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100+ Free Tanzania ERB Mechanical Practice Questions
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Key Facts: Tanzania ERB Mechanical Exam
100
Practice Questions
OpenExamPrep
50%
Pass Score
Official Guidelines
3.0 hours
Time Limit
Exam Rules
Tanzania ERB Professional Engineers Examination - Mechanical Engineering prep course featuring 100 high-quality practice questions and detailed explanations.
Sample Tanzania ERB Mechanical Practice Questions
Try these sample questions to test your Tanzania ERB Mechanical exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Which of the following statements correctly defines a closed thermodynamic system?
A.A system that allows both energy and mass transfer across its boundaries.
B.A system that allows mass transfer but not energy transfer across its boundaries.
C.A system that allows neither energy nor mass transfer across its boundaries.
D.A system that allows energy transfer but not mass transfer across its boundaries.
Explanation: A closed system is characterized by the transfer of energy (heat and/or work) across its boundaries, while the mass within the system remains constant, meaning no mass can enter or leave. This is a fundamental concept in thermodynamics.
2A heat engine operates between a high-temperature reservoir at 800 K and a low-temperature reservoir at 300 K. What is the maximum possible thermal efficiency of this engine?
A.75%
B.50%
C.37.5%
D.62.5%
Explanation: The maximum possible thermal efficiency for any heat engine operating between two temperatures is given by the Carnot efficiency formula: η_carnot = 1 - (T_L / T_H), where T_L and T_H are the absolute temperatures of the cold and hot reservoirs, respectively. Substituting the given values: η_carnot = 1 - (300 K / 800 K) = 1 - 0.375 = 0.625 or 62.5%.
3Heat transfer through a solid material by molecular vibration and collision without any net movement of the material itself is known as:
A.Convection
B.Advection
C.Radiation
D.Conduction
Explanation: Conduction is the transfer of thermal energy between adjacent atoms or molecules in a substance due to a temperature gradient. It typically occurs in solids and stationary fluids where there is no macroscopic movement of the material.
4Steam enters a turbine at 5 MPa, 400 °C, and leaves at 10 kPa as saturated vapor. If the turbine produces 2 MW of power, and heat loss from the turbine is negligible, what is the mass flow rate of the steam? (Assume h1 = 3195.7 kJ/kg, h2 = 2583.9 kJ/kg for inlet and outlet respectively).
A.0.33 kg/s
B.3.26 kg/s
C.2.89 kg/s
D.1.95 kg/s
Explanation: For a steady-flow turbine with negligible heat loss (Q=0) and kinetic/potential energy changes, the First Law of Thermodynamics simplifies to W_out = m_dot * (h1 - h2). Given W_out = 2000 kW, h1 = 3195.7 kJ/kg, h2 = 2583.9 kJ/kg. So, m_dot = W_out / (h1 - h2) = 2000 kJ/s / (3195.7 - 2583.9) kJ/kg = 2000 / 611.8 = 3.269 kg/s, approximately 3.26 kg/s.
5An adiabatic steam turbine operates with an inlet state of 4 MPa, 500 °C (h1 = 3446 kJ/kg, s1 = 7.090 kJ/kg·K) and an outlet pressure of 10 kPa. If the actual specific work output is 850 kJ/kg, what is the isentropic efficiency of the turbine? (At 10 kPa, h_f = 191.8 kJ/kg, h_fg = 2392.1 kJ/kg, s_f = 0.649 kJ/kg·K, s_fg = 7.501 kJ/kg·K).
A.70.8%
B.81.3%
C.75.0%
D.65.2%
Explanation: First, calculate the isentropic specific work. For an isentropic process, s2s = s1 = 7.090 kJ/kg·K. At 10 kPa, s2s = s_f + x_s * s_fg. So, x_s = (7.090 - 0.649) / 7.501 = 0.8586. Then, h2s = h_f + x_s * h_fg = 191.8 + 0.8586 * 2392.1 = 2245.9 kJ/kg. The isentropic work output is W_s = h1 - h2s = 3446 - 2245.9 = 1200.1 kJ/kg. The isentropic efficiency is η_isen = W_actual / W_s = 850 kJ/kg / 1200.1 kJ/kg = 0.7082 or 70.8%.
6A 1 cm thick plate of steel (k = 50 W/m·K) has one surface maintained at 100 °C and the other at 50 °C. What is the rate of heat transfer per unit area through the plate?
A.2.5 kW/m²
B.250 kW/m²
C.50 kW/m²
D.5 kW/m²
Explanation: For steady-state conduction through a plane wall, Fourier's Law states Q/A = k * (T1 - T2) / L. Given k = 50 W/m·K, T1 = 100 °C, T2 = 50 °C, and L = 1 cm = 0.01 m. So, Q/A = 50 * (100 - 50) / 0.01 = 50 * 50 / 0.01 = 2500 / 0.01 = 250,000 W/m² = 250 kW/m².
