EES has built in property tables that follow the thermodynamics rule that for saturated vapor the value of x is 1 and for saturated liquid the value is 0. 2 You should notice there is an additional property given to us in the problem statement, the saturated vapor or the vapor quality of the fluid at state one, which can be represented as x1. Step 1: Enter the problem information as shown in Fig. EES Solution VV 2 = 1.2 kkkk ss mm kkkk = ss mm 3 Lastly, we solve for the ideal work done by the system where WW Д±Д±Д±Д±Д±Д±Д±Д±Д±Д± = WW aaaaaaaaaaaa О·О· cc = kkkk 0.7 = kkww Now, let s solve using EES to see how the program can help speed up the process and also help solve for multiple variables. We know an equation for the mass flow rate: Solving for VV yields, mm = ПЃПЃПЃПЃПЃПЃ = ПЃПЃvv VV 2 = mm ПЃПЃ 2 From the same superheated table we used to find h 2 we can also find that vv 2 = mm3 kkkk which means that ПЃПЃ 2 = 1 vv 2 = Hence, mm3 kkkk = kkkk mm 3. Hence, our energy equation can be now written as, WW ss = 1.2 kkkk ss kkkk kkkk kkkk = kkkk kkkkДў Now, we need to find the volumetric flow rate at the exit and what the ideal power output is for the efficiency given. WW ss = mm(h 2 h 1 ) From our thermodynamics property tables for R-134a as a saturated vapor at -24 C, the enthalpy can be found as h 1 = kkkk, and for state two where the pressure is 0.8 kkkk MPa and the temperature is 100 C, the enthalpy can be found as h 2 = kkkk found kkkk from the superheated tables because TT 2 > TT ssssss for 0.8 MPa. Where we can assume steady flow, the change in potential and kinetic energy is equal to zero, steady state and adiabatic, which will set pp 1 ee pp 2 ee = 0, kk 1 ee kk 2 ee = 0, dddd dddd cccc = 0, and QQ = 0, giving us a our simplified energy equation that we can use for this problem. Energy Equation for the compressor shown in Figure 1: mm 1(h 1 + pp 1 ee + kk 1 ee) + WW ss = QQ + mm 2(h 2 + pp 2 ee + kk 2 ee) + dddd dddd cccc Fig. Solution: First let s solve the problem by hand so we can compare to the EES results. No more than three sig figs for results computed for EES. Then, set up a parametric table that re-solves for both the power input and volumetric outflow rate for outlet temperatures: 180, 160, 100, and 80 C. Find: The power input of R-134 by the compressor, the volumetric flow rate at the exit and how much power must be provided by an electric motor if the compressor s efficiency is 70%. The compressor outlet state is at 0.8 MPa and 100 C. Problem Solution Given: A compressor takes in 1.2 kg/s of R-134 that is in a saturated vapor state at -24 C. The solution to the problem is shown below to help the reader better understand the problem before it is solved in EES.

Paul Dellenback in the fall semester of 2014) to better understand how the program EES can be used to help solve problems.

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1 Engineering Equation Solver (EES) Tutorial In this tutorial, we will use a thermodynamics problem (courtesy of ES2310 taught by Dr.

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