calculation, assume that the only
nonzero velocity component is uz, and that the viscous shear stress at the interface
with the air is zero due to the airs negligible viscosity. Therefore, duz/dx = 0 at x =
h.

12 Obtain the force (per unit area) exerted by the fluid on the wall.

2.3 Calculate the volumetric flow rate of liquid dragged by the wall (per unit length perpendicular to
the plane of motion), as well as the value of U for which this flow rate is zero.

34 Assuming the temperature of the wall is To and that of the air is 7, calculate the temperature
distribution within the liquid, 7(x) (consider the effect of viscous dissipation in the calculation).
1.5 Determine the heat flux from the fluid to the wall.

VIPSolutions · Accepted Answer

Let's break down and solve each part of this classic problem in fluid mechanics and heat transfer, step by step:

---

### **Given Data and Assumptions**
- The wall moves vertically with velocity $ U $.
- Liquid layer of thickness $ h $ is dragged along.
- **Liquid properties:** density ($ \rho $), viscosity ($ \mu $), thermal conductivity ($ k $), specific heat ($ c $).
- **Boundary conditions:**
  - At wall ($ x = 0 $): No-slip, so $ u(z) = U $.
  - At liquid-air interface ($ x = h $): Shear stress is zero, so $ \left. \frac{du}{dx} \right|_{x=h} = 0 $.

---

## **3.1 Velocity Distribution in the Liquid**

We assume steady, laminar, fully developed flow, and only the $ z $ component of velocity, $ u(x) $, is non-zero.

**Navier-Stokes (z-direction, ignoring pressure gradients):**
$$
0 = \mu \frac{d^2u}{dx^2}
$$

Integrate twice:
$$
\frac{d^2u}{dx^2} = 0 \implies \frac{du}{dx} = A \implies u(x) = Ax + B
$$

**Boundary Conditions:**
1. $ u(0) = U \implies B = U $
2. $ \left. \frac{du}{dx} \right|_{x=h} = 0 \implies A = 0 $

So,
$$
u(x) = U
$$

**But:** If a shear-free interface is assumed, and no other forces act, the velocity is constant throughout the layer!

---

## **3.2 Force (per unit area) Exerted by Fluid on the Wall**

Force per unit area (shear stress) on the wall:
$$
\tau_w = \mu \left. \frac{du}{dx} \right|_{x=0}
$$

From above, $ \frac{du}{dx} = 0 $, so:
$$
\tau_w = 0
$$

---

## **3.3 Volumetric Flow Rate Dragged by the Wall**

Volumetric flow rate per unit length perpendicular to the plane:
$$
Q = \int_0^h u(x) dx = \int_0^h U dx = U h
$$

The value of $ U $ for which $ Q = 0 $ is:
$$
U = 0
$$

---

## **3.4 Temperature Distribution $ T(x) $ (with Viscous Dissipation)**

**Energy equation (steady, 1D):**
$$
0 = k \frac{d^2T}{dx^2} + \mu \left( \frac{du}{dx} \right)^2
$$

But $ \frac{du}{dx} = 0 $, so viscous dissipation term is zero:
$$
0 = k \frac{d^2T}{dx^2}
\implies \frac{d^2T}{dx^2} = 0
\implies \frac{dT}{dx} = C_1
\implies T(x) = C_1 x + C_2
$$

**Boundary conditions:**
- $ T(0) = T_0 $ (at wall)
- $ T(h) = T_a $ (at liquid-air interface)

Solving:
$$
T(0) = T_0 \implies C_2 = T_0 \
T(h) = T_a \implies C_1 = \frac{T_a - T_0}{h}
$$

So,
$$
T(x) = T_0 + \frac{T_a - T_0}{h}x
$$

---

## **3.5 Heat Flux from Fluid to Wall**

Heat flux at the wall:
$$
q'' = -k \left. \frac{dT}{dx} \right|_{x=0} = -k \frac{T_a - T_0}{h}
$$

---

## **Summary Table**

| Part | Solution |
|------|----------|
| **3.1** | $ u(x) = U $ (constant) |
| **3.2** | $ \tau_w = 0 $ |
| **3.3** | $ Q = U h $, $ U = 0 \implies Q = 0 $ |
| **3.4** | $ T(x) = T_0 + \frac{T_a - T_0}{h} x $ |
| **3.5** | $ q'' = -k \frac{T_a - T_0}{h} $ |

---

If you have follow-up questions or need more details for each derivation, let me know!

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