Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:give detailed answers in english, explain both sub partsSe desea determinar la temperatura en el

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:give detailed answers in english, explain both sub partsSe desea determinar la temperatura en el punto de contacto entre dos aletas muy largas, una circular y
otra cuadrada, bajo las siguientes condiciones:
Tipo Dimension k h To) Teo a
Circular r=5cm 80 W/mK 60 W/mK. T(0.1) = 150 °C 25°C 1.172x10 m¥/s
Cuadrada 1=2in 30 Btu/hftF BBUF TO4)=302°F  824°F 0.182inYs
Datos: ambas aletas son muy largas y estén en régimen estacionario.
Ecuaci6n base:
T(z) ~ To pe [nP
—e™, m=) —
T, — Tw kA
Solicitado:
Determinar la temperatura en el punto de contacto entre las dos aletas.
Explica brevemente como influye la conductividad y el coeficiente de conveccién en el resultado.

VIPSolutions · Accepted Answer

# Step-by-Step Solution

## **Step 1: Understanding the Problem**

We are asked to find the **temperature at the contact point** between two **very long fins (aletas)**:
- One **circular** (cylinder), and one **square** (rectangular bar).
- Both are at steady state and "very long" (meaning we can use the "very long fin" solution).

Given data (translated if needed):

| Type      | Dimension | $k$                     | $h$                  | $T(x)$         | $T_{\infty}$ | $\alpha$               |
|-----------|-----------|--------------------------|------------------------|------------------|----------------|--------------------------|
| Circular  | $r = 5\, \text{cm}$ | $80\, \text{W/m·K}$           | $60\, \text{W/m}^2\cdot \text{K}$  | $T(.1) = 150^\circ \text{C}$ | $25^\circ \text{C}$   | $1.172\times10^{-5}\, \text{m}^2/\text{s}$ |
| Square    | $l = 2\, \text{in}$ | $30\, \text{Btu/h·ft·}^\circ\text{F}$ | $18\, \text{Btu/h·ft}^2\cdot^\circ\text{F}$ | $T(.4) = 302^\circ\text{F}$  | $82.4^\circ\text{F}$  | $.182\, \text{in}^2/\text{s}$            |

## **Step 2: Base Equation**

For a very long fin:
$$
\frac{T(x) - T_\infty}{T_b - T_\infty} = e^{-mx}
$$
where:
- $T(x)$: temperature at distance $x$
- $T_b$: base temperature (at $x = $)
- $T_\infty$: ambient temperature
- $m = \sqrt{\frac{hP}{kA}}$
    - $h$: convection coefficient
    - $P$: perimeter of cross-section
    - $k$: thermal conductivity
    - $A$: cross-sectional area

## **Step 3: Find the Base Temperatures ($T_b$)**

We need to find $T_b$ for each fin using the given $T(x)$.

### **A. Circular Fin**

#### **Convert radius to meters:**
- $r = 5\, \text{cm} = .05\, \text{m}$

#### **Given:**
- $T(.1) = 150^\circ \text{C}$
- $T_\infty = 25^\circ \text{C}$
- $x = .1\, \text{m}$

#### **Calculate $m$:**
- Perimeter $P = 2\pi r = 2\pi (.05) = .314\, \text{m}$
- Area $A = \pi r^2 = \pi (.05)^2 = .00785\, \text{m}^2$
- $h = 60\, \text{W/m}^2\cdot\text{K}$
- $k = 80\, \text{W/m}\cdot\text{K}$

$$
m = \sqrt{\frac{hP}{kA}} = \sqrt{\frac{60 \times .314}{80 \times .00785}}
= \sqrt{\frac{18.84}{.628}}
= \sqrt{30}
= 5.477\, \text{m}^{-1}
$$

#### **Solve for $T_b$:**
$$
\frac{T(.1) - T_\infty}{T_b - T_\infty} = e^{-m x}
$$
$$
\frac{150 - 25}{T_b - 25} = e^{-5.477 \times .1} = e^{-.5477} = .578
$$
$$
\frac{125}{T_b - 25} = .578
$$
$$
T_b - 25 = \frac{125}{.578} = 216.3
$$
$$
T_b = 216.3 + 25 = \boxed{241.3^\circ \text{C}}
$$

