Technology & Engineering
Power in Mechanics
Power in mechanics refers to the rate at which work is done or energy is transferred. It is a measure of how quickly a force can do work. In the context of technology and engineering, power is a crucial concept for understanding the performance of machines and systems, as well as for designing and optimizing their efficiency.
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7 Key excerpts on "Power in Mechanics"
- Paul Anthony Russell(Author)
- 2021(Publication Date)
- Reeds(Publisher)
vertical height is raised mm 0.2342 m G GO GA = - = - = = 309 2 75 234 2 . . work done weight vertical lift of J = × = × = G 1040 0 2342 243 5 . . Ans. (ii) Power POWER is the rate of doing work, or the quantity of work completed in a given time. The unit of power is the watt, which is equal to the rate 1 joule of work being done every second. power J/s Nm/s) force velocity work done (for small ( W F ds = = = × = displacement) power (for small time interval) (i = = F ds dt P Fv nstantaneous velocity ) v 84 • Applied Mechanics When describing the power generated in marine engineering applications, either mechanical, electrical or hydraulic, the kilowatt (kW) is usually a more convenient unit of measure, due to the size of the power that is being generated. Example 4.2. A pump lifts fresh water from one tank to another through an effective height of 12 m. If the mass flow of water is 40 t/h, find the output power of the pump. 40 40 10 9 81 3 3 t kg weight of water lifted per second 40 10 3 = × = × × . 600 9 81 1 N power weight lifted per second height = 40 10 3 = × × × × . 2 3600 1308 W W or 1.308 kW P = MECHANICAL EFFICIENCY is the ratio of power that comes out of a machine compared to the power that is put in, therefore: mechanical efficiency output power input power = This will always give a fraction less than unity; however, it is common practice to multiply the result by 100 and express the efficiency as a percentage. For example, if the input power of the pump in Example 4.2 was 1.75 kW, then the efficiency would be ℑ= = = output power input power or 74.74% 1 308 1 75 0 7474 . . . PRESSURE is defined as the force per unit area. It is measured in units of N/m 2 and is used in calculating power. POWER OF RECIPROCATING ENGINES. When a gas or a liquid acts on a piston in a closed cylinder, the pressure multiplied by the piston area on which the pressure acts gives the total force on the piston.
eBook - PDFEngineering Mechanics
Problems and Solutions
- Arshad Noor Siddiquee, Zahid A. Khan, Pankul Goel(Authors)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
13.4. If the particle is infinitesimally displaced by displacement ‘ ds ’ during time ‘ dt ’ then the work done by pushing force will be given by δ δ w F ds w m a ds a dv dt = = = . . . where Substituting the value of a in equation (1), δ δ δ w m dv dt ds w m ds dt dv w m v dv = = = . . . . . . ..... (1) 594 Engineering Mechanics If the particle is finally displaced by distance ‘s’ from position 1 to position 2 (not shown), then work done can be determine by integrating the above equation, δ δ w m v dv w m v dv W m v v 1 2 1 2 1 2 1 2 2 2 1 2 1 2 ∫ ∫ ∫ ∫ = = = -( ) . . . Work done = Change in kinetic energy of the particle Thus according to this principle, the work done on a moving particle is equal to the change in its kinetic energy. This principle is useful when kinetic problems are required to be exclusively dealt with in variation of velocities. 13.6 Power Power is a term which designates the work performed with minimum time. For example, a person wants to travel a distance of 5 km. Suppose he travels the same distance by three different modes, i.e., by walking, bicycle and motor cycle, and takes time 70 min, 25 min, and 8 min, respectively. Thus the work performed in the third case is most meaningful as it is conducted with the least time. Hence power is the capacity of a machine to perform work with respect to time or it is the rate at which work is performed. If a machine performs work W in total time duration t then, the average power of such machine will be given by P W t avg = F mg ds v 1 v 2 Fig. 13.4 Work and Energy 595 If a machine performs work δw for instant time dt then the instantaneous power of machine will be given by P w dt w F ds P F ds dt P F dv inst inst inst = = = = δ δ as . . . Hence, Power P = F. v Power is expressed in watt or joule per second in S.I. units. However, in electrical and mechanical equipment devices like motor and engine, the power is defined in terms of horsepower (hp).
