![]() Solution 1 Here, the system is defined for us – no details are given as to its nature. Determine the specific work transfer during the compression stroke. Example 2 In an air compressor, the compression takes place at constant internal energy and 100 kJ of energy are rejected to the cooling water per kg of air. At a third point there is a further work transfer. At two points in the cycle, work is transferred to the surroundings of 96 kJ and 20 kJ. Example 1 During a complete cycle, a system is subjected to the following heat transfers: 800 kJ from the surroundings and 500 kJ to the surroundings. ![]() Where U2 is the internal energy in state 2, after the process and U1 is the internal energy in state 1, before the process. So, the non-flow energy equation (NFEE) becomes, simply, Q + W = ∆U or There will, however, be a change in the internal energy, ∆U. Similarly, there is no significant change in potential energy, so ∆Ep = 0. The gas inside the cylinder is the control mass.įrom the First Law, Q + W = ∆E Q + W = ∆Ek + ∆Ep + ∆Uīut for a non-flowing gas, its velocity will be 0 to start with, and when it has settled down after the process, its velocity will again be 0, so the change in kinetic energy, ∆Ek = 0. A typical closed system is a gas enclosed in a cylinder by means of a piston. ![]() I hope these ideas will become clearer when we consider some examples. If we are considering a flow situation, then a control volume through which the fluid flows is more useful. When dealing with a non-flow situation, then the system will be of fixed mass - no matter crosses the boundary - so it is useful to define a control mass. To do this we construct an imaginary boundary around what we are interested in – for example, the cricket ball (struck by Nasser Hussein) or the water in the kettle). It is necessary, therefore, that before we start any analysis we define the system that we are looking at. Non-Flow Energy Equation (NFEE) You may have noticed that the term “system” keeps cropping up.
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