Let's solve the question step by step: --- ### **Step 1: Understanding the Problem** - **Given:** - Initial heat transfer coefficient of the walls and ceiling: \( U_1 = 0.1282 \) kJ m\(^{-2}\)°C\(^{-1}\)min\(^{-1}\) - Improved heat transfer coefficient: \( U_2 = 0.06 \) kJ m\(^{-2}\)°C\(^{-1}\)min\(^{-1}\) - The ON/OFF time cycles for the air conditioner before improvement are given in the table. - **Required:** - Calculate the new ON/OFF cycle times after the improvement in insulation. - **Duty factor:** Fraction of time the air conditioner is ON. --- ### **Step 2: Analyze the ON/OFF Cycles** #### **Before Insulation (Given Data)** | Cycle | ON (min) | OFF (min) | |-------|----------|-----------| | 1 | 8.96 | 1.74 | | 2 | 2.30 | 1.74 | | 3 | 2.30 | 1.74 | | 4 | 2.30 | 1.74 | | ... | ... | ... | - **Initial ON time:** 8.96 min (first cycle), then 2.30 min for subsequent ON cycles. - **OFF time:** 1.74 min (constant). --- ### **Step 3: Relationship Between Heat Transfer Coefficient and ON/OFF Time** The heat loss (or gain) rate is proportional to the heat transfer coefficient \( U \): \[ Q = U \cdot A \cdot \Delta T \] If insulation improves (\( U \) decreases), it will take longer for the indoor temperature to rise to the set point (longer OFF period), and the ON period for cooling will be shorter, as less heat enters. For a first-order system (as typically used for buildings), the time constant \( \tau \) is inversely proportional to \( U \): \[ \tau \propto \frac{1}{U} \] So, the ON and OFF times will increase as \( U \) decreases: \[ t_2 = t_1 \times \frac{U_1}{U_2} \] --- ### **Step 4: Calculate the New ON and OFF Times** \[ \frac{U_1}{U_2} = \frac{0.1282}{0.06} = 2.1367 \] #### **New OFF time:** \[ \text{New OFF time} = \text{Old OFF time} \times 2.1367 = 1.74 \times 2.1367 = 3.72 \text{ min} \] #### **New ON time:** \[ \text{New ON time} = \text{Old ON time} \times 2.1367 \] - For the first ON cycle: \(8.96 \times 2.1367 = 19.13\) min - For subsequent ON cycles: \(2.30 \times 2.1367 = 4.91\) min --- ### **Step 5: Fill in the Table** | ON/OFF | ON | OFF | ON | OFF | ON | OFF | ON | OFF | ... | |--------|----------|----------|---------|---------|---------|---------|---------|---------|-----| | **Before (U = 0.1282)** | 8.96 | 1.74 | 2.30 | 1.74 | 2.30 | 1.74 | 2.30 | 1.74 | ... | | **After (U = 0.06)** | 19.13 | 3.72 | 4.91 | 3.72 | 4.91 | 3.72 | 4.91 | 3.72 | ... | --- ### **Step 6: Calculate the Duty Factor** The **duty factor** is the fraction of time the air conditioner is ON: \[ \text{Duty factor} = \frac{\text{ON time}}{\text{ON time} + \text{OFF time}} \] For cycles **after the first** (using subsequent ON/OFF times): - **Before improvement:** \[ \frac{2.30}{2.30 + 1.74} = \frac{2.30}{4.04} = 0.569 \quad (\approx 56.9\%) \] - **After improvement:** \[ \frac{4.91}{4.91 + 3.72} = \frac{4.91}{8.63} = 0.569 \quad (\approx 56.9\%) \] So the **duty factor remains the same**, but the **cycle time increases** (the air conditioner turns ON and OFF less frequently). --- ## **Final Answers** ### **The completed table:** | ON/OFF | ON | OFF | ON | OFF | ON | OFF | ON | OFF | ... | |--------|-------|-------|------|------|------|------|------|------|-----| | **U = 0.1282** | 8.96 | 1.74 | 2.30 | 1.74 | 2.30 | 1.74 | 2.30 | 1.74 | ... | | **U = 0.06** | 19.13 | 3.72 | 4.91 | 3.72 | 4.91 | 3.72 | 4.91 | 3.72 | ... | ### **The new duty factor:** \[ \boxed{0.569 \text{ or } 56.9\%} \] --- **Explanation:** By improving the insulation (lower \( U \)), the air-conditioner cycles less frequently with longer ON and OFF periods, but the fraction of time it is ON (duty factor) remains the same.