(a) Calculate the power loss in the transmission system given in the following figure. The numerical values of transmission…

(a) Power loss in the transmission system

Currents. At bus 3, I₁ and I₂ enter and I₃ leaves towards the load (KCL): I₃ = I₁ + I₂ = 0.75∠0° + 0.8∠0° = 1.55∠0° pu

Direct I²R loss. Line 1: 0.75² × 0.07 = 0.0394 pu Line 2: 0.80² × 0.06 = 0.0384 pu Line 3: 1.55² × 0.05 = 0.1201 pu P_L = 0.1979 pu The reactive loss is Q_L = 0.75² × 0.15 + 0.8² × 0.20 + 1.55² × 0.06 = 0.3565 pu.

Loss-formula (B-coefficient) method. V₃ is given so that the plant voltages and outputs, and hence the loss coefficients, can be found. V₁ = V₃ + I₁Z₁ = 1.2 + 0.75(0.07 + j0.15) = 1.2525 + j0.1125 = 1.2575∠5.13° pu V₂ = V₃ + I₂Z₂ = 1.2 + 0.8(0.06 + j0.20) = 1.248 + j0.160 = 1.2582∠7.31° pu P₁ = Re(V₁I₁*) = 0.9394 pu; P₂ = Re(V₂I₂*) = 0.9984 pu Plant 1's current flows through lines 1 and 3, and plant 2's through lines 2 and 3. Since both currents are at 0°, |V_i| cos φ_i = Re V_i: B₁₁ = (R₁ + R₃)/(|V₁| cos φ₁)² = 0.12/1.2525² = 0.0765 pu⁻¹ B₂₂ = (R₂ + R₃)/(|V₂| cos φ₂)² = 0.11/1.248² = 0.0706 pu⁻¹ B₁₂ = R₃/(|V₁| cos φ₁ |V₂| cos φ₂) = 0.05/(1.2525 × 1.248) = 0.0320 pu⁻¹ P_L = B₁₁P₁² + 2B₁₂P₁P₂ + B₂₂P₂² = 0.0675 + 0.0600 + 0.0704 = 0.1979 pu

Power-balance check. V₄ = V₃ − I₃Z₃ = 1.1225 − j0.093 pu, so P_load = Re(V₄I₃*) = 1.7399 pu. Then P₁ + P₂ − P_load = 1.9378 − 1.7399 = 0.1979 pu, which agrees. On a 100 MVA base the loss is 19.8 MW.

(b) Economic dispatch and saving over equal sharing

Incremental costs. dF₁/dP₁ = 0.6P₁ + 35 Rs/MWh; dF₂/dP₂ = 0.4P₂ + 30 Rs/MWh

Note on the printed limits. With an upper limit of 90 MW on each unit, the two units together can supply only 180 MW, so a 200 MW demand cannot be met at all. The question clearly intends the limits not to bind (for example, with an upper limit of 125 MW or more). The equal-λ schedule is therefore worked out below and checked against 20 MW ≤ P ≤ 125 MW.

Equal incremental cost. 0.6P₁ + 35 = 0.4P₂ + 30 with P₁ + P₂ = 200: 0.6P₁ + 35 = 0.4(200 − P₁) + 30 ⇒ P₁ = 75 MW P₁ = 75 MW, P₂ = 125 MW, λ = 80 Rs/MWh

Cost of the economic schedule. F₁ = 0.3(75)² + 35(75) + 125 = 4437.5 Rs/h F₂ = 0.2(125)² + 30(125) + 140 = 7015 Rs/h Total = 11 452.5 Rs/h

Equal sharing (100 MW each). F₁ = 0.3(100)² + 35(100) + 125 = 6625 Rs/h F₂ = 0.2(100)² + 30(100) + 140 = 5140 Rs/h Total = 11 765 Rs/h

Saving = 11 765 − 11 452.5 = 312.5 Rs/h (7500 Rs per day) Check by integration: moving 25 MW from unit 1 to unit 2 saves ∫ from 75 to 100 of (0.6P + 35) dP = 2187.5 Rs/h on unit 1 and costs ∫ from 100 to 125 of (0.4P + 30) dP = 1875 Rs/h on unit 2, a net saving of 312.5 Rs/h.

With equal sharing, unit 1 runs at 95 Rs/MWh and unit 2 at 70 Rs/MWh. Shifting load to the unit with the lower incremental cost reduces the total until the two incremental costs are equal.

(c) Maximum message amplitude in delta modulation

No slope overload. The staircase can rise by at most one step δ per sampling interval, so the message slope must not exceed δf_s: max |dm(t)/dt| ≤ δf_s

Sinusoidal message. For m(t) = A sin 2πf_mt, the maximum slope is 2πf_mA.

Sampling rate. Nyquist rate = 2f_m, so f_s = 10 × 2f_m = 20f_m.

Result. 2πf_mA ≤ δ × 20f_m ⇒ A ≤ 20δ/(2π) = 20 × 0.25/(2π) A_max = 2.5/π ≈ 0.796 V The limit is independent of the message frequency, because f_s is tied to f_m. A larger step would allow a larger amplitude but would increase granular noise when the signal changes slowly, which is the trade-off that adaptive delta modulation removes.