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Adiabatic total work

Consider a two-stage compression in which the intermediate gas is cooled down to the initial temperature. The total work for a two-stage adiabatic gas compression of an ideal gas is given by ... [Pg.659]

Finally, to get the total work, we do an overall energy balance (system = two tanks adiabatic, closed, constant volume). [Pg.63]

A portable engine of nineteenth-century design used a tank of compressed air and an evacuated" tank as its power source. The first tank had a capacity of 0.3 m and was initially filled with air at 14 bar and a temperature of 700 C. The evacuated" tank had a capacity of 0.75 m- . Unfortunately, nineteenth-century vacuum techniques were not very efficient. sO the "evacuated" tank contained air at 0.35 bar and 25 C. What is the ma.ximum total work that could be obtained from an air-driven engine connected between the two tanks if the process is adiabatic What would be the temperature and pressure in each tank at the end of the process You may assume that air is an ideal gas with Cp = 7/ /2. [Pg.143]

It was then shown, as a result of Joule s experiments, that the total work done by a body in an adiabatic process depends only on... [Pg.46]

A piston reversibly and adiabatically contracts 3.88 moles of ideal gas to one-tenth of its original volume, then expands back to the original conditions. It does this a total of five times. If the initial and final temperatures both are 27.5°C, calculate (a) the total work and (b) the total AL/for the overall process. [Pg.72]

For all of these adiabatic processes, the total (net) work is exactly the same. [Pg.330]

Measurement of Performance The amount of useful work that any fluid-transport device performs is the product of (1) the mass rate of fluid flowthrough it ana (2) the total pressure differential measured immediately before and after the device, usually expressed in the height of column of fluid equivalent under adiabatic conditions. The first of these quantities is normally referred to as capacity, and the second is known as head. [Pg.900]

This only depends on the probability of the termini, the total adiabatic works, and the total weight of stochastic transitions. The first of these is for uncorrelated motion and is the one that occurs in Glauber or Kawasaki dynamics [75-78]. The last term is, of course, very sensitive to the specified trajectory and the degree to which it departs from the adiabatic motion. However, the stochastic transitions are the same on the forward and on the reverse trajectory, and the ratio of the probabilities of these is... [Pg.50]

The work for a real adiabatic compression can also be calculated from the difference between the total enthalpy of the outlet and inlet flows ... [Pg.657]

Most of the theoretical works concerning dynamical aspects of chemical reactions are treated within the adiabatic approximation, which is based on the assumption that the solvent instantaneously adjusts itself to any change in the solute charge distribution. However, in certain conditions, such as sudden perturbations or long solvent relaxation times, the total polarization of the solvent is no longer equilibrated with the actual solute charge distribution and cannot be properly described by the adiabatic approximation. In such a case, the reacting system is better described by nonequilibrium dynamics. [Pg.335]

The temperature T0 corresponding to zero velocity is known as the adiabatic stagnation temperature it is the temperature that the flowing gas would attain if it were brought to rest adiabatically without doing any shaft work. It is sometimes called the total temperature. The temperature difference is small for air flowing at a speed of 100 m/s, T0 - T = 5 K. [Pg.205]

Figure 4.3 Reversible Camot cycle, showing steps (1) reversible isothermal expansion at th (2) reversible adiabatic expansion and cooling from th to tc (3) reversible isothermal compression at tc (4) reversible adiabatic compression and heating back to the original starting point. The total area of the Camot cycle, P dV, is the net useful work w performed in the cyclic process (see text). Figure 4.3 Reversible Camot cycle, showing steps (1) reversible isothermal expansion at th (2) reversible adiabatic expansion and cooling from th to tc (3) reversible isothermal compression at tc (4) reversible adiabatic compression and heating back to the original starting point. The total area of the Camot cycle, P dV, is the net useful work w performed in the cyclic process (see text).
The area enclosed by the solid line is the total or actual compression work. The area enclosed by the dotted line is ideal or useful compression work. The areas between the dotted line and the solid line represent compression work lost to heat. The area inside the dotted line, divided by the area inside the solid line, is called adiabatic compressor efficiency. [Pg.383]

