Let's break down the problem step by step: ### **Given Data** - **Plastic sheet dimensions:** 1 mm thick, 1 m wide, 1 m long (cooled section) - **Sheet velocity:** \( 9 \, \text{m/min} = 0.15 \, \text{m/s} \) - **Surface temperature of sheet:** \( T_s = 127^\circ \text{C} = 400.15 \, \text{K} \) - **Air temperature:** \( T_\infty = 27^\circ \text{C} = 300.15 \, \text{K} \) - **Air velocity:** \( u_\infty = 30 \, \text{m/s} \) - **Width of the sheet:** \( w = 1 \, \text{m} \) - **Length of cooled section:** \( L = 1 \, \text{m} \) - **Assume boundary layer is laminar** ### **Assumptions** - Surface temperature is uniform. - Laminar boundary layer. - Properties of air (from tables at film temperature): - \( T_f = \frac{T_s + T_\infty}{2} = \frac{400.15 + 300.15}{2} = 350.15 \, \text{K} \approx 77^\circ \text{C} \) - At \( T_f \approx 77^\circ \text{C} \): - \( k \approx 0.028 \, \text{W/m}\cdot\text{K} \) - \( \nu \approx 22.39 \times 10^{-6} \, \text{m}^2/\text{s} \) - \( Pr \approx 0.7 \) --- ## **Step 1: Calculate Reynolds Number** \[ Re_L = \frac{u_\infty L}{\nu} \] \[ Re_L = \frac{30 \times 1}{22.39 \times 10^{-6}} = 1.34 \times 10^6 \] --- ## **Step 2: Calculate Nusselt Number (Laminar Flow Over Flat Plate)** For laminar flow (\( Re_L < 5 \times 10^5 \)), use: \[ Nu_L = 0.664 \, Re_L^{1/2} Pr^{1/3} \] \[ Nu_L = 0.664 \times (1.34 \times 10^6)^{1/2} \times (0.7)^{1/3} \] \[ Nu_L = 0.664 \times 1157 \times 0.887 \] \[ Nu_L \approx 682 \] --- ## **Step 3: Calculate the Heat Transfer Coefficient** \[ h = \frac{Nu_L \cdot k}{L} \] \[ h = \frac{682 \times 0.028}{1} = 19.1 \, \text{W/m}^2\cdot\text{K} \] --- ## **Step 4: Calculate the Rate of Heat Transfer** \[ \dot{Q} = h \cdot A \cdot (T_s - T_\infty) \] Since both sides are cooled: \[ A = 2 \times (1 \times 1) = 2 \, \text{m}^2 \] \[ \dot{Q} = 19.1 \times 2 \times (127 - 27) \] \[ \dot{Q} = 19.1 \times 2 \times 100 = 3820 \, \text{W} \] --- ## **Final Answer** ### **The rate of heat transfer from the 1-m-long section of the plastic sheet is:** \[ \boxed{3820 \, \text{W}} \] --- ### **Summary of Steps** 1. Found relevant air properties at film temperature. 2. Calculated Reynolds number. 3. Used laminar Nusselt number correlation. 4. Found heat transfer coefficient. 5. Calculated total rate of heat transfer. --- **If you need a more detailed explanation for any step, let me know!**

**Step 1: Conceptual Introduction** To estimate the heat transfer rate from the plastic sheet during forced convection cooling, we need to analyze how heat is transferred from the hot plastic surface to the cooler air flowing over it. The process involves understanding the flow characteristics of air over the sheet, determining the convective heat transfer coefficient, and then calculating the total heat transfer based on the temperature difference and surface area. Since the boundary layer initially remains laminar, we will use laminar flow correlations to estimate the heat transfer rate over the initial 1-meter cooled section. **Explanation Block:** In this problem, the plastic sheet is hot at 127°C and is being cooled by a fast-moving air