eBook - PDF- Michael Brumbach(Author)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
Power is a measurement of the rate of doing work. The amount of work can be calculated by the formula, Work 5 Force 3 Dis-tance (W 5 FD), where the force is measured in pounds (or kilograms) and the distance is measured in feet (or meters). To determine the power a machine must de-liver, it is necessary to know the speed at which the work must be accomplished. The formula for power is Power 5 Work/Time (P 5 W/T), where work is measured in foot-pounds and time is measured in minutes (min). Horsepower Horsepower (hp) is the common unit of measure-ment of mechanical power in the English system. In early times, water was pumped from mines by horses pulling and turning a wheel at the end of a shaft to drive a pump. Later, horses were replaced by the steam engine. Steam engines are rated as having the power of a certain number of horses (horsepower). It was determined that an aver-age horse could do work at the rate of 33,000 foot-pounds (44,740 joules) per minute. This means that the average horse can move 1000 pounds (453.6 ki-lograms) through 33 feet (10.058 meters) in 1 min-ute. The mathematical formula for horsepower is hp 5 ft 1b / min 33,000 (Eq. 3.1) Example 1 A machine can place 50,000 lb (22,680 kg) of scrap metal onto a truck 10 ft (3.048 m) high in 5 min. What horsepower is the machine capable of delivering? P I E FIGURE 3–1 Power law triangle. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 30 CHAPTER 3 Combining like terms gives: P 5 E 2 R This results in Equation 3.3. Equation 3.4 is also a combination of two for-mulas.
eBook - PDF- William Bolton(Author)
- 2016(Publication Date)
- Newnes(Publisher)
Hence Figure 12.1 Work being done as a result of rotation work done per second = T x 2 / = 7 where is the angular velocity. 182 Engineering Science Power The rate at which energy is transferred is called power, hence where the energy is transferred as a result of work, the power is the rate of doing work. The unit of power is the watt (W), where 1 W = 1 J/s. For an object on which work is done for a time / and results in a displace-ment s in that time, W Fs _ power = — = — = Fv t t where v is the velocity. Example 1 The locomotive of a train exerts a constant force of 120 kN on a train while pulling it at 40 km/h along a level track. What is (a) the work done in 15 minutes and (b) the power? (a) In 15 minutes the train covers a distance of 10 km. Hence work done = 120 x 10 3 x 10 x 10 3 = 1.2 x 10 9 J (b) Power is the rate of work being done; hence 120 x 10 3 x 40 x 10 3 Λ . 1 6 117 Λ . ..... power = = 1.3 x 10 6 W = 1.3 MW 60 x 60 Example 2 Calculate the power a car must exert if it is to maintain a constant velocity of 30 m/s when the resisting forces amount to 4.0 kN. As above, power = Fv = 4.0 x 10 3 x 30 = 120 x 10 3 W = 120 kW Example 3 What power must be developed by the motor of a tractor of mass 800 kg which pulls a trailer of mass 300 kg up a 10° incline at 10 m/s if the resistance to motion amounts to 500 N? The component of the weight of tractor plus trailer down the plane is 1100 x 9.8 x sin 10°. Thus the total force acting down the plane is F= 1100x9.8xsin 10° + 500 = 2372 N Hence power = Fv = 2372 x 10 = 23.72 x 10 3 W = 23.72 kW Energy 183 Example 4 What power will be required for a pump to extract water from a mine at 5 mVs and pump it through a vertical height of 20 m? Water has a density of 1000 kg/m 3 . As power is the rate of doing work power = y = 5 x 1000 x 9.8 x 20 = 980 x 10 3 W = 980 kW Example 5 A constant torque of 50 N m is used to keep a flywheel rotating at a constant 5 revolutions per second.
