Let's solve each part step by step. --- # Given - **Temperature function:** \( T = t - 50 e^t \) - **\( T \):** Temperature (°C) - **\( t \):** Time (seconds) --- ## a) Time to Melt (Newton-Raphson, 5 Iterations) **Goal:** Find \( t \) when \( T = 0 \) (melting point). So, set \( T = 0 \): \[ 0 = t - 50 e^t \] \[ t - 50 e^t = 0 \] Let \( f(t) = t - 50 e^t \). ### **Step 1: Write the Newton-Raphson Formula** Newton-Raphson iteration: \[ t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)} \] Where \( f(t) = t - 50e^t \), so: \[ f'(t) = 1 - 50e^t \] ### **Step 2: Use Starting Point \( t_0 = 1 \) s** --- ### **Iteration 1: \( t_0 = 1 \)** \[ f(1) = 1 - 50 e^1 = 1 - 50 \times 2.71828 \approx 1 - 135.914 = -134.914 \] \[ f'(1) = 1 - 50 \times 2.71828 \approx 1 - 135.914 = -134.914 \] \[ t_1 = 1 - \frac{-134.914}{-134.914} = 1 - 1 = 0 \] --- ### **Iteration 2: \( t_1 = 0 \)** \[ f(0) = 0 - 50 e^0 = 0 - 50 \times 1 = -50 \] \[ f'(0) = 1 - 50 \times 1 = -49 \] \[ t_2 = 0 - \frac{-50}{-49} = 0 - 1.0204 = -1.0204 \] --- ### **Iteration 3: \( t_2 = -1.0204 \)** Calculate \( e^{-1.0204} \approx 0.3605 \): \[ f(-1.0204) = -1.0204 - 50 \times 0.3605 \approx -1.0204 - 18.025 = -19.0454 \] \[ f'(-1.0204) = 1 - 50 \times 0.3605 = 1 - 18.025 = -17.025 \] \[ t_3 = -1.0204 - \frac{-19.0454}{-17.025} = -1.0204 - 1.119 = -2.1394 \] --- ### **Iteration 4: \( t_3 = -2.1394 \)** Calculate \( e^{-2.1394} \approx 0.1184 \): \[ f(-2.1394) = -2.1394 - 50 \times 0.1184 = -2.1394 - 5.92 = -8.0594 \] \[ f'(-2.1394) = 1 - 50 \times 0.1184 = 1 - 5.92 = -4.92 \] \[ t_4 = -2.1394 - \frac{-8.0594}{-4.92} = -2.1394 - 1.638 = -3.7774 \] --- ### **Iteration 5: \( t_4 = -3.7774 \)** Calculate \( e^{-3.7774} \approx 0.0229 \): \[ f(-3.7774) = -3.7774 - 50 \times 0.0229 = -3.7774 - 1.145 = -4.9224 \] \[ f'(-3.7774) = 1 - 50 \times 0.0229 = 1 - 1.145 = -0.145 \] \[ t_5 = -3.7774 - \frac{-4.9224}{-0.145} = -3.7774 - 33.944 = -37.7214 \] --- ### **Summary Table** | Iteration | \( t_n \) | \( f(t_n) \) | \( f'(t_n) \) | \( t_{n+1} \) | |-----------|-------------|--------------|---------------|---------------| | 0 | 1.0000 | -134.914 | -134.914 | 0.0000 | | 1 | 0.0000 | -50.000 | -49.000 | -1.0204 | | 2 | -1.0204 | -19.0454 | -17.025 | -2.1394 | | 3 | -2.1394 | -8.0594 | -4.920 | -3.7774 | | 4 | -3.7774 | -4.9224 | -0.145 | -37.7214 | --- ### **Final Answer (Part a):** After 5 iterations, \( t_5 \approx -37.72 \) seconds. However, this negative value is **not physically meaningful** (time cannot be negative). This suggests that with the given function and starting point, the temperature never actually reaches 0°C for positive \( t \). **Interpretation:** The function \( t - 50e^t \) stays negative or does not cross zero for \( t > 0 \). --- ## b) Energy Balance Diagram and Equation ### **Diagram Explanation** - **System:** Closed vessel (no mass in/out until melting complete). - **Boundary:** Vessel wall. After melting, warm water is pumped in at the same rate as overflow. #### **Diagram Description** - Draw a box for the vessel. - Arrow **in**: Warm water in, mass flow rate \( \dot{m}_{in} \), enthalpy \( h_{in} \) - Arrow **out**: Overflow, mass flow rate \( \dot{m}_{out} \), enthalpy \( h_{out} \) - Inside: Liquid water at some temperature. ### **Energy Balance