jet. The air's velocity and temperature influence the boundary layer development on the sheet's surface, which in turn affects how efficiently heat is removed. By modeling the flow as laminar at the leading edge, we can approximate the heat transfer using established correlations for laminar boundary layers over flat surfaces. This approach simplifies the complex heat transfer process into calculable parameters, allowing us to estimate the rate of cooling accurately for the initial section. --- **Step 2: Relevant Formulae** 1. **Reynolds Number (Re):** \[ Re_L = \frac{u_\infty L}{\nu} \] where - \( u_\infty \) = free stream air velocity - \( L \) = characteristic length (here, length of the cooled section) - \( \nu \) = kinematic viscosity of air 2. **Nusselt Number for laminar flow over a flat plate:** \[ Nu_L = 0.664 \, Re_L^{1/2} Pr^{1/3} \] where - \( Pr \) = Prandtl number of air 3. **Convective Heat Transfer Coefficient (h):** \[ h = \frac{Nu_L \cdot k}{L} \] where - \( k \) = thermal conductivity of air 4. **Heat Transfer Rate (\( \dot{Q} \)):** \[ \dot{Q} = h \cdot A \cdot (T_s - T_\infty) \] where - \( A \) = surface area exposed to convection (both sides if cooled on top and bottom) - \( T_s \), \( T_\infty \) = surface and ambient temperatures --- **Step 3: Step-by-Step Complete Calculation** **Given Data:** - \( u_\infty = 30\, \text{m/s} \) - \( L = 1\, \text{m} \) - \( T_s = 127^\circ C = 400.15\, K \) - \( T_\infty = 27^\circ C = 300.15\, K \) - Sheet width \( w = 1\, \text{m} \) - Properties of air at film temperature (~77°C): - \( k \approx 0.028\, \text{W/m·K} \) - \( \nu \approx 22.39 \times 10^{-6}\, \text{m}^2/\text{s} \) - \( Pr \approx 0.7 \) **Step 3.1: Calculate Reynolds Number** \[ Re_L = \frac{u_\infty \times L}{\nu} = \frac{30\, \text{m/s} \times 1\, \text{m}}{22.39 \times 10^{-6}\, \text{m}^2/\text{s}} \] \[ Re_L = \frac{30}{22.39 \times 10^{-6}} \approx 1.34 \times 10^6 \] **Step 3.2: Calculate Nusselt Number** \[ Nu_L = 0.664 \times Re_L^{1/2} \times Pr^{1/3} \] Calculate each term: \[ Re_L^{1/2} = \sqrt{1.34 \times 10^6} \approx 1157 \] \[ Pr^{1/3} = 0.7^{1/3} \approx 0.887 \] Now, \[ Nu_L = 0.664 \times 1157 \times 0.887 \approx 0.664 \times 1026 \approx 682 \] **Step 3.3: Calculate Heat Transfer Coefficient \( h \)** \[ h = \frac{Nu_L \times k}{L} = \frac{682 \times 0.028}{1} \approx 19.1\, \text{W/m}^2\cdot \text{K} \] **Step 3.4: Calculate Total Heat Transfer Rate \( \dot{Q} \)** - Surface area for both sides: \[ A = 2 \times (w \times L) = 2 \times (1\, \text{m} \times 1\, \text{m}) = 2\, \text{m}^2 \] - Temperature difference: \[ \Delta T = T_s - T_\infty = 127^\circ C - 27^\circ C = 100^\circ C \] - Heat transfer rate: \[ \dot{Q} = h \times A \times \Delta T = 19.1\, \text{W/m}^2\cdot \text{K} \times 2\, \text{m}^2 \times 100\, \text{K} = 3820\, \text{W} \] --- **Final Answer:** \[ \boxed{\text{The rate of heat transfer via convection from the 1-m-long section is approximately } 3820\, \text{Watts}.} \] This estimate considers laminar boundary layer assumptions and uses standard correlations for convective heat transfer over a flat plate.

