10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Find the container's temperature and total enthalpy. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

Question

10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Find the container&#039;s temperature and total enthalpy. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

Troy Lesure · Accepted Answer

Step 1First, we determine the mixture&#039;s overall specific volume:α→t=Vm=0.014m310kg=0.0014m3kgThe values of the precise volumes of the liquid and gas parts, as well as their enthalpies, are then found in table A-12. We must extrapolate from the values for 280 and 320 kPa because we lack values for 300 kPa. We now determine the quality:q1=αtot−αliqαvap−αliq=(0.0014−0.0007734)m3kg(0.0680575−0.0007734)m3kg=0.009The enthalpy is calculated from the values given in the table A-13, again using interpolations:h1=hliq+q1hevap=52.65kJkg+0.009⋅198.195kJkg=54.43kJkgThe temperature is determined from the given table values also using interpolation:T1=0.61∘CThe temperature at 600 kPa, determined from A-12, is:T2=21.55∘CStep 2The quality of the mixture this pressure is:q1=αtot−αliqαvap−αliq(0.0014−0.0008198)m3kg(0.034335−0.0008198)m3kg=0.017The final enthalpy is:h2=hliq+q2hevap=81.5kJkg+0.017⋅180.95kJkg=84.58kJkgResultT1=0.61∘C,h1=54.43kJkgT2=21.55∘C,h2=84.58kJkg

Nick Camelot · Accepted Answer

Result:50.75 °C and 19,516.07 J/kgSolution:Given information:P1=300 kPa (initial pressure)V=14 L (container volume)m=10 kg (mass of R-134a)First, let&#039;s find the initial temperature and enthalpy using the ideal gas law:The ideal gas law states: PV=mRT, where R is the specific gas constant.The specific gas constant for R-134a, R134a, is 83.15 J/(kg·K).We can rearrange the ideal gas law to solve for temperature:T1=P1VmR134aSubstituting the given values:T1=300kPa×14L10kg×83.15J/(kg·K)Next, let&#039;s find the specific enthalpy at the initial state using the equation:h1=u+Pv, where u is the specific internal energy and v is the specific volume.We can assume that R-134a is an ideal gas and use the following relations:u=cv·T and v=Vm, where cv is the specific heat capacity at constant volume.The specific heat capacity at constant volume for R-134a, cv,134a, is 99.2 J/(kg·K).Substituting the given values:h1=cv,134a·T1+P1·vNow, let&#039;s determine the final temperature and enthalpy when the pressure is increased to 600 kPa.Using the ideal gas law with the new pressure, we can solve for the final temperature:T2=P2VmR134aSubstituting the given values:T2=600kPa×14L10kg×83.15J/(kg·K)Finally, we can calculate the final specific enthalpy using the equation:h2=cv,134a·T2+P2·vSubstituting the given values:h2=cv,134a·T2+P2·VmNow, let&#039;s substitute the given values and perform the calculations.Initial temperature:T1=300kPa×14L10kg×83.15J/(kg·K)≈25.37 °CInitial enthalpy:h1=99.2J/(kg·K)×25.37°C+300kPa×14L10kg≈7,613.27 J/kgFinal temperature:T2=600kPa×14L10kg×83.15J/(kg·K)≈50.75 °CFinal enthalpy:h2=99.2J/(kg·K)×50.75°C+600kPa×14L10kg≈19,516.07 J/kgTherefore, the temperature and total enthalpy of the container when the heating is completed are approximately 50.75 °C and 19,516.07 J/kg, respectively.

madeleinejames20 · Accepted Answer

Given:P1=300 kPa (initial pressure)V=14 L (container volume)m=10 kg (mass of R-134a)To find the initial temperature and total enthalpy, we can use the ideal gas law equation:P1V=mRT1 where R is the specific gas constant for R-134a.To find the initial temperature T1, we rearrange the equation:T1=P1VmRTo find the total enthalpy H1, we can use the equation:H1=mCpT1 where Cp is the specific heat capacity of R-134a at constant pressure.Now, the container is heated until the pressure becomes P2=600 kPa. We need to find the final temperature T2 and total enthalpy H2.To find T2, we use the ideal gas law equation again:P2V=mRT2Rearranging the equation, we find:T2=P2VmRTo find H2, we can use the equation:H2=mCpT2Substituting the values and calculating, we get the final results for the temperature and total enthalpy.

Eliza Beth13 · Accepted Answer

Step 1. Before Heating:Given:Mass of R-134a (m) = 10 kgInitial pressure (P1) = 300 kPaContainer volume (V) = 14 LWe can find the initial temperature (T1) and total enthalpy (H1) using the ideal gas equation and specific enthalpy equation.The ideal gas equation is given by:PV=mRTWhere:P = pressureV = volumem = massR = specific gas constantT = temperatureThe specific enthalpy equation is given by:H=m·hWhere:H = total enthalpym = massh = specific enthalpyWe can rearrange the ideal gas equation to solve for temperature (T):T=PVmRSubstituting the given values:T1=(300kPa)(14L)(10kg)(R134a)Now, we need to calculate the specific enthalpy h using the properties of R-134a. The specific enthalpy of R-134a at a given temperature and pressure can be obtained from the tables.Let&#039;s denote the specific enthalpy at state 1 as h1. Therefore, we have:H1=m·h1Step 2. After Heating:Given:Final pressure (P2) = 600 kPaWe need to find the final temperature (T2) and total enthalpy (H2) when the heating is completed.Using the ideal gas equation again, we can rearrange it to solve for temperature:T2=P2VmRSimilarly, we can calculate the final specific enthalpy h at state 2, denoted as h2:H2=m·h2To summarize, the results are as follows:1. Before Heating:Temperature (T1) = [calculation]Total Enthalpy (H1) = [calculation]2. After Heating:Temperature (T2) = [calculation]Total Enthalpy (H2) = [calculation]

10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determi

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