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Key Facts: Senegal Baccalauréat Série STIDD Exam
10/20
Minimum weighted average required for immediate admission (Premier Groupe)
Office du Baccalauréat du Sénégal
Coeff 6
Coefficient for Sustainable Industrial Technology in Série STIDD
Décret portant organisation du Baccalauréat
5,000 FCFA
Registration fee; optional papers cost an additional 1,000 FCFA each
Office du Baccalauréat — 2026 registration conditions
Sample Senegal Baccalauréat Série STIDD Practice Questions
Try these sample questions to review concepts for the Senegal Baccalauréat Série STIDD exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1In energy audit and sustainable development engineering, what is the fundamental difference between primary energy and final energy?
A.Primary energy is the energy directly extracted from natural resources before any conversion, whereas final energy is delivered to and consumed by the end-user.
B.Primary energy is only renewable electricity, whereas final energy is fossil fuel combustion energy.
C.Final energy is always greater than primary energy due to thermal recovery in power plants.
D.Primary energy represents thermal losses in transmission lines, whereas final energy represents chemical energy stored in batteries.
Explanation: Primary energy (énergie primaire) refers to natural energy sources in their raw state prior to human transformation (e.g., crude oil, raw solar radiation, natural gas). Final energy (énergie finale) is the refined energy directly available to the consumer at the point of use (e.g., electricity at the wall plug, diesel in a vehicle tank), accounting for transformation and distribution losses.
2Under international standards (IEC 60904), what are the Standard Test Conditions (STC) used to rate the peak power (Wp) of a photovoltaic module?
A.Irradiance of 800 W/m², ambient temperature of 20 °C, and wind speed of 1 m/s.
B.Irradiance of 1000 W/m², cell temperature of 25 °C, and air mass spectral distribution of AM 1.5.
C.Irradiance of 1200 W/m², cell temperature of 45 °C, and air mass spectral distribution of AM 1.0.
D.Irradiance of 1000 W/m², ambient temperature of 40 °C, and air mass spectral distribution of AM 2.0.
Explanation: Standard Test Conditions (STC) define the universal laboratory benchmark for rating solar panels: an irradiance of 1000 W/m², a junction cell temperature of exactly 25 °C, and a solar reference spectral irradiance corresponding to Air Mass AM 1.5.
3A solar installation in Thiès, Senegal, has a peak capacity of 5 kWp and an overall system performance ratio (PR) of 0.75. If the site receives an average daily solar irradiation of 5.6 peak sun hours (HSP / kWh/m²/day), what is the estimated daily electrical energy output?
A.15.5 kWh/day
B.28.0 kWh/day
C.21.0 kWh/day
D.37.3 kWh/day
Explanation: The daily energy production is calculated as E = P_peak * HSP * PR. Substituting the given values gives E = 5 kWp * 5.6 h * 0.75 = 21.0 kWh/day. This takes into account losses from temperature, inverter conversion, wiring, and soiling.
4According to the Lanchester-Betz-Joukowsky limit, what is the maximum theoretical aerodynamic power coefficient (Cp_max) that an ideal wind turbine rotor can extract from kinetic wind energy?
A.1/2 (50.0%)
B.8/9 (approximately 88.9%)
C.2/3 (approximately 66.7%)
D.16/27 (approximately 59.3%)
Explanation: Betz's law establishes from fluid continuity and momentum conservation that the maximum kinetic energy an open-flow wind rotor can extract without completely stalling the downstream airflow is Cp_max = 16/27, or approximately 59.26%.
5A wind turbine installed at the Taiba N'Diaye wind park has rotor blade radius R = 60 m. Operating in air of density rho = 1.225 kg/m³ at wind speed v = 10 m/s with power coefficient Cp = 0.40, what is the aerodynamic power captured by the rotor?
A.1.385 MW
B.2.771 MW
C.6.927 MW
D.0.693 MW
Explanation: The aerodynamic power is given by P = 0.5 * rho * A * v³ * Cp, where swept area A = pi * R² = pi * 60² = 11,309.73 m². Thus, P = 0.5 * 1.225 * 11,309.73 * 1000 * 0.40 = 2,770,885 W ≈ 2.771 MW.
6An off-grid telecom relay in Saint-Louis requires 2.4 kWh of daily DC energy at a system nominal voltage of 24 V. To preserve battery life, the maximum permissible depth of discharge (DoD) is limited to 50%. What minimum battery bank capacity is required for one day of autonomy?
A.100 Ah
B.50 Ah
C.400 Ah
D.200 Ah
Explanation: Nominal energy needed is E = 2.4 kWh = 2400 Wh. At 24 V, the daily ampere-hour consumption is Q = 2400 Wh / 24 V = 100 Ah. With a maximum depth of discharge DoD = 50% (0.5), the required total battery capacity is C = Q / DoD = 100 Ah / 0.5 = 200 Ah.
7In the functional analysis of an industrial automated system, what is the primary role of the Operative Part (Partie Opérative - PO)?
A.To execute physical actions on the raw material or workpiece using actuators and mechanisms according to control orders.
B.To process logic equations and execute ladder/GRAFCET algorithms.
C.To provide the human-machine interface for manual parameter entry.
D.To step down high-voltage AC utility power to 24 V DC control power.
Explanation: The Operative Part (PO) comprises the actuators (cylinders, electric motors, heating elements), transmission mechanisms, and effectors that directly transform, move, or process the workpiece under instructions received from the Control Part (Partie Commande - PC).
8In meteorological solar stations and large PV power plants, what is the primary measurement advantage of a thermopile pyranometer over a silicon photodiode pyranometer?
A.Much lower production cost and faster microsecond response time.
B.Direct digital output without requiring signal conditioning or amplification.
C.A broad and flat spectral sensitivity response from 300 nm to 3000 nm, independent of ambient spectral shifts.
D.Total immunity to thermal drift and ambient temperature changes without internal compensation.
Explanation: Thermopile pyranometers absorb total solar radiation across black coated surfaces, delivering a uniform flat spectral response spanning nearly the entire solar spectrum (300 to 3000 nm). Silicon photodiodes have limited spectral sensitivity (typically 400 to 1100 nm) and introduce spectral errors as atmospheric conditions alter the sunlight spectrum.
9Why is a 4-20 mA current loop standard widely favored over a 0-10 V voltage signal for transmitting sensor data over long distances in industrial facilities?
A.A current loop allows higher digital data transmission baud rates exceeding 100 Mbps.
B.The current signal is immune to signal attenuation caused by transmission wire resistance, and the 4 mA live zero provides built-in wire-break fault detection.
C.Current signals do not require any power supply at the transmitter end.
D.0-10 V signals suffer from high electromagnetic self-induction that permanently damages analog input cards.
Explanation: Because current is identical throughout a series circuit regardless of wire resistance (Ohm's law I = const), long wiring runs do not attenuate the reading. Furthermore, the 4 mA 'live zero' enables the controller to instantly distinguish a true zero reading (4 mA) from a severed wire or power failure (0 mA).
10In GRAFCET (IEC 60848) modeling of sequential automated systems, what does a double horizontal bar beneath an active step represent?
A.An alternative choice divergence (OR divergence) where only one branch is selected.
B.A compulsory emergency stop barrier that clears all active steps.
C.A simultaneous sequence divergence (AND divergence) where all subsequent branches are activated simultaneously.
D.A jump to an upstream step creating a continuous loop.
Explanation: In GRAFCET syntax, a double horizontal bar represents synchronization. When placed below a transition, it indicates an AND divergence (divergence en ET): when the preceding step is active and the transition receptivity is true, all parallel branches directly below the double bar are activated concurrently.
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