**Step 1: Conceptual Introduction** To estimate the heat transfer rate from the plastic sheet during forced convection cooling, we analyze how heat moves from the hot plastic surface to the cooler air flowing over it. This involves understanding the flow characteristics of air over the sheet, calculating how effectively heat is transferred via convection, and then determining the total heat loss based on the temperature difference and the surface area. Since the boundary layer at the leading edge initially remains laminar, we can apply laminar flow correlations to find the convective heat transfer coefficient, which simplifies the problem into a manageable calculation. **Explanation Block:** In this problem, the plastic sheet is hot at 127°C and is cooled by a high-velocity air jet. The flow over the sheet creates a boundary layer—a thin region of fluid near the surface where velocity and temperature change from their surface values to the free stream values. The rate of heat transfer depends on how thick this boundary layer is and how effectively heat is transported through it. Because the flow starts laminar at the leading edge, we can use laminar boundary layer correlations to estimate the convective heat transfer coefficient. This approach helps us approximate the cooling rate during the initial section of cooling. --- **Step 2: Relevant Formulae** 1. **Reynolds Number (Re):** \[ Re_L = \frac{u_\infty L}{\nu} \] - Represents the ratio of inertial to viscous forces in the flow, indicating whether the flow is laminar or turbulent over the length \(L\). 2. **Nusselt Number (Nu) for laminar flow over a flat plate:** \[ Nu_L = 0.664 \, Re_L^{1/2} \, Pr^{1/3} \] - Relates convective heat transfer to conductive heat transfer; higher Nu means more efficient heat transfer. 3. **Convective Heat Transfer Coefficient (h):** \[ h = \frac{Nu_L \times k}{L} \] - Quantifies the heat transfer per unit area per degree of temperature difference. 4. **Heat Transfer Rate (\(\dot{Q}\)):** \[ \dot{Q} = h \times A \times (T_s - T_\infty) \] - Total heat transferred from the surface to the air. **Explanation Block:** The Reynolds number tells us whether the flow is laminar or turbulent. Since the problem states the boundary layer is initially laminar, we use the laminar flow correlation for Nusselt number, which depends on Re and Prandtl number. Once we find Nu, we calculate \(h\), the heat transfer coefficient, which, when multiplied by the surface area and temperature difference, gives us the total heat transfer rate. --- **Step 3: Complete Calculation with Explanation** **Given Data:** - Air velocity: \(u_\infty = 30\, \text{m/s}\) - Length of cooled section: \(L = 1\, \text{m}\) - Surface temperature: \(T_s = 127^\circ C = 400.15\, K\) - Ambient air temperature: \(T_\infty = 27^\circ C = 300.15\, K\) - Air properties at film temperature (~77°C): - Thermal conductivity: \(k \approx 0.028\, \text{W/m·K}\) - Kinematic viscosity: \(\nu \approx 22.39 \times 10^{-6}\, \text{m}^2/\text{s}\) - Prandtl number: \(Pr \approx 0.7\) --- **Step 3.1: Calculate Reynolds Number** \[ Re_L = \frac{u_\infty \times L}{\nu} = \frac{30\, \text{m/s} \times 1\, \text{m}}{22.39 \times 10^{-6}\, \text{m}^2/\text{s}} \] Calculating numerator: \[ 30 \times 1 = 30 \] Calculating denominator: \[ 22.39 \times 10^{-6} = 2.239 \times 10^{-5} \] Thus, \[ Re_L = \frac{30}{2.239 \times 10^{-5}} \approx 1.34 \times 10^6 \] This high Reynolds number confirms the flow is turbulent, but since the problem states the boundary layer is initially laminar, we're considering the laminar region at the leading edge. --- **Step 3.2: Calculate Nusselt Number** Using the laminar boundary layer correlation: \[ Nu_L = 0.664 \times Re_L^{1/2} \times Pr^{1/3} \] Calculate \(Re_L^{1/2}\): \[ \sqrt{1.34 \times 10^{6}} \approx 1157 \] Calculate \(Pr^{1/3}\): \[ 0.7^{1/3} \approx 0.887 \] Now, \[ Nu_L = 0.664 \times 1157 \times 0.887 \approx 0.664 \times 1026 \approx 682 \] This Nusselt number indicates the convective heat transfer efficiency for the boundary layer. --- **Step 3.3: Calculate Heat Transfer Coefficient \(h\)** \[ h = \frac{Nu_L \times k}{L} = \frac{682 \times 0.028}{1} \approx 19.1\, \text{W/m}^2\cdot \text{K} \] --- **Step 3.4: Calculate Total Heat Transfer Rate \(\dot{Q}\)** - Surface area: \[ A = 2 \times (w \times L) = 2 \times (1\, \text{m} \times 1\, \text{m}) = 2\, \text{m}^2 \] - Temperature difference: \[ \Delta T = T_s - T_\infty = 127^\circ C - 27^\circ C = 100^\circ C \] - Heat transfer rate: \[ \dot{Q} = h \times A \times \Delta T = 19.1\, \text{W/m}^2\cdot \text{K} \times 2\, \text{m}^2 \times 100\, \text{K} = 3820\, \text{W} \] --- **Final Answer:** The estimated rate of heat transfer from the 1-meter section of the plastic sheet during forced convection cooling is approximately **3820 Watts**. This calculation combines flow physics, heat transfer correlations, and material properties to arrive at an approximate cooling rate under the given conditions.