By Michael Clifford
An advent to Mechanical Engineering: half 2 is an important textual content for all second-year undergraduate scholars in addition to these learning origin levels and HNDs. The textual content presents thorough insurance of the next center engineering topics:
- Fluid dynamics
- Solid mechanics
- Control thought and techniques
- Mechanical strength, rather a lot and transmissions
- Structural vibration
As good as mechanical engineers, the textual content should be hugely suitable to automobile, aeronautical/aerospace and normal engineering students.
The fabric during this booklet has complete pupil and lecturer help on an accompanying site at http://cw.tandf.co.uk/mechanicalengineering/, which includes:
- worked ideas for exam-style questions
- multiple-choice self-assessment
- revision material
The textual content is written via an skilled workforce of teachers on the across the world popular college of Nottingham.
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Additional info for An Introduction to Mechanical Engineering: Part 2
5 3 10–5 m2/s, respectively. 2 for a sphere when the flow is turbulent, where the critical Reynolds number is Re 5 3 3 105. 91 [N] 2 The dimples on a golf ball can reduce the pressure drag by making the ball surface rough. This reduces the transition Reynolds number by artificially forcing (tripping) the boundary layer to turbulent flow at low Reynolds numbers. 21. Although the friction drag is increased in this case, the total drag of the golf ball is reduced. This is because the golf balls are bluff (non-streamlined) bodies, where Dpres is much greater than Dfric.
The system has to do with the fluid itself rather than the system containing it. The control volume is often specified by reference to an artefact, but it is important to note the distinction. In Part 1, both closed systems – such as pistons in closed cylinders – and open systems were considered although no specific open systems were discussed. The open system is of significant importance in this unit, and the first law in the form of the SFEE (steady flow energy equation) dominates the analysis.
We can easily see that these two cannot form a non-dimensional quantity. Since the p theorem After some calculation, we find that p1 5 ____ Q n D2 cannot give a functional form of a non-dimensional quantity, we take the liberty of Q nD3 . making an assumption on its form. Conventionally, we use ____ 3 rather then ____ nD Q Therefore, we set Q 1 5 ____ 3 p195 ___ p1 nD The performance of the centrifugal pump can be expressed as p2 5 fn (p19): ( nD ) gH Q _____ 5 fn ____ 3 2 n D2 or, CH 5 fn(CQ) where, CH and CQ are the head coefficient and the capacity coefficient, respectively.