North-West Power Generation Company Limited (NWPGCL)- 2026

Solution: $${\eta }_{overall}={\eta }_{boiler}\times {\eta }_{turbine}\times {\eta }_{generator}\times {\eta }_{cycle}\times {\eta }_{aux}$$

$${\eta }_{aux}=\frac{0.33}{0.92\times 0.95\times 0.97\times 0.42}=\frac{0.33}{0.3561}=0.9268$$

$$\text{Fraction used for auxiliaries}=1-0.9268=0.0732$$

∴ Electricity used for running the auxiliaries = 7.32% (Answer)

Solution: The performance of a cooling tower is expressed by:

Range: Difference between the hot water inlet temperature and the cold water outlet temperature.

Approach: Difference between the cold water outlet temperature and the ambient wet-bulb temperature. A lower approach indicates better performance; it is the better indicator of cooling tower performance.

Effectiveness (efficiency): Ratio of the range to the ideal range (range + approach).

$$\text{Range}=36-31=5^{\circ }\text{C},\text{Approach}=31-26=5^{\circ }\text{C}$$

$${\eta }_{CT}=\frac{\text{Range}}{\text{Range}+\text{Approach}}\times 100=\frac{5}{5+5}\times 100=50\%$$

The dry-bulb temperature (28°C) is not required.

∴ Cooling tower efficiency = 50% (Answer)

Solution:

$${\eta }_{Carnot}=1-\frac{T_{\min }}{T_{\max }},{\eta }_{Rankine}=1-\frac{T_{\min }}{T_{m,add}}$$

In the Carnot cycle, all the heat is added at the maximum temperature. In the Rankine cycle, a large part of the heat is added while the feed water is heated from the pump outlet temperature to the saturation temperature (and in the superheater), so the mean temperature of heat addition T_m,add is lower than T_max. Hence, for the same temperature limits,

$${\eta }_{Carnot}>{\eta }_{Rankine}$$

The Rankine cycle is still used in practice because it avoids compression of a wet mixture and needs only a small pump work.

Answer: [Repeated question: BPDB-2026] [Repeated question: Q4, BPDB-2026 (this Supplement)]

Solution: Assuming a reversed Carnot cycle, T_H = 293 K, T_L = 253 K.

$$COP=\frac{T_L}{T_H-T_L}=\frac{253}{293-253}=6.325$$

$$Q_L=COP\times W=6.325\times 3.5=22.14\text{kW}=\frac{22.14}{3.5}=6.325\text{TR}$$

(With 1 TR = 3.517 kW, the capacity is 6.30 TR.)

∴ COP = 6.325; Cooling capacity ≈ 6.3 TR (Answer)

Solution: $$\text{Energy saving}=(45-35)\times 6000=60,000\text{kWh/year}$$

$$\text{Cost saving}=60,000\times 0.01=\$600\text{per year}$$

∴ Energy saving = 60,000 kWh/year; Cost saving = $600/year (Answer)

Solution: The heat transfer rate is determined by

(i) Energy balance of either fluid:

$$\dot{Q}={\dot{m}}_h{c}_{ph}\left(T_{h,in}-T_{h,out}\right)={\dot{m}}_c{c}_{pc}\left(T_{c,out}-T_{c,in}\right)$$

For a condensing or evaporating fluid, Q̇ = ṁh_fg.

(ii) LMTD method:

$$\dot{Q}=UAF\Delta T_{lm},\Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln \left(\Delta T_1/\Delta T_2\right)}$$

where F = 1 for parallel or counter flow and F < 1 for multipass or cross flow. (When outlet temperatures are unknown, the ε–NTU method is used.)

Factors affecting the heat transfer rate: overall heat transfer coefficient U (fluid velocity, properties, wall material and thickness); heat transfer area A; temperature difference and flow arrangement (counter flow gives the highest ΔT_lm); mass flow rates and specific heats; fouling and scaling; and phase change of the fluids.

