Question

A system consisting of 1.0 kg of water and undergoes a thermodynamic power cycle composed of the following four processes: Process 1 – 2: Constant pressure heating at 1 MPa from saturated vapor Process 2 – 3: Constant volume cooling to pz = 500 kPa, Tz = 160°C Process 3 – 4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant volume heating (a) Sketch the cycle on T - v and p - v diagrams. (b) Is this a power or refrigeration cycle? Explain. (b) Determine the work and heat transfer for each process. (c) Determine the thermal efficiency of the cycle. (d) Is energy conserved during this cycle? Explain.

Answer 1

Solution:

Inputs given are :- 1) fluid – water . Molecular weight (n) = 18 g/kmol 4 Mass of water (m. 1 kg of Py diagram & TV diagram - TI Po live ... Qout. V=V4 V2: 3 v drawn in PV and TV diagrams Given processes are as shown above. saturation point. Note that Gnstant point 1 an TV digram is pressure lines are also shown 6 W = PAN will be positive as during this constant pressure process volume will increase due to heating which will cause work done by the system kl. = lahus =0 as these are constant volume procestes -Pa 434 = Pol en (P) will be negative as at constant temperature and constant internal energy, all the werkdane is on the system due to compression.

This positive work in por diagram is area under 1.2 while negative work is area under us. This net work which is area enclosed by cycle 1-2-3-4-1 is positive. Thus, giun cycle is a power cycle. producing As a thumb rule & every clockwise cycle is a power cap de whereas anticlockwise cycle is power consuming te refrigeration cycle.

& for 309 wcle 34 we can write P3 - MRT – & V3- MR Q 8.314 komol. K x 433 K = 1 kg x 18 Tegkomol - 500 sooo - Inspirat 439 - V3 = 0.4 m3 from pv diagram, (1= = 0.4 m3 process change in internal As for constant temperature energy is zero - 9= DE+W put law changes to T= WF – 815. 8 kJ Here re sign signifies heat lost & work done on the system. As workedone for isothermal uncle is Water Pole (1) = m RT3 da (3) Ww=57605 min. 0-4 m. encom - - 815.8 x en C -8080 -4.079 solering we will get P= 29 543 lepa 12 not possible Nm - 815.8 -4.079 200

Note that heat transfer (and hence workdone) value during 3-4 seems to be giving an out-of-range value of P4. from P-V diagram it is clear that P1>P4>P3. For the given thermodynamic cycle this much heat loss is not possible.

Rest of the calculations are very easy as it only reacquire 1st law equation i.e. $q = \Delta E + W$ and $W = \int p dv$

Kindly check the heat transfer value and revert back. Feel free to comment and I will be more than happy to answer your questions in order to enhance your understanding.