Let's break down the problem step by step: ### 1. **Given Data** - **Fuel Input Equations**: \[ F_1 = 0.3P_1^2 + 35P_1 + 125 \quad \text{(x/h)} \] \[ F_2 = 0.2P_2^2 + 30P_2 + 140 \quad \text{(x/h)} \] - **Load Limits:** \[ 20 \leq P_1 \leq 90 \quad \text{MW} \] \[ 20 \leq P_2 \leq 90 \quad \text{MW} \] - **Total Demand:** \[ P_1 + P_2 = 200 \quad \text{MW} \] However, note that the sum of the maximums is only \(90 + 90 = 180\) MW, but the demand is 200 MW. This is not possible unless there is a typo. But let's check if the limits were on minimum/maximum outputs per unit, and perhaps the max per unit is 120 MW. But with 90 MW max and 20 MW min, max possible = 180 MW. Let's assume **the maximum per unit is 120 MW** (common in such problems), otherwise the demand is not feasible. Let's proceed with: \[ 20 \leq P_1 \leq 120,\quad 20 \leq P_2 \leq 120 \] \[ P_1 + P_2 = 200 \] --- ### 2. **Equal Incremental Cost Criterion** For economic dispatch (neglecting transmission losses), **incremental cost** (marginal fuel cost) must be equal for both units: \[ \frac{dF_1}{dP_1} = \frac{dF_2}{dP_2} \] Calculate derivatives: - \( F_1 = 0.3P_1^2 + 35P_1 + 125 \) \[ \frac{dF_1}{dP_1} = 0.6P_1 + 35 \] - \( F_2 = 0.2P_2^2 + 30P_2 + 140 \) \[ \frac{dF_2}{dP_2} = 0.4P_2 + 30 \] Set equal and substitute \( P_2 = 200 - P_1 \): \[ 0.6P_1 + 35 = 0.4P_2 + 30 \] \[ 0.6P_1 + 35 = 0.4(200 - P_1) + 30 \] \[ 0.6P_1 + 35 = 80 - 0.4P_1 + 30 \] \[ 0.6P_1 + 35 = 110 - 0.4P_1 \] \[ 0.6P_1 + 0.4P_1 = 110 - 35 \] \[ 1.0P_1 = 75 \] \[ P_1 = 75 \text{ MW} \] \[ P_2 = 200 - 75 = 125 \text{ MW} \] Check if these are within limits (\(20 \leq P_1 \leq 120\), \(20 \leq P_2 \leq 120\)). Here, \( P_2 = 125 \) MW, which is **above the maximum**. So, set \( P_2 = 120 \) MW (max limit), then \( P_1 = 200 - 120 = 80 \) MW. --- ### 3. **Recompute Incremental Costs at Limits** Compute incremental costs at these points: - \( P_1 = 80 \) MW - \( P_2 = 120 \) MW \[ \frac{dF_1}{dP_1} = 0.6 \times 80 + 35 = 48 + 35 = 83 \] \[ \frac{dF_2}{dP_2} = 0.4 \times 120 + 30 = 48 + 30 = 78 \] They are not equal, but since \( P_2 \) can't exceed 120 MW, this is the closest feasible allocation. --- ### 4. **Calculate Total Fuel Cost (Economic Schedule)** Plug into the original fuel cost functions: #### For \( P_1 = 80 \): \[ F_1 = 0.3 \times (80)^2 + 35 \times 80 + 125 \] \[ = 0.3 \times 6400 + 2800 + 125 \] \[ = 1920 + 2800 + 125 = 4845 \text{ x/h} \] #### For \( P_2 = 120 \): \[ F_2 = 0.2 \times (120)^2 + 30 \times 120 + 140 \] \[ = 0.2 \times 14,400 + 3600 + 140 \] \[ = 2880 + 3600 + 140 = 6620 \text{ x/h} \] #### **Total Cost (Economic):** \[ F_{\text{total,econ}} = 4845 + 6620 = 11,465 \text{ x/h} \] --- ### 5. **Equal Sharing (Each Unit = 100 MW)** \( P_1 = 100, P_2 = 100 \) #### For \( P_1 = 100 \): \[ F_1 = 0.3 \times (100)^2 + 35 \times 100 + 125 \] \[ = 0.3 \times 10,000 + 3500 + 125 \] \[ = 3000 + 3500 + 125 = 6625 \text{ x/h} \] #### For \( P_2 = 100 \): \[ F_2 = 0.2 \times (100)^2 + 30 \times 100 + 140 \] \[ = 0.2 \times 10,000 + 3000 + 140 \] \[ = 2000 + 3000 + 140 = 5140 \text{ x/h} \] #### **Total Cost (Equal Share):** \[ F_{\text{total,eq}} = 6625 + 5140 = 11,765 \text{ x/h} \] --- ### 6. **Savings Calculation** \[ \text{Savings} = 11,765 - 11,465 = 300 \text{ x/h} \] --- ## **Final Answers** ### **(a) Economical Operating Schedule:** - **Unit 1:** \( P_1 = 80 \) MW - **Unit 2:** \( P_2 = 120 \) MW ### **(b) Corresponding Cost of Generation:** \[ \boxed{11,465 \;\; x/h} \] ### **(c) Cost if Load is Equally Shared:** \[ \boxed{11,765 \;\; x/h} \] ### **(d) Savings by Economic Dispatch:** \[ \boxed{300 \;\; x/h} \] --- ## **Summary Table** | Case | \(P_1\) (MW) | \(P_2\) (MW) | Total Cost (x/h) | Savings (x/h) | |----------------------|:------------:|:------------:|:----------------:|:-------------:| | Economic Dispatch | 80 | 120 | 11,465 | 300 | | Equal Sharing | 100 | 100 | 11,765 | – | --- **Note:** If the original unit maximums are indeed only 90 MW, the total demand of 200 MW is not possible, and you should clarify the problem statement. The above solution assumes the likely intended maximum is 120 MW per unit.