Bangladesh Power Development Board (BPDB)- 2026

Solution: [Repeated question: BGFCL-2021]

Processes: 1–2 condensate pump (P_C → P_H); 2–3 mixing with bled steam in the open feedwater heater (state 3 is saturated liquid at P_H); 3–4 feed pump (P_H → P_B); 4–5 heating, evaporation and superheating at constant P_B; 5–6 expansion to P_H where a fraction y is extracted; 6–7 expansion of the remaining (1 − y) to P_C; 7–1 condensation.

Per kg of steam through the boiler:

$$y=\frac{h_3-h_2}{h_6-h_2}$$

$$w_T=\left(h_5-h_6\right)+(1-y)\left(h_6-h_7\right),w_P=(1-y)\left(h_2-h_1\right)+\left(h_4-h_3\right)$$

$$q_{in}=h_5-h_4,{\eta }_{th}=\frac{w_T-w_P}{q_{in}}$$

Regeneration raises the mean temperature of heat addition, so the cycle efficiency increases. The open feedwater heater also acts as a deaerator.

Solution: Given, T₁ = 30 + 273 = 303 K, r_p = 8, η_C = 0.85, η_T = 0.80, T₃ = 800 + 273 = 1073 K, γ = 1.4, c_p = 1.005 kJ/kg·K.

Compressor outlet temperature:

$$T_{2s}=T_1{\left(r_p\right)}^{(\gamma -1)/\gamma }=303\times 8^{0.2857}=303\times 1.8114=548.9\text{K}$$

$${\eta }_C=\frac{T_{2s}-T_1}{T_2-T_1}\Rightarrow T_2=303+\frac{548.9-303}{0.85}=592.3\text{K}={319.3}^{\circ }\text{C}$$

Cycle efficiency:

$$T_{4s}=\frac{T_3}{8^{0.2857}}=\frac{1073}{1.8114}=592.3\text{K}$$

$$T_4=T_3-{\eta }_T\left(T_3-T_{4s}\right)=1073-0.80(1073-592.3)=688.5\text{K}$$

$$w_T=c_p\left(T_3-T_4\right)=1.005(1073-688.5)=386.4\text{kJ/kg}$$

$$w_C=c_p\left(T_2-T_1\right)=1.005(592.3-303)=290.7\text{kJ/kg}$$

$$w_{net}=386.4-290.7=95.7\text{kJ/kg},q_{in}=c_p\left(T_3-T_2\right)=1.005(1073-592.3)=483.1\text{kJ/kg}$$

$${\eta }_{cycle}=\frac{w_{net}}{q_{in}}=\frac{95.7}{483.1}=0.198$$

∴ Compressed air temperature, T₂ = 592.3 K (319.3°C); Cycle efficiency = 19.8% (Answer)

Solution: Given, L = 98.1 m, D = 0.1 m, Q = 0.0314 m³/s, ν = 800 mm²/s = 8 × 10⁻⁴ m²/s.

$$A=\frac{\pi }{4}(0.1{)}^2=7.854\times 10^{-3}{\text{m}}^2,V=\frac{Q}{A}=\frac{0.0314}{7.854\times 10^{-3}}=4.0\text{m/s}$$

$$Re=\frac{VD}{\nu }=\frac{4.0\times 0.1}{8\times 10^{-4}}=500$$

Since Re < 2000, the flow is laminar.

Major loss: For laminar flow, f = 64/Re.

$$f=\frac{64}{500}=0.128,\frac{V^2}{2g}=\frac{{4.0}^2}{2\times 9.81}=0.815\text{m}$$

$$h_f=f\frac{L}{D}\frac{V^2}{2g}=0.128\times \frac{98.1}{0.1}\times 0.815=102.3\text{m}$$

Minor losses:

$$\Sigma K=8\times 2+5\times 6=46,h_m=\Sigma K\frac{V^2}{2g}=46\times 0.815=37.5\text{m}$$

$$h_{total}=h_f+h_m=102.3+37.5=139.8\text{m}$$

∴ Re = 500 (laminar); Total head loss = 139.8 m (Answer)

Solution: [Repeated question: NWPGCL-2018, NWPGCL-2026]

• Pump H–Q curve (constant speed): head is maximum at zero discharge (shut-off head) and falls as discharge increases.
• System curve: H_sys = H_s + kQ², i.e. static head plus friction and fitting losses which vary with Q².
• Design point (D): the best-efficiency point (BEP), at which the pump efficiency is maximum; the pump is selected for this duty.
• Operating (duty) point (O): the intersection of the pump H–Q curve and the system curve; the pump actually runs at this point.
• Ideally O ≈ D. Throttling, speed change or impeller trimming shifts the operating point.

Solution: Given, d = 40 mm, L = 200 mm, P = 40 kN, δl = 0.0304 mm, L_f = 249 mm. (The yield and maximum loads are not required for the two answers asked.)

$$A=\frac{\pi }{4}\times 40^2=1256.6{\text{mm}}^2$$

$$\sigma =\frac{P}{A}=\frac{40000}{1256.6}=31.83{\text{N/mm}}^2,\epsilon =\frac{\delta l}{L}=\frac{0.0304}{200}=1.52\times 10^{-4}$$

$$E=\frac{\sigma }{\epsilon }=\frac{31.83}{1.52\times 10^{-4}}=2.094\times 10^5{\text{N/mm}}^2$$

$$\text{Percentage elongation}=\frac{L_f-L}{L}\times 100=\frac{249-200}{200}\times 100=24.5\%$$

∴ E = 2.094 × 10⁵ N/mm² (209.4 GPa); Percentage elongation = 24.5% (Answer)

Solution: [Repeated question: DPDC-2020] $$\frac{N_1}{N_2}=\frac{d_2}{d_1}=\frac{360}{120}=3\Rightarrow d_2=3d_1$$

$$x=\frac{d_1+d_2}{2}=2d_1=600\Rightarrow d_1=300\text{mm},d_2=900\text{mm}$$

$$T_1=\frac{\pi d_1}{p_c}=\frac{\pi \times 300}{25}=37.7\approx 38,T_2=3T_1=114$$

Since the number of teeth must be a whole number, the pitch circle diameters are recalculated:

$$d_1=\frac{T_1{p}_c}{\pi }=\frac{38\times 25}{\pi }=302.4\text{mm},d_2=\frac{114\times 25}{\pi }=907.2\text{mm}$$

$$\text{Exact }\text{centre}\text{ distance}=\frac{302.4+907.2}{2}=604.8\text{mm}$$

∴ Pinion: 38 teeth, d = 302.4 mm; Gear: 114 teeth, d = 907.2 mm (Answer)

Solution: Assume silicon devices: V_BE = 0.7 V, diode drop V_D = 0.7 V, transistor in the active region (I_C = βI_B, I_E = (β + 1)I_B).

KVL around the base loop:

$$15=I_B{R}_B+V_{BE}+V_D+(\beta +1)I_B{R}_E$$

$$I_B=\frac{15-0.7-0.7}{1000+121\times 1}\text{mA}=\frac{13.6}{1121\text{k}\Omega }=12.13\mu \text{A}$$

$$I_C=\beta I_B=120\times 12.13\mu \text{A}=1.456\text{mA},I_E=121\times 12.13\mu \text{A}=1.468\text{mA}$$

Node voltages (with respect to ground):

$$V_X\left(\text{top of }{R}_E\right)=I_E{R}_E=1.468\text{V},V_E=V_X+V_D=2.168\text{V},V_B=V_E+V_{BE}=2.868\text{V}$$

$$V_C=15-I_C{R}_C=15-1.456\times 2=12.09\text{V},V_{CE}=V_C-V_E=9.92\text{V}$$

Since V_C > V_B > V_E and V_CE = 9.92 V ≫ V_CE(sat), the transistor operates in the active region.

Solution: Over the rising half of a period, i = (2I_m/T)t. The falling half is symmetrical.

$$I_{rms}^2=\frac{1}{T/2}{\int }_0^{T/2}{\left(\frac{2I_{m}t}{T}\right)}^{2}dt=\frac{I_m^2}{3}$$

$$I_{rms}=\frac{I_m}{\sqrt{3}}=\frac{8}{\sqrt{3}}=4.62\text{A}$$

$$P_{avg}=I_{rms}^{2}R=\frac{8^2}{3}\times 18=384\text{W}$$

∴ (a) I_rms = 4.62 A; (b) Average power = 384 W (Answer)

X and Y are both 40 sin45° ≈ 28.3 m south of Z; X lies to the west and Y to the east. Hence Y is due East of X.

2d + 4g = 2(d + g) + 16 ⇒ 2g = 16 ⇒ g = 8.

A’s father’s wife is A’s mother; the only child of her father is A’s mother herself (B). So A is B’s child.

ⁿC₂ = n(n − 1)/2 = 36 ⇒ n² − n − 72 = 0 ⇒ (n − 9)(n + 8) = 0 ⇒ n = 9.

At 7:00 the angle between the hands is 210°. For the hands to be opposite (180°): 210 − 5.5m = 180 ⇒ m = 30/5.5 = 5$\frac{5}{11}$ min past 7.