Act like a helpful tutor and exlain me :
I need exact and accurate answers clearly for chegg without aiA rectangular plate (L = 200mm in length and W = 50mm in width) is subjected to
convective heat

Question

Act like a helpful tutor and exlain me :
I need exact and accurate answers clearly for chegg without aiA rectangular plate (L = 200mm in length and W = 50mm in width) is subjected to
convective heat transfer with ambient air at 300K. The convective heat transfer coefficients
S50W/m?K . The plate has a thermal conductivity of 15 W/mK and an internal heat
2x,
generation at a rate of ¢ = 700e‘z) kW/m? , where x is in the length direction. The
temperatures at two sides of the plate (x = 0 and y = 0) are fixed at 300K. The problem is
considered two-dimensional. Radiation heat transfer is neglected.
(1) Derive the finite difference equations governing the heat transfer in the plate. (4 points)
* Show work by following the steps used in the examples in the book and lecture
notes. (Known, Find, Schematic, Properties, Analysis, etc.).
* Clearly state assumptions, simplifications and the logic behind them.
* Present the governing equations, derivations, and finite difference equations.

VIPSolutions · Accepted Answer

# Finite Difference Equations for Heat Transfer in a Plate

Let's break down the problem step by step as asked:

---

## **1. Known (Given Data)**

- **Plate dimensions:**  
  $ L = 200\, \text{mm} = .2\, \text{m} $ (length, x-direction)  
  $ W = 50\, \text{mm} = .05\, \text{m} $ (width, y-direction)  
- **Convective boundary:**  
  Ambient temperature: $ T_\infty = 300\, \text{K} $  
  Heat transfer coefficient: $ h = 50\, \text{W/m}^2\text{K} $  
- **Thermal conductivity:**  
  $ k = 15\, \text{W/mK} $
- **Internal heat generation:**  
  $ \dot{q} = 700e^{\frac{2x}{L}} $ kW/m³ (convert to W/m³: $ 700,000e^{\frac{2x}{L}} $ W/m³)
- **Boundary temperatures:**  
  At $ x= $ and $ y= $: $ T = 300\, \text{K} $ (Dirichlet BC)
- **Radiation is neglected.**
- **2D steady-state conduction with generation.**

---

## **2. Find**

**Derive the finite difference equations governing the steady-state temperature distribution in the plate.**

---

## **3. Schematic**

```
    y ^
      |
      |------------------- (x=L, y=) (T=?)
      |                   |
      |                   |  Plate
      |                   |
      |-------------------|-----> x
     (x=, y=) (T=300K)
```
- Bottom edge ($ y= $) and left edge ($ x= $) at fixed $ T=300K $.
- Top edge ($ y=W $) and right edge ($ x=L $): Convective boundary.

---

## **4. Properties**

- **Thermal conductivity:** $ k = 15\, \text{W/mK} $ (constant)
- **Internal heat generation:** $ \dot{q} = 700,000e^{\frac{2x}{L}} $ W/m³

---

## **5. Assumptions and Simplifications**

- Steady-state, 2D conduction ($ \frac{\partial T}{\partial t} =  $)
- Constant thermal conductivity.
- No radiation.
- Internal heat generation is a function of $ x $ only.

---

## **6. Governing Equation**

For 2D steady-state conduction with generation:

$$
\frac{\partial^2 T}{\partial x^2}
+ \frac{\partial^2 T}{\partial y^2} 
+ \frac{\dot{q}(x)}{k} = 
$$

$$
\frac{\partial^2 T}{\partial x^2}
+ \frac{\partial^2 T}{\partial y^2} 
= -\frac{700,000\, e^{\frac{2x}{L}}}{15}
$$

---

## **7. Finite Difference Formulation**

Let $ \Delta x $ and $ \Delta y $ be the grid spacings. Let $ T_{i,j} $ be the temperature at grid point $ (i,j) $:

$$
\frac{T_{i+1, j} - 2T_{i,j} + T_{i-1, j}}{(\Delta x)^2}
+ \frac{T_{i, j+1} - 2T_{i,j} + T_{i, j-1}}{(\Delta y)^2}
= -\frac{700,000\, e^{\frac{2x_i}{L}}}{15}
$$

If $ \Delta x = \Delta y = h $:

$$
T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1} - 4T_{i,j} = -h^2 \frac{700,000\, e^{\frac{2x_i}{L}}}{15}
$$

Or,

$$
T_{i,j} = \frac{1}{4} \left[ T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1} + h^2 \frac{700,000\, e^{\frac{2x_i}{L}}}{15} \right]
$$

---

## **8. Boundary Conditions**

- $ x= $ and $ y= $: $ T = 300\, \text{K} $ (Dirichlet)
- $ x=L $ and $ y=W $: Convective boundary (Newton’s law of cooling)

For convective boundary (e.g., at $ x=L $):
$$
-k \left. \frac{\partial T}{\partial x} \right|_{x=L} = h(T_{L,j} - T_\infty)
$$

Using finite difference (forward/backward difference as needed):
$$
\frac{T_{N,j} - T_{N-1,j}}{\Delta x} = -\frac{h}{k}(T_{N,j}-T_\infty)
$$

---

## **Summary Table**

| Node Type         | Finite Difference Equation                                                                 |
|-------------------|-------------------------------------------------------------------------------------------|
| **Interior**      | $ T_{i,j} = \frac{1}{4} [ T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1} + h^2 S_{i,j} ] $ |
| **Dirichlet**     | $ T_{i,j} = 300 $                                                                        |
| **Convective BC** | $ T_{N,j} = \frac{ k T_{N-1,j} + h \Delta x T_\infty }{ k + h \Delta x } $                |

where $ S_{i,j} = \frac{700,000\, e^{\frac{2x_i}{L}}}{15} $

---

### **In summary:**

- Write out the node equations for interior and boundary points.
- Apply Dirichlet condition at fixed-T boundaries.
- Use the convective boundary equation for boundaries with convection.

---

**If you need the step-by-step derivation for each type of node (corner, edge, interior), let me know!**

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers