# Step-by-Step Solution: Heat Transfer in a Fin-and-Wall Heat Exchanger Let's break down the problem and solve it systematically. --- ## 1. **Given Data** - **Hot water temperature, \( T_{h} \):** \( 90^\circ C \) - **Cold air temperature, \( T_{c} \):** \( 25^\circ C \) - **Wall thickness, \( L \):** \( 1 \) cm \( = 0.01 \) m - **Wall thermal conductivity, \( k_{wall} \):** \( 20 \) W/m·K - **Wall area, \( A \):** \( 1 \) m² - **Heat transfer coefficient (water side), \( h_{w} \):** \( 100 \) W/m²·K - **Heat transfer coefficient (air side), \( h_{a} \):** \( 10 \) W/m²·K **Fins:** - **Number of fins, \( N \):** \( 500 \) - **Diameter, \( d \):** \( 6 \) mm \( = 0.006 \) m - **Length, \( L_{fin} \):** \( 30 \) mm \( = 0.03 \) m - **Fin material:** Same as wall (\( k_{fin} = 20 \) W/m·K) - **Fin tips are insulated (adiabatic tip)** --- ## 2. **Thermal Resistances Without Fins** Heat transfer occurs in series: - Water side convection (\( R_{w} \)) - Wall conduction (\( R_{wall} \)) - Air side convection (\( R_{a} \)) ### Formulas: \[ R_{w} = \frac{1}{h_{w}A} \] \[ R_{wall} = \frac{L}{k_{wall}A} \] \[ R_{a} = \frac{1}{h_{a}A} \] ### Calculations: - \( R_{w} = \frac{1}{100 \times 1} = 0.01 \) K/W - \( R_{wall} = \frac{0.01}{20 \times 1} = 0.0005 \) K/W - \( R_{a} = \frac{1}{10 \times 1} = 0.1 \) K/W **Total resistance (without fins):** \[ R_{total,\,plain} = R_{w} + R_{wall} + R_{a} = 0.01 + 0.0005 + 0.1 = 0.1105 \text{ K/W} \] --- ## 3. **With Fins on Air Side** Fins increase the effective area and reduce the resistance on the air side. ### **Fin Calculations** #### **Step 1: Fin Surface Area** - **Perimeter of fin base (\( P \)):** \( \pi d = \pi \times 0.006 = 0.01885 \) m - **Cross-section area (\( A_c \)):** \( \frac{\pi}{4} d^2 = \frac{\pi}{4} \times (0.006)^2 = 2.827 \times 10^{-5} \) m² - **Lateral (side) surface area per fin:** \( A_{fin,side} = P \times L_{fin} = 0.01885 \times 0.03 = 0.0005655 \) m² **Total finned area (excluding tips, assuming insulated):** \[ A_{fin,\,total} = N \times A_{fin,side} = 500 \times 0.0005655 = 0.28275 \text{ m}^2 \] #### **Step 2: Unfinned Area Left** Each fin base area: \( A_{base,fin} = A_c = 2.827 \times 10^{-5} \) m² Total base area covered by fins: \( 500 \times 2.827 \times 10^{-5} = 0.014135 \) m² Uncovered area on wall: \( 1 - 0.014135 = 0.98587 \) m² --- #### **Step 3: Fin Efficiency** For a pin (cylindrical) fin with insulated tip: \[ \eta_{fin} = \frac{\tanh(mL_{fin})}{mL_{fin}} \] Where: - \( m = \sqrt{\frac{h_a P}{k_{fin} A_c}} \) Calculate \( m \): \[ m = \sqrt{\frac{10 \times 0.01885}{20 \times 2.827 \times 10^{-5}}} = \sqrt{\frac{0.1885}{0.0005654}} = \sqrt{333.66} = 18.27 \text{ m}^{-1} \] Now, \( mL_{fin} = 18.27 \times 0.03 = 0.548 \) \[ \eta_{fin} = \frac{\tanh(0.548)}{0.548} = \frac{0.499}{0.548} = 0.911 \] --- #### **Step 4: Fin Effectiveness and Total Surface Area** **Effective fin area:** \[ A_{fin,eff} = \eta_{fin} \times A_{fin,side} \times N = 0.911 \times 0.0005655 \times 500 = 0.2577 \text{ m}^2 \] **Total effective area on air side:** \[ A_{total,air,eff} = \text{Uncovered area} + \text{Effective fin area} = 0.98587 + 0.2577 = 1.2436 \text{ m}^2 \] --- #### **Step 5: Air Side Resistance with Fins** \[ R_{a,fin} = \frac{1}{h_a \times A_{total,air,eff}} = \frac{1}{10 \times 1.2436} = 0.0804 \text{ K/W} \] --- #### **Step 6: Total Resistance with Fins** \[ R_{total,fin} = R_{w} + R_{wall} + R_{a,fin} = 0.01 + 0.0005 + 0.0804 = 0.0909 \text{ K/W} \] --- ## 4. **Final Step: Heat Transfer Rate** \[ Q = \frac{\Delta T}{R_{total,fin}} = \frac{90 - 25}{0.0909} = \frac{65}{0.0909} = 715.6 \text{ W} \] --- ## ## **Final Answer** ### **The total heat transfer rate achieved with 500 pin fins is:** \[ \boxed{716 \text{ W}} \] (Rounded to nearest integer.) --- ### **Summary Table** | Parameter | Value | Units | |------------------------|-----------------|------------| | Hot Water Temp | \(90\) | °C | | Cold Air Temp | \(25\) | °C | | Wall Area | \(1\) | m² | | Wall Thickness | \(0.01\) | m | | Wall Conductivity | \(20\) | W/m·K | | Water Side \(h_w\) | \(100\) | W/m²·K | | Air Side \(h_a\) | \(10\) | W/m²·K | | Fins (Number) | \(500\) | | | Fins (Diameter) | \(0.006\) | m | | Fins (Length) | \(0.03\) | m | | Fin Efficiency | \(0.91\) | | | Final Heat Transfer | \(\approx 716\) | W | --- ## **Diagram**  *Illustration of a heat exchanger wall with pin fins attached on the air side to enhance heat transfer (fins not to scale).* --- **Key Steps Recap:** - Calculated thermal resistances for each part (water, wall, air). - Determined effective area added by fins and their efficiency. - Computed new air-side resistance with fins. - Found overall heat transfer rate using combined resistance. If you need step-by-step substitution with all numbers or have any doubts about a particular step, let me know!