7Which of the following equations represents the ideal gas law using the universal gas constant?
A.PV = nRT
B.P/V = nRT
C.P = nRT/V²
D.PV = C
Explanation: The ideal gas law, in its most common form using the universal gas constant (R_u or R), relates pressure (P), volume (V), number of moles (n), and absolute temperature (T) by the equation PV = nRT. This equation describes the behavior of ideal gases under various conditions.
8A counter-flow heat exchanger heats water from 20 °C to 80 °C using hot oil that cools from 120 °C to 60 °C. What is the Log Mean Temperature Difference (LMTD) for this heat exchanger?
A.30 °C
B.40 °C
C.50 °C
D.60 °C
Explanation: For a counter-flow heat exchanger, the temperature differences at the two ends are ΔT1 = T_h,in - T_c,out = 120 - 80 = 40 °C and ΔT2 = T_h,out - T_c,in = 60 - 20 = 40 °C. Since ΔT1 = ΔT2 = 40 °C, the LMTD formula simplifies to LMTD = ΔT1 = ΔT2 = 40 °C.
9A blackbody at 1000 K radiates heat to its surroundings. If the temperature of the blackbody is doubled to 2000 K, by what factor does the total emitted radiation increase?
A.4 times
B.2 times
C.16 times
D.8 times
Explanation: According to the Stefan-Boltzmann Law, the total emissive power of a blackbody is directly proportional to the fourth power of its absolute temperature (E_b = σT^4). If the temperature T is doubled to 2T, the new emissive power will be E_b' = σ(2T)^4 = σ(16T^4) = 16 * (σT^4). Thus, the radiation increases by a factor of 16.
10In an ideal Rankine cycle, steam enters the turbine at 10 MPa, 600 °C (h1 = 3625.8 kJ/kg, s1 = 6.9034 kJ/kg·K) and condenses at 10 kPa. Assuming isentropic expansion, what is the specific work output of the turbine? (At 10 kPa, h_f = 191.81 kJ/kg, h_fg = 2392.1 kJ/kg, s_f = 0.6493 kJ/kg·K, s_fg = 7.5009 kJ/kg·K).
A.1510 kJ/kg
B.1440 kJ/kg
C.1350 kJ/kg
D.1600 kJ/kg
Explanation: For an isentropic expansion in the turbine, the entropy remains constant, so s2 = s1 = 6.9034 kJ/kg·K. At the outlet pressure of 10 kPa, we find the quality x2 using s2 = s_f + x2 * s_fg: x2 = (6.9034 - 0.6493) / 7.5009 = 0.83378. Then, calculate the outlet enthalpy h2 = h_f + x2 * h_fg = 191.81 + 0.83378 * 2392.1 = 2185.45 kJ/kg. The specific work output of the turbine is W_t = h1 - h2 = 3625.8 - 2185.45 = 1440.35 kJ/kg, approximately 1440 kJ/kg.
About the Tanzania ERB Mechanical Exam
The ERB Tanzania mechanical engineering professional registration exam covers thermodynamics, hydraulic pumps/turbines, machinery design, safety, and professional code of conduct.
Assessment
Multiple-choice computerised exam administered by Engineers Registration Board (ERB) Tanzania.
Time Limit
3.0 hours
Passing Score
50%
Exam Fee
150,000 TZS (Engineers Registration Board (ERB) Tanzania)
Tanzania ERB Mechanical Exam Content Outline
25%
Thermodynamics And Heat Transfer
Practice questions covering the domain: thermodynamics and heat transfer.
25%
Fluid Mechanics And Hydraulic Machinery
Practice questions covering the domain: fluid mechanics and hydraulic machinery.
25%
Mechanical Design And Materials Science
Practice questions covering the domain: mechanical design and materials science.
25%
Manufacturing Processes And Erb Ethics
Practice questions covering the domain: manufacturing processes and erb ethics.
How to Pass the Tanzania ERB Mechanical Exam
What You Need to Know
- Passing score: 50%
- Assessment: Multiple-choice computerised exam administered by Engineers Registration Board (ERB) Tanzania.
- Time limit: 3.0 hours
- Exam fee: 150,000 TZS
Keys to Passing
- Complete 500+ practice questions
- Score 80%+ consistently before scheduling
- Focus on highest-weighted sections
- Use our AI tutor for tough concepts
Tanzania ERB Mechanical Study Tips from Top Performers
1Carefully study all regulatory and legislative requirements.
2Practice sample calculations and review real-world scenario items.
3Review the explanations for all incorrect practice questions to build core conceptual clarity.
Frequently Asked Questions
What is the passing score for Tanzania ERB Mechanical?
The passing score is 50%.
Who administers the Tanzania ERB Mechanical exam?
The exam is administered by the Engineers Registration Board (ERB) Tanzania.