Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:give complete and detailed answerProblem 2.8. Consider the flow of a fluid parallel to a flat and

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:give complete and detailed answerProblem 2.8. Consider the flow of a fluid parallel to a flat and smooth plate.

(a) Assume that the fluid is air at 100-kPa pressure, with Ux, = 1 m/s. The
entire system is at 350 K. The surface of the plate is slightly wet, such that
water vapor is transferred from the surface to the fluid. Estimate the thick-
ness of the velocity and concentration boundary layers at 1-, 10-, and 30-cm
distances from the leading edge of the plate.

(b) Repeat part (a), this time assuming that Uy. = 2.5 cm/s, the fluid is water
and the transferred species is chlorine.

VIPSolutions · Accepted Answer

Let's solve **Problem 2.8** step by step.

---

## **Given Data**

### **Part (a):**
- Fluid: Air
- Pressure: 100 kPa
- Free stream velocity ($U_\infty$): 1 m/s
- Temperature: 350 K
- Plate is slightly wet (water vapor transferred to air)
- Find: Thickness of **velocity boundary layer** ($\delta$) and **concentration boundary layer** ($\delta_c$) at:
  - $x = 1$ cm = .01 m
  - $x = 10$ cm = .10 m
  - $x = 30$ cm = .30 m

---

## **Step 1: Calculate Properties of Air at 350 K**

From tables (approximate values at 350 K, 100 kPa):
- **Kinematic viscosity ($\nu$)**: $ \approx 2.4 \times 10^{-5} $ m$^2$/s
- **Diffusion coefficient of water vapor in air ($D_{AB}$)**: $ \approx 2.6 \times 10^{-5} $ m$^2$/s

---

## **Step 2: Boundary Layer Thickness Formulae**

### **(a) Velocity Boundary Layer ($\delta$)**

For laminar flow over a flat plate:
$$
\delta(x) \approx \frac{5.x}{\sqrt{Re_x}}
$$
where:
$$
Re_x = \frac{U_\infty x}{\nu}
$$

### **(b) Concentration Boundary Layer ($\delta_c$)**

$$
\delta_c(x) \approx \frac{\delta(x)}{\sqrt{Sc}}
$$
where:
$$
Sc = \frac{\nu}{D_{AB}}
$$
is the **Schmidt number**.

---

## **Step 3: Calculate at Each Location**

### **Calculate Reynolds Number at Each $x$:**
$$
Re_x = \frac{U_\infty x}{\nu}
$$

| $x$ (m) | $Re_x$ |
|---|---|
| .01 | $\frac{1 \times .01}{2.4 \times 10^{-5}} = 417$ |
| .10 | $\frac{1 \times .10}{2.4 \times 10^{-5}} = 4167$ |
| .30 | $\frac{1 \times .30}{2.4 \times 10^{-5}} = 12500$ |

---

### **Compute Schmidt Number**
$$
Sc = \frac{\nu}{D_{AB}} = \frac{2.4 \times 10^{-5}}{2.6 \times 10^{-5}} \approx .92
$$

---

### **Compute Boundary Layer Thicknesses**

#### **For $x = .01$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .01}{\sqrt{417}} = \frac{.05}{20.4} = .00245 \text{ m} = 2.45 \text{ mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{\delta}{\sqrt{Sc}} = \frac{.00245}{\sqrt{.92}} = \frac{.00245}{.96} = .00255 \text{ m} = 2.55 \text{ mm}
$$

---

#### **For $x = .10$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .10}{\sqrt{4167}} = \frac{.5}{64.5} = .00775 \text{ m} = 7.75 \text{ mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{.00775}{.96} = .00807 \text{ m} = 8.07 \text{ mm}
$$

---

#### **For $x = .30$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .30}{\sqrt{12500}} = \frac{1.5}{111.8} = .0134 \text{ m} = 13.4 \text{ mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{.0134}{.96} = .01396 \text{ m} = 14. \text{ mm}
$$

---

## **Summary Table (Part a)**
| $x$ (cm) | $\delta$ (mm) | $\delta_c$ (mm) |
|---|---|---|
| 1 | 2.45 | 2.55 |
| 10 | 7.75 | 8.07 |
| 30 | 13.4 | 14. |

---

## **Part (b): Water and Chlorine**

### **Given:**
- $U_\infty = 2.5$ cm/s = .025 m/s
- Fluid: **Water**
- Transferred species: **Chlorine**

### **Typical Properties for Water at 350 K:**
- $\nu \approx .4 \times 10^{-6}$ m$^2$/s
- $D_{AB}$ (chlorine in water) $ \approx 1.5 \times 10^{-9} $ m$^2$/s

---

### **Calculate $Re_x$:**

| $x$ (m) | $Re_x$ |
|---|---|
| .01 | $\frac{.025 \times .01}{.4 \times 10^{-6}} = 625$ |
| .10 | $\frac{.025 \times .10}{.4 \times 10^{-6}} = 625$ |
| .30 | $\frac{.025 \times .30}{.4 \times 10^{-6}} = 18750$ |

---

### **Calculate Schmidt Number:**

$$
Sc = \frac{\nu}{D_{AB}} = \frac{.4 \times 10^{-6}}{1.5 \times 10^{-9}} \approx 267
$$

---

### **Compute Thicknesses**

#### **For $x = .01$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .01}{\sqrt{625}} = \frac{.05}{25} = .002 \ \text{m} = 2. \ \text{mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{.002}{\sqrt{267}} = \frac{.002}{16.34} = .000122 \ \text{m} = .122 \ \text{mm}
$$

---

#### **For $x = .10$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .10}{\sqrt{625}} = \frac{.5}{79.1} = .00632 \ \text{m} = 6.32 \ \text{mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{.00632}{16.34} = .000387 \ \text{m} = .387 \ \text{mm}
$$

---

#### **For $x = .30$ m:**

**Velocity boundary layer:**
$$
\delta = \frac{5 \times .30}{\sqrt{18750}} = \frac{1.5}{136.9} = .01095 \ \text{m} = 10.95 \ \text{mm}
$$

**Concentration boundary layer:**
$$
\delta_c = \frac{.01095}{16.34} = .00067 \ \text{m} = .67 \ \text{mm}
$$

---

## **Summary Table (Part b)**

| $x$ (cm) | $\delta$ (mm) | $\delta_c$ (mm) |
|---|---|---|
| 1 | 2. | .12 |
| 10 | 6.3 | .39 |
| 30 | 11. | .67 |

---

## **Final Answers**

### (a) **Air/Water Vapor System** (at 350 K, $U_\infty = 1$ m/s):
| Distance from Leading Edge | Velocity BL Thickness ($\delta$) | Concentration BL Thickness ($\delta_c$) |
|:---:|:---:|:---:|
| 1 cm | 2.45 mm | 2.55 mm |
| 10 cm | 7.75 mm | 8.07 mm |
| 30 cm | 13.4 mm | 14. mm |

### (b) **Water/Chlorine System** (at 350 K, $U_\infty = 2.5$ cm/s):
| Distance from Leading Edge | Velocity BL Thickness ($\delta$) | Concentration BL Thickness ($\delta_c$) |
|:---:|:---:|:---:|
| 1 cm

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