Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer: Consider a
1
.
2
-
m
-
high and
2
-
m
-
wide glass window whose thickness is
6
mm and therm

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer: ﻿Consider a 
1
.
2
-
m
-
high and 
2
-
m
-
wide glass window whose thickness is 
6
 ﻿mm and thermal conductivity is k 
=
 
0
.
7
8
 ﻿W
/
m
-
°
C
.
 ﻿Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 
2
4
°
C while the temperature of the outdoors is 
-
5
°
C
.
 ﻿Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be hi 
=
 
1
0
 ﻿W
/
m
^
2
°
C and h
2
 
=
 
2
5
 ﻿W
/
m
^
2
°
C
,
 ﻿and disregard any heat transfer by radiation.

VIPSolutions · Accepted Answer

# Step-by-Step Solution: Heat Transfer Through a Glass Window

Let's break down the problem and solve it step by step.

---

## **Given Data**

- **Window height, $ h $:** 1.2 m  
- **Window width, $ w $: 2 m  
- **Thickness of glass, $ L $:** 6 mm = .006 m  
- **Thermal conductivity of glass, $ k $:** .78 W/m·°C  
- **Indoor temperature, $ T_{\text{in}} $:** 24°C  
- **Outdoor temperature, $ T_{\text{out}} $:** -5°C  
- **Convection coefficient (indoor), $ h_i $:** 10 W/m²·°C  
- **Convection coefficient (outdoor), $ h_o $:** 25 W/m²·°C

---

## **1. Calculate the Area of the Window**

$$
A = \text{height} \times \text{width} = 1.2~\text{m} \times 2~\text{m} = 2.4~\text{m}^2
$$

---

## **2. Identify the Thermal Resistances**

The heat transfer path consists of three resistances in series:

1. **Indoor convective resistance ($ R_{\text{conv, in}} $):**
   $$
   R_{\text{conv, in}} = \frac{1}{h_i A}
   $$

2. **Glass conduction resistance ($ R_{\text{glass}} $):**
   $$
   R_{\text{glass}} = \frac{L}{k A}
   $$

3. **Outdoor convective resistance ($ R_{\text{conv, out}} $):**
   $$
   R_{\text{conv, out}} = \frac{1}{h_o A}
   $$

---

## **3. Calculate Each Resistance**

### a. **Indoor Convection**
$$
R_{\text{conv, in}} = \frac{1}{10 \times 2.4} = \frac{1}{24} = .0417~\text{°C/W}
$$

### b. **Glass Conduction**
$$
R_{\text{glass}} = \frac{.006}{.78 \times 2.4} = \frac{.006}{1.872} \approx .00321~\text{°C/W}
$$

### c. **Outdoor Convection**
$$
R_{\text{conv, out}} = \frac{1}{25 \times 2.4} = \frac{1}{60} = .0167~\text{°C/W}
$$

---

## **4. Total Thermal Resistance**

$$
R_{\text{total}} = R_{\text{conv, in}} + R_{\text{glass}} + R_{\text{conv, out}}
$$
$$
R_{\text{total}} = .0417 + .00321 + .0167 = .0616~\text{°C/W}
$$

---

## **5. Calculate the Steady Rate of Heat Transfer**

The temperature difference is:
$$
\Delta T = T_{\text{in}} - T_{\text{out}} = 24 - (-5) = 29~\text{°C}
$$

The heat transfer rate:
$$
\dot{Q} = \frac{\Delta T}{R_{\text{total}}} = \frac{29}{.0616} \approx 471~\text{W}
$$

---

## **6. Find the Inner Surface Temperature of the Glass**

Let $ T_s $ be the inner surface temperature.

The heat flow through the indoor convection resistance:
$$
\dot{Q} = \frac{T_{\text{in}} - T_s}{R_{\text{conv, in}}}
$$

$$
471 = \frac{24 - T_s}{.0417}
$$
$$
24 - T_s = 471 \times .0417 \approx 19.6
$$
$$
T_s = 24 - 19.6 = 4.4~\text{°C}
$$

---

## **Final Answers**

### **Steady Rate of Heat Transfer**
$$
\boxed{471~\text{W}}
$$

### **Inner Surface Temperature of the Glass**
$$
\boxed{4.4~\text{°C}}
$$

---

### **Summary Table**

| Quantity                | Value             |
|-------------------------|------------------|
| Window Area (A)         | 2.4 m²           |
| Total Resistance        | .0616 °C/W      |
| Heat Transfer Rate ($ \dot{Q} $) | 471 W             |
| Inner Surface Temp ($ T_s $)     | 4.4 °C            |

---

**Diagram (for understanding, not mandatory):**

![Glass window heat transfer diagram showing resistances in series: indoor convection, glass conduction, and outdoor convection.](https://i.imgur.com/7R9v5hC.png)

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