NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersOil hydraulics on tractors and machines: pumps, motors, cylinders, valves and circuits.
In the exams: NCEES PE Machinery Systems (hydraulic components and circuits, explicit); elective in the ICAR 6th Deans' Committee B.Tech
NCEES PE Agricultural and Biological Engineering — Machinery Systems
Explicit fluid-power content: pump and motor sizing from displacement, speed and efficiencies; cylinder force, speed and cycle time; hydraulic power and engine power required; pressure drop in lines and heat load; selecting valves and reading circuits drawn with standard symbols. Calculations dominate; unit consistency in psi, gpm, in³ and hp is tested.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Hydraulic Drives & Controls. Difficulty: 16 easy, 44 medium, 21 hard.
| Pressure relief and flow control valves | 16 | |
| Hydraulic cylinders | 14 | |
| Pressure drop and heat generation | 11 | |
| Principles of fluid power | 9 | |
| Directional control valves | 9 | |
| Hydraulic pumps: gear, vane and piston | 8 | |
| Hydraulic motors | 8 | |
| Open-centre, closed-centre and load-sensing circuits | 4 | |
| Tractor hydraulic systems | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A gear pump has a displacement of 40 cm³/rev, runs at 1800 rev/min and has a volumetric efficiency of 92 %. Its actual delivery is
B. 66.2 L/min
Q = V × n × ηv = 40 cm³ × 1800 /min × 0.92 = 66240 cm³/min = 66.2 L/min. 72.0 L/min is the theoretical flow; dividing by ηv (78.3) treats the pump as a motor; 6.62 is a units slip.
Further reading: Esposito, Fluid Power with Applications
A log splitter uses a 100 mm bore, 70 mm rod cylinder at 200 bar. When the cylinder extends in regenerative mode, the maximum force available is
C. 77.0 kN
Pressure acts on both sides of the piston, so the net area is the rod area: F = 200 × 10⁵ × π(0.07)²/4 = 77.0 kN. The full piston area gives 157.1 kN, available only in normal extension; the annulus gives 80.1 kN, the retraction force; adding the areas has no basis. This is why regenerative circuits switch to normal extension when the wedge meets the log. Adding the areas gives 234.0 kN.
Further reading: Esposito, Fluid Power with Applications
A pump supplies 50 L/min to the rod end of a cylinder with a 100 mm bore and a 70 mm rod to retract it. The flow returning from the cap end to tank during retraction is
A. 98.0 L/min
Return flow = Q × Ap/Aa = 50 × (π0.1²/4)/(π(0.1² − 0.07²)/4) = 50 × 0.01/0.0051 = 98.0 L/min. The return line, valve and filter must handle this, almost double the pump flow. Assuming it equals pump flow gives 50.0; inverting the ratio gives 25.5; using the rod area gives 102.0.
Further reading: Esposito, Fluid Power with Applications
A cylinder must exert 50 kN on extension with a supply pressure of 140 bar, ignoring friction and back pressure. The minimum bore is about
C. 67.4 mm
A = F/p = 50000/(140 × 10⁵) = 3.571 × 10⁻³ m²; D = √(4A/π) = 67.4 mm. √(A/π) is the radius (33.7 mm); 3.57 is the area in a wrong unit; 213 mm comes from treating 140 bar as 14 bar. In practice the next larger standard bore and a friction allowance are chosen.
Further reading: Esposito, Fluid Power with Applications
The cushions built into the ends of many hydraulic cylinders are intended to
A. slow the piston near the end of stroke by restricting the outgoing oil
A cushion spear or collar enters a pocket near the end cap, forcing the exhaust oil through a small adjustable orifice so the piston decelerates instead of striking the cap. It does not add force, lock the piston or filter the oil.
Further reading: Esposito, Fluid Power with Applications
A sequence valve differs from a simple relief valve in that the sequence valve
D. has a pressurised working outlet, so it must be drained externally
Both are normally closed and opened by inlet pressure, but a relief valve's outlet goes to tank while a sequence valve's outlet is a working line at pressure, so its spring chamber must be drained externally. Normally open with outlet sensing describes a reducing valve; external pilot opening describes an unloading valve.
Further reading: Esposito, Fluid Power with Applications
A circuit uses an accumulator and a fixed pump. When the accumulator is charged to its set pressure, an external pilot signal from the accumulator line opens a valve that returns pump flow to tank at low pressure. That valve is
A. an unloading valve
An unloading valve is normally closed and opened by an external pilot signal, letting the pump idle at low pressure while a check valve keeps the accumulator charged. Sequence and counterbalance valves are opened by their own inlet pressure, and a reducing valve is normally open and senses its outlet.
Further reading: Esposito, Fluid Power with Applications
Oil of 46 cSt flows at 60 L/min in a 20 mm bore line. The Reynolds number, and hence the flow regime, is
C. about 1384, laminar
v = (60/60000)/(π × 0.02²/4) = 3.18 m/s; Re = vD/ν = 3.18 × 0.020/46 × 10⁻⁶ = 1384, below about 2000, so laminar. Using 46 × 10⁻⁹ for cSt gives 1383956; using the radius instead of the diameter as the length in Re gives 692; 5536 comes from using the radius in the velocity area but the diameter in Re.
Further reading: Esposito, Fluid Power with Applications
Why does a hydraulic line carry a much higher pressure loss on a cold morning than at normal working temperature, for the same flow?
B. Flow is laminar, so friction loss rises with viscosity, which is far higher when cold
With f = 64/Re in laminar flow, Δp = 32μLv/D², proportional to dynamic viscosity, which can rise many times at low temperature. Density changes little with temperature; cold, thick oil makes flow more laminar, not turbulent; and hose bores do not shrink appreciably.
Further reading: Esposito, Fluid Power with Applications
In the design of a stationary industrial hydraulic power unit, a commonly used rule of thumb for an open-loop reservoir is a volume of about
D. two to three times the pump flow per minute
Fluid-power texts commonly suggest a reservoir of about two to three times the pump's flow per minute for industrial units, giving time for air release, settling and cooling. Mobile machines use much smaller tanks and rely on coolers. The largest cylinder's volume sets only the minimum working reserve, not the tank size.
Further reading: Esposito, Fluid Power with Applications
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