The use of steam has a number of other advantages in the styrene process. The most important of these is that it acts as a source of internal heat supply so that the reactor can be operated adiabatic-ally. The dehydrogenation reaction is strongly endothermic, the heat of reaction at 560°C being (- AH) - 125,000 kJ/kmoL It is instructive to look closely at the conditions which were originally worked out for this process (Fig. 1.7). Most of the.steam, 90 per cent of the total used, is heated separately from the ethylbenzene stream, and to a higher temperature (7I0SC) than is required at the inlet to the... [Pg.13]

Note that diSC is independent of the total kinetic energy of the projectile provided the adiabatic criterion is fulfilled, namely that the magnetic field does not change appreciably over the axial distance travelled by the particle during one turn of its helical trajectory. A more detailed discussion of this point can be found in the work of Kruit and Read (1983). [Pg.62]

Consider any number of systems that may do work on each other and also transfer heat from one to another by reversible processes. The changes of state may be of any nature, and any type of work may be involved. This collection of systems is isolated from the surroundings by a rigid, adiabatic envelope. We assume first that the temperatures of all the systems between which heat is transferred are the same, because of the requirements for the reversible transfer of heat. For any infinitesimal change that takes place within the isolated system, the change in the value of the entropy function for the ith system is dQJT, where Qt is the heat absorbed by the ith system. The total entropy change is the sum of such quantities over all of the subsystems in the isolated system, so... [Pg.42]

We start by studying the steady-state design and economics of a process with a single adiabatic reactor. The design considers the entire plantwide process reactor, heat exchangers, gas recycle compressor, preheat furnace, condenser, and separator. The economic objective function is total annual cost, which includes annual capital cost (reactor, catalyst, compressor, and heat exchangers) and energy cost (compressor work and furnace fuel). [Pg.265]

It can be seen that the electrolytic cell must accept 11 679 cal. from the surroundings per each mole of decomposed water to keep a constant temperature (when working adiabatically the electrolytic cell would cool down). This heat is also covered by electric energy. Therefore, to achieve an isothermal decomposition of a mole of water a total amount of not only 56 693 cal. but 68 372 cal. in the form of electrical energy is necessary which corresponds to the minimum terminal voltage across the electrolytic cell ... [Pg.203]

However, according to statement 1 a of the second law, Q v cannot be directed into the system, for the cycle would then be a process for the complete conversion of heat into work. Thus, j dQnv is negative, and it follows that SA - SB is also negative whence SB > SA. Since the original irreversible process is adiabatic, the total entropy change as a result of this process is AStota, = SB - SA > 0. [Pg.88]

Suppose an insulated container, partitioned into two equal volumes, contains Avogadro s number N0 of molecules of an ideal gas in one section and no molecules in the other. When the partition is withdrawn, the molecules quickly distribute themselves uniformly throughout the total volume. The process is an adiabatic expansion that accomplishes no work. Therefore... [Pg.415]

The problems which arise are perhaps best illustrated by reference to Fig. 1.8.1, which depicts an adiabatically insulated enclosure S equipped with a movable piston P of mass Mp. The latter rests on release pins and r2 the container is also furnished with stops sx and s2 which ultimately arrest the downward motion of the piston under the action of the earth s gravitational field. Let the space in the enclosure be totally evacuated and let the release pins then be retracted the piston now is accelerated through a vertical distance h, until arrested by the lower stops, sx and s2. The volume of the enclosure is thereby reduced from to Vf. The work involved in moving P through the distance h is Mpgh (for the problem of sign, see Exercise 1.8.5). If the system is considered to be... [Pg.57]


See other pages where Adiabatic total work is mentioned: [Pg.475]    [Pg.58]    [Pg.138]    [Pg.10]    [Pg.266]    [Pg.100]    [Pg.191]    [Pg.10]    [Pg.258]    [Pg.176]    [Pg.357]    [Pg.2048]    [Pg.49]    [Pg.54]    [Pg.245]    [Pg.139]    [Pg.49]    [Pg.377]    [Pg.165]    [Pg.51]    [Pg.112]    [Pg.132]    [Pg.42]    [Pg.43]    [Pg.296]    [Pg.167]    [Pg.262]    [Pg.66]   
See also in sourсe #XX -- [ Pg.46 ]




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