Solution: With h₁ = 200 kJ/kg (condenser outlet), h₃ = 3200 kJ/kg, h₄ = 2200 kJ/kg and w_p = 10 kJ/kg:

$$h_2=h_1+w_p=200+10=210\text{kJ/kg}$$

$$w_{net}=\left(h_3-h_4\right)-w_p=1000-10=990\text{kJ/kg},q_{in}=h_3-h_2=3200-210=2990\text{kJ/kg}$$

$${\eta }_{th}=\frac{w_{net}}{q_{in}}=\frac{990}{2990}=0.3311$$

∴ Thermal efficiency = 33.11% (Answer)

Solution: $$S_A=0.30(9)+0.30(7)+0.25(8)+0.15(6)=7.70$$

$$S_B=0.30(5)+0.30(9)+0.25(9)+0.15(7)=7.50$$

$$S_C=0.30(8)+0.30(7)+0.25(9)+0.15(6)=7.65$$

Pump A has the highest weighted score and is selected.

The result cannot be called statistically significant. The matrix is a deterministic decision (weighted scoring) model: single ratings are given without any sample, variance or repeated observations, so no significance test (t-test, ANOVA) can be applied. Moreover, the difference between A and C is only 0.05, so a sensitivity analysis of the weights would be more meaningful.

∴ Best pump = Pump A (score 7.70); statistical significance cannot be established (Answer)

Solution: $$Q=\frac{150}{3600}=0.04167{\text{m}}^3/\text{s},P_w=\rho gQH=1000\times 9.81\times 0.04167\times 450=183.9\text{kW}$$

$$P_{shaft}=\frac{P_w}{{\eta }_p}=\frac{183.9}{0.78}=235.8\text{kW}$$

(Electrical input to the motor = 235.8/0.92 = 256.3 kW. The speed is not needed.)

∴ Shaft power required = 235.8 kW (Answer)

Solution: The heating value of coal is evidently 25 MJ/kg (25 kJ/g); this value is used.

$${\dot{Q}}_{in}=\frac{P}{\eta }=\frac{50}{0.4}=125\text{MW},{\dot{m}}_{coal}=\frac{125\text{MJ/s}}{25\text{MJ/kg}}=5\text{kg/s}$$

$$\text{Coal consumption}=5\times 3600=18,000\text{kg/h}$$

∴ Coal consumption = 18,000 kg/h = 18 tonnes/h (Answer)

Solution: The first interval is taken as 0–6 h so that the schedule covers the full day.

$$E=30\times 6+90\times 6+60\times 2+100\times 4+50\times 6=1540\text{MWh}$$

$$\text{Average load}=\frac{1540}{24}=64.17\text{MW},\text{Load factor}=\frac{64.17}{100}=0.642$$

∴ Load factor = 64.2% (Answer)

Answer: When the load demand becomes greater than the generation.

Answer: Higher current for the same power (higher losses and voltage drop).

Answer: Natural gas.

Answer: Hydroelectric power plant.

Answer: Gas turbine power plant.

Answer: Rankine cycle.

উত্তর: ৩ ডিসেম্বর ১৯৫৫।

উত্তর: ষট্ + দশ (নিপাতনে সিদ্ধ)।

Answer: Present everywhere; omnipresent.

Answer: Something that seems unfortunate at first but later turns out to be good.

উত্তর: নিশাত মজুমদার (২০১২)।

Answer: Neil Armstrong (1969).

Answer: Mars.

Answer: The Pacific Ocean.

Answer: The Nile.

উত্তর: মাতারবাড়ী গভীর সমুদ্রবন্দর (কক্সবাজার)।

উত্তর: চীন।

উত্তর: যুক্তরাষ্ট্র।

উত্তর: লিরা।

উত্তর: বৃক্ক (কিডনি)।

উত্তর: ভূমিকম্প এবং পৃথিবীর অভ্যন্তরে ভূকম্পন তরঙ্গের (seismic waves) সঞ্চালন নিয়ে বিজ্ঞানের যে শাখা আলোচনা করে, তাকে সিসমোলজি বলে।