eBook - PDF- W. C. Bolton(Author)
- 2013(Publication Date)
- Wiley-Blackwell(Publisher)
Chapter 13 Power transmission 13.1 Machines and some basic definitions A machine is a mechanical device which can enable: • A force to be perhaps magnified, e.g. a screw jack, or reduced. • A force to be applied in a more convenient direction. • The displacement of the point of application of a force to be magnified or reduced, e.g. a lever. They are power transmission devices. Some basic terms used in describing the operation of machines are: 1 The effort is the input force to the machine. 2 The load is the output force from a machine. 3 The force ratio or mechanical advantage (MA) is the ratio load/effort. 4 The movement ratio or velocity ratio (VR) is the ratio (distance moved by effort)/(distance moved by load). 5 The efficiency of a machine is the fraction: (energy transferred from the machine to the load)/(energy transferred from the effort to the machine) or, if we consider the energy input and output per second, (power input)/(power output). For an ideal machine where there are no frictional forces, mechanical energy is conserved and so the work done by the effort must equal the work done by the load, thus effort × distance moved by effort = load × distance moved by load This can be rearranged to give load effort = distance moved by effort distance moved by load and so we have MA = VR. For a non-ideal, simple machine, the relationship between the effort E and the load L is of the form E = aL + b 324 Power transmission 325 where a and b are constant. In the ideal machine b is zero; in the non-ideal machine it can be considered to be the effort needed to overcome friction. Hence MA = F E = F aL + b = 1 a + b/L If the load is large, then b/L becomes small and so the mechanical advantage approximates to 1 /a .
eBook - PDFTechnology
Made Simple
- Don McCloy(Author)
- 2014(Publication Date)
- Made Simple(Publisher)
2.7 Transmission of Power In some engines, such as the aircraft gas turbine, the power is generated at the place where it is needed. In many other cases, this is not so. For example, the power developed in the car engine has to be transmitted to the wheels. An even greater problem is the transmission of the power of the steam turbine from the power station to the home or the factory. We shall look briefly at transmission of power by mechanical, by fluid and by electrical means. Mechanical transmission Belt drives were used to transmit power around the early factories. Now-adays they can be found in vacuum cleaners, motor mowers and lathes, but perhaps their best-known application is found under the bonnet of the car where a belt is used to drive the fan and the alternator from the crank shaft. As well as transmitting power from A to B, the belt drive can also be used to transform the power (see earlier). Fig. 2.21(a) shows a belt connecting two wheels A and B. Wheel A of radius r is the input and wheel B of radius R is the output. One clockwise revolution of A will move the belt through a distance of 2nr. Since it takes a belt movement of 2nR to turn wheel B through one revolution, then a movement of 2nr will turn wheel B through (2nr)/ (2nR) = r/R revolutions. The transmission factor is thus r/R, and the output will rotate at r/R times the speed of the input. Now power is the product of torque and angular velocity,* so, if there is no power loss in the transmission, the output torque will be R/r times the input torque. For example, if the diameter of wheel B is twice that of A, then the output speed will be one half of the input and the output torque will be twice that of the input. In Fig. 2.21(a) the output and the input wheels rotate in the same direc-tion. A reversal can be achieved by using a crossed belt, as in Fig. 2.2 (b). Since belts rely on friction for their effective operation, there is a limit to the force that they can transmit.
eBook - PDFGeneral Engineering Science in SI Units
The Commonwealth and International Library: Mechanical Engineering Division
- G. W. Marr, N. Hiller(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
To the question What has happened to the mechanical energy expended by the motor?, the reader will no doubt reply that the mechani-93 GENERAL ENGINEERING SCIENCE IN SI UNITS cal energy has been converted into heat energy. This seems quite obvious. Suppose, however, that the motor had been used to hoist a load. The motor then does work in lifting the load against the gravitational pull exerted on the load. What, however, be-comes of the energy expended in lifting the load? It is not simply used up, for one of the fundamental principles of engineering science is that we can neither create nor destroy energy; we can only change energy from one form to another. In order to answer this question it is necessary to introduce the idea of stored energy. This idea is not really so novel as it may appear at first sight. Our definition of the energy possessed by a body was simply that it was a measure of the capability of the body to do work. We are quite familiar with the idea that a body which is allowed to fall can do work. A simple example of this is the work done in driving a pile into the ground when the pile is struck by the falling hammer. Experience indicates that the pile is driven further into the ground when the height from which the hammer is dropped is increased. The work done is also increased if the hammer is replaced by one which is heavier. We thus associate the ability of the hammer to do work, that is the energy possessed by the hammer, with its weight and its height. Put another way, we consider that the hammer possesses or stores an amount of energy by virtue of its position. This form of stored energy is known as potential energy. We can now say that the energy expended by a motor in raising a body is converted into potential energy which is stored by the body. The work done in raising the body = force overcome X dis-tance = gravitational pull on body X height to which it is raised — W.h.
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