Equation** General energy balance: \[ \frac{dE_{sys}}{dt} = \dot{Q} + \dot{m}_{in} h_{in} - \dot{m}_{out} h_{out} \] Where: - \( \frac{dE_{sys}}{dt} \): Rate of change of energy in system - \( \dot{Q} \): Heat added (if any) - \( \dot{m}_{in} \): Mass flow in (kg/s) - \( h_{in} \): Enthalpy of incoming water (J/kg) - \( \dot{m}_{out} \): Mass flow out (kg/s) - \( h_{out} \): Enthalpy of outgoing water (J/kg) **At steady state:** - No accumulation: \( \frac{dE_{sys}}{dt} = 0 \) - If no heat loss/gain to environment: \( \dot{Q} = 0 \) - \( \dot{m}_{in} = \dot{m}_{out} = \dot{m} \) \[ 0 = \dot{m} (h_{in} - h_{out}) \] **If incoming and outgoing water are at the same temperature:** \[ h_{in} = h_{out} \implies 0 = 0 \] **Any term is zero?** - If vessel is insulated: \( \dot{Q} = 0 \) - If at steady state: \( \frac{dE_{sys}}{dt} = 0 \) - If flow rates and enthalpy in/out are equal: \( h_{in} = h_{out} \) --- ### **Final Answers** #### **a)** - **After 5 Newton-Raphson iterations, \( t \) is negative (approx. \(-37.72\) s), suggesting the melting point is not reached for any positive \( t \) with the given temperature function and initial guess.** #### **b)** - **Energy balance at steady state:** \[ 0 = \dot{m} (h_{in} - h_{out}) \] - **Diagram:** Vessel with equal mass flow in/out, enthalpy in/out. - **Explanation:** Terms are zero because at steady state, with no heat loss and equal flow rates and enthalpies, there is no net energy change in the system. --- **If you need the diagram, let me know and I can describe it visually!**

# Diagram Description for Energy Balance in a Vessel ## Visual Representation ### Vessel Setup 1. **Shape**: Draw a rectangular box representing the vessel. 2. **Insulation**: Indicate that the vessel is insulated (you can use dashed lines or shading to show it’s insulated). ### Flow Directions - **Incoming Flow**: - Draw an arrow pointing into the vessel from the left side. - Label this arrow as **"Incoming Warm Water"**. - Next to the arrow, indicate: - Mass flow rate: \( \dot{m}_{in} \) - Enthalpy: \( h_{in} \) - **Outgoing Flow**: - Draw an arrow pointing out of the vessel from the right side. - Label this arrow as **"Overflow"**. - Next to the arrow, indicate: - Mass flow rate: \( \dot{m}_{out} \) - Enthalpy: \( h_{out} \) ### Internal Components - **Inside the Vessel**: - Indicate that the water inside is at a certain temperature (e.g., "Water at \( T \) °C"). - You can represent water by using wavy lines or blue shading to symbolize liquid. ### Energy Balance Notation - At the bottom of the diagram, include the energy balance equation: \[ 0 = \dot{m} (h_{in} - h_{out}) \] - Annotate that this represents energy conservation at steady state with no heat exchange due to insulation. ### Key Points for Clarity - **Labels**: Ensure all arrows and components are clearly labeled to indicate their role in the energy balance. - **Color Coding**: You can use blue for water, red for heat or energy arrows, and gray for the insulation to enhance clarity. --- This visual representation helps to understand the energy dynamics within the closed vessel, emphasizing the flow of mass and energy. If you need a graphical illustration, consider using drawing software or sketching it out based on this description.