Act like a helpful tutor and exlain me :

Give step-by-step solution with explanation and final answer:
Water is heated by submerging
5
0
-
mm
-
diameter, thin
-
walled copper tubes in a tank and pas

Question

Act like a helpful tutor and exlain me :

Give step-by-step solution with explanation and final answer:
Water is heated by submerging 
5
0
-
mm
-
diameter, thin
-
walled copper tubes in a tank and passing hot combustion gases 
(
T
_
g 
=
 
7
5
0
 ﻿K
)
through the tubes. To enhance heat transfer to the water, four straight fins of uniform cross section, which form a cross, are inserted in
each tube. The fins are 
5
 ﻿mm thick and are also made of copper 
(
k 
=
 
4
0
0
 ﻿W
/
m
•
K
)
.
If the tube surface temperature is T
_
s 
=
 
3
5
0
 ﻿K and the gas
-
side convection coefficient is h 
_
g 
=
 
2
6
 ﻿W
/
m
^
2
•
K
,
 ﻿what is the rate of heat transfer to the water per meter of pipe length?

VIPSolutions · Accepted Answer

# Step-by-Step Solution: Heat Transfer from Gas to Water via Finned Tube

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

---

## **Given Data**

- Tube diameter, $ D = 50 $ mm $ = 0.05 $ m
- Tube is thin-walled copper (high $ k $), so wall resistance is negligible
- Hot gas temperature, $ T_g = 750 $ K
- Tube surface temperature, $ T_s = 350 $ K
- Gas-side convection coefficient, $ h_g = 26 $ W/m²•K
- Fins: 4 straight (cross), each 5 mm thick ($ t = 0.005 $ m), also copper ($ k = 400 $ W/m•K)
- Find: **Total heat transfer rate from gas to water per meter tube**

---

## **Step 1: Find Tube Surface Area (per meter)**

Circumference:  
$$
P = \pi D = \pi \times 0.05 = 0.1571~\text{m}
$$

Per meter length, the area:
$$
A_{\text{tube}} = P \times 1~\text{m} = 0.1571~\text{m}^2
$$

---

## **Step 2: Fin Details**

Each fin is a straight plate, sticking radially out from the tube. There are 4 fins, each with thickness $ t = 0.005~\text{m} $.  
**Fin length** is not specified, but typically it extends to the wall of the tube. Since the wall is thin, let's assume the fin length is the radius:

$$
L_f = \frac{D}{2} = 0.025~\text{m}
$$

Each fin has two sides exposed to the gas.

- **Fin perimeter (for heat transfer):** The two wide faces, each with length $ L_f $ and height $ 1~\text{m} $ (length of tube), so per fin:
    $$
    A_{\text{fin, one side}} = L_f \times 1
    $$
    For two sides:
    $$
    A_{\text{fin, total per fin}} = 2 \times 0.025 \times 1 = 0.05~\text{m}^2
    $$
    For 4 fins:
    $$
    A_{\text{fins, total}} = 4 \times 0.05 = 0.20~\text{m}^2
    $$

---

## **Step 3: Fin Efficiency Calculation**

For a straight fin of uniform cross-section (thickness $ t $),  
Fin parameter:
$$
m = \sqrt{\frac{2 h_g}{k t}}
$$
Given:
- $ h_g = 26 $ W/m²•K
- $ k = 400 $ W/m•K
- $ t = 0.005 $ m

$$
m = \sqrt{\frac{2 \cdot 26}{400 \cdot 0.005}}
= \sqrt{\frac{52}{2}}
= \sqrt{26}
= 5.1~\text{(approx)}
$$

Wait, let's do it properly:

$$
m = \sqrt{\frac{2 \cdot 26}{400 \cdot 0.005}}
= \sqrt{\frac{52}{2}}
= \sqrt{26}
$$
But,
$$
400 \cdot 0.005 = 2
$$
So,
$$
m = \sqrt{\frac{52}{2}} = \sqrt{26} \approx 5.1~\text{m}^{-1}
$$

- **Fin length:** $ L_f = 0.025~\text{m} $
- $ m L_f = 5.1 \times 0.025 = 0.1275 $

**Fin efficiency for an adiabatic tip (long fin):**
$$
\eta_f = \frac{\tanh(m L_f)}{m L_f}
$$
$$
\eta_f = \frac{\tanh(0.1275)}{0.1275}
$$
$$
\tanh(0.1275) \approx 0.1268
$$
$$
\eta_f = \frac{0.1268}{0.1275} \approx 0.995
$$

So, the fin is very efficient (as expected for copper).

---

## **Step 4: Total Heat Transfer Area (Effective Area)**

$$
A_{\text{total, effective}} = A_{\text{tube, bare}} + \eta_f \cdot A_{\text{fins}}
$$

But the area covered by the fins is not part of the bare tube area. The fins are thin, so the area "lost" is negligible compared to the tube circumference. We'll proceed with the full circumference as the bare area, but be aware in precision calculations you'd subtract the fin base area.

$$
A_{\text{bare}} = \text{Tube circumference} \times 1~\text{m} = 0.1571~\text{m}^2
$$

$$
A_{\text{fins}} = 0.20~\text{m}^2
$$
$$
\eta_f \approx 0.995
$$
$$
A_{\text{total, effective}} = 0.1571 + 0.995 \times 0.20 = 0.1571 + 0.199 = 0.3561~\text{m}^2
$$

---

## **Step 5: Total Heat Transfer Rate**

$$
\dot{Q} = h_g \cdot A_{\text{total, effective}} \cdot (T_g - T_s)
$$

- $ h_g = 26 $ W/m²•K
- $ A_{\text{total, effective}} = 0.3561~\text{m}^2 $
- $ (T_g - T_s) = 750 - 350 = 400~\text{K} $

$$
\dot{Q} = 26 \cdot 0.3561 \cdot 400
$$
$$
= 26 \cdot 142.44
$$
$$
= 3703.44~\text{W}
$$

---

## **Final Answer**

### **The rate of heat transfer to the water per meter of pipe is:**

$$
\boxed{3700~\text{W/m}}
$$
(Rounded to two significant digits)

---

## **Summary Table**

| Parameter                       | Value           |
|----------------------------------|-----------------|
| Tube diameter ($D$)           | 0.05 m          |
| Hot gas temperature ($T_g$)    | 750 K           |
| Tube surface temperature ($T_s$) | 350 K         |
| Convection coefficient ($h_g$) | 26 W/m²•K       |
| Fin thickness ($t$)            | 0.005 m         |
| Number of fins                   | 4               |
| Fin material ($k$)             | 400 W/m•K       |
| Effective area ($A_{eff}$)     | 0.356 m²        |
| Heat transfer ($\dot{Q}$)      | 3700 W/m        |

---

## **Diagram (for clarity)**

```
      |      ^
      |     /|

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