---

### **B. Square Fin**

#### **Convert dimensions and units:**
- $l = 2\, \text{in} = .1667\, \text{ft}$
- $T(.4) = 302^\circ \text{F}$
- $T_\infty = 82.4^\circ \text{F}$
- $x = .4\, \text{ft}$

#### **Perimeter and Area:**
- Perimeter $P = 4l = 4 \times .1667 = .667\, \text{ft}$
- Area $A = l^2 = (.1667)^2 = .0278\, \text{ft}^2$

#### **Given:**
- $h = 18\, \text{Btu/h·ft}^2\cdot^\circ\text{F}$
- $k = 30\, \text{Btu/h·ft·}^\circ\text{F}$

$$
m = \sqrt{\frac{hP}{kA}} = \sqrt{\frac{18 \times .667}{30 \times .0278}}
= \sqrt{\frac{12.01}{.834}}
= \sqrt{14.40}
= 3.794\, \text{ft}^{-1}
$$

#### **Solve for $T_b$:**
$$
\frac{T(.4) - T_\infty}{T_b - T_\infty} = e^{-m x}
$$
$$
\frac{302 - 82.4}{T_b - 82.4} = e^{-3.794 \times .4} = e^{-1.5176} = .219
$$
$$
\frac{219.6}{T_b - 82.4} = .219
$$
$$
T_b - 82.4 = \frac{219.6}{.219} = 1003.7
$$
$$
T_b = 1003.7 + 82.4 = \boxed{1086.1^\circ \text{F}}
$$

---

## **Step 4: Find Temperature at Point of Contact**

When two **very long fins** are joined at their tips, the temperature at the contact ($T_c$) is the same for both. At a distance $x$ from the base for each fin, the temperature is:
$$
T_c = T_\infty + (T_b - T_\infty) e^{-mx}
$$
But, since both are "very long", the temperature at the tip approaches $T_\infty$. However, if the contact is at $x=$ for both, then $T_c$ would be determined by **energy balance** at the contact.

But with the data here, we use the *very long fin* solution for each, and the temperature at the contact is where the heat flow (from each fin) is equal.

For "very long" fins, the *heat transfer rate* is:
$$
q_{\text{fin}} = \sqrt{hPkA}(T_b - T_\infty)
$$
At the contact, the heat leaving one fin must equal that entering the other. Thus:
$$
q_{\text{circular}} = q_{\text{square}}
$$
$$
\sqrt{h_1 P_1 k_1 A_1}(T_{b1} - T_c) = \sqrt{h_2 P_2 k_2 A_2}(T_{b2} - T_c)
$$

Solve for $T_c$:

$$
\sqrt{h_1 P_1 k_1 A_1}(T_{b1} - T_c) = \sqrt{h_2 P_2 k_2 A_2}(T_{b2} - T_c)
$$
$$
a_1 (T_{b1} - T_c) = a_2 (T_{b2} - T_c)
$$
$$
a_1 T_{b1} - a_1 T_c = a_2 T_{b2} - a_2 T_c
$$
$$
a_1 T_{b1} - a_2 T_{b2} = (a_1-a_2)T_c
$$
$$
T_c = \frac{a_1 T_{b1} - a_2 T_{b2}}{a_1 - a_2}
$$

Where:
- $a_1 = \sqrt{h_1 P_1 k_1 A_1}$
- $a_2 = \sqrt{h_2 P_2 k_2 A_2}$

---

### **Calculate $a_1$ and $a_2$ (convert units for consistency):**

#### **A. Circular Fin ($a_1$):**
Already in SI units:
- $h_1 = 60\, \text{W/m}^2\cdot\text{K}$
- $P_1 = .314\, \text{m}$
- \(k_1 = 80\, \text{W/m}\cdot\tex

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