ENGR 224 - Thermodynamics HW #2 Problem : 3.26 ... ... Baratuci HW #2 Problem : 3.26 - Steam NIST

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Transcript of ENGR 224 - Thermodynamics HW #2 Problem : 3.26 ... ... Baratuci HW #2 Problem : 3.26 - Steam NIST

  • Baratuci HW #2

    Problem : 3.26 - Steam NIST / TFT Fundamentals - 2 pts 14-Apr-11

    T (oC) P (kPa) H (kJ/kg) x (kg vap/kg) 200 0.7

    140 1800 950 0

    80 500 800 3162.2

    Read :

    It also tests your understanding and ability to use quality, x.

    Given : Two pieces of data for each part, (a) through (e).

    Find : Complete the table.

    Assumption: The system is in an equilibrium state.

    Solution :

    Part a.) Given : P 200 kPa x 0.7 kg vap/kg

    NIST

    Results : T 120.21 oC Hsat liq 504.70 kJ/kg Hsat vap 2706.2 kJ/kg

    Eqn 1

    H 2045.8 kJ/kg

    TFT In order to determine Tsat at 200 kPa, use the following cell formula :

    =TFProp("Water","SI_C","P",200000,"X",0.7,"T") Eqn 2

    T 120.23 oC In order to determine H at 200 kPa and x = 0.7 use the following cell formula :

    =TFProp("Water","SI_C","P",200000,"X",0.7,"H") Eqn 3 H 2045.7 kJ/kg

    Part b.) Given : T 140 oC H 1800 kJ/kg

    NIST

    The key results are : Sat. Liq Sat. Vap

    140 361.54 589.16 2733.4

    Since :

    P= Psat 361.54 kPa

    Eqn 4

    Plugging values into Eqn 4 yields : x 0.5647 kg vap/kg

    ENGR 224 - Thermodynamics

    Complete the following table for water using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default reference state for both the NIST and TFT.

    Because we are given a quality we know that vapor and liquid both exist in the system in equilibrium. Therefore, we use the Saturation properties — pressure increments option.

    Because we are given a temperature and we need to determine which phases are present, the first thing we need to do is generate a saturation temperature table using the NIST Webbook and the Saturation properties — temperature increments option.

    Phase Description

    This problem is designed to test how well you understand how to use the NIST Webbook tables of thermodynamic properties and/or the Thermal/Fluid Properties (TFP) plug-in for Excel.

    Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent.

    the system contains a saturated mixture and the pressure must be equal to the saturation pressure.

    Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:

    Temp. (oC)

    Pressure (kPa)

    H (kJ/kg)

     sat sat sat mix vap liq

    ˆ ˆ ˆH x H 1 x H  

    sat sat liq vap

    ˆ ˆ ˆH H H 

    sat liq

    sat vap sat liq

    ˆ ˆH H x ˆ ˆH H

     

    Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 4/12/2011

  • TFT

    Hsat liq =TFProp("Water","SI_C","T",140,"X",0,"H") Eqn 5 Hsat vap =TFProp("Water","SI_C","T",140,"X",1,"H") Eqn 6

    Hsat liq 588.72 kJ/kg Hsat vap 2733.40 kJ/kg

    Since :

    P =TFProp("Water","SI_C","T",140,"X",1,"P") Eqn 7

    P= Psat 361.29 kPa

    x =TFProp("Water","SI_C","T",140,"H",1800,"X") Eqn 7 x 0.5648 kg vap/kg

    Part c.) Given : P 950 kPa x 0 kg vap/kg

    NIST Because the quality is zero, we immediately know the system contains a saturated liquid. This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given). So, we can use 950 kPa in the Saturation properties — pressure increments option.

    Here is the relevant data : Sat. Liq Sat. Vap

    950 177.66 752.74 2775.1

    T = Tsat 177.66 oC H = Hsat liq 752.74 kJ/kg

    TFT Because the quality is zero, we immediately know the system contains a saturated liquid. This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given).

    x =TFProp("Water","SI_C","P",950000,"X",0,"T") Eqn 8 T 177.69 oC H =TFProp("Water","SI_C","P",950000,"X",0,"H") Eqn 9 H 752.68 kJ/kg

    Part d.) Given : T 80 oC P 500 kPa

    NIST

    Tsat(500 kPa) = 151.83 oC

    Here is the relevant data :

    80 500 335.37

    H 335.37 kJ/kg

    TFT Here we still need to determine Tsat associated with P.

    Tsat(500 kPa) =TFProp("Water","SI_C","P",500000,"X",0,"T") Eqn 10

    Tsat(500 kPa) = 151.86 oC

    H =TFProp("Water","SI_C","P",500000,"T",80,"H") Eqn 11 H 334.78 kJ/kg

    Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:

    H (kJ/kg)

    Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.

    the system contains a saturated mixture and the pressure must be equal to the saturation pressure.

    Pressure (kPa)

    Temp. (oC)

    H (kJ/kg)

    Here we are given both the T and P. Let's use the given P, and the Saturation properties- pressure increments option, to determine the Tsat associated with P. Then, we can determine the phases present by comparing the given T to Tsat(P).

    Since T = 80oC is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! So, we need to look at either the Isothermal properties option or the Isobaric properties option in the NIST Webbook. Either way, no interpolation is required because we know and enter botht he T and P values !

    Temp. (oC)

    Pressure (kPa)

    Since T = 80oC is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! But it is easy to use the TFT to determine H.

    sat sat liq vap

    ˆ ˆ ˆH H H 

    Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 4/12/2011

  • Part e.) Given : P 800 kPa H 3162.2 kJ/kg

    NIST

    Here is the relevant data : Sat. Liq Sat. Vap

    800 170.41 720.86 2768.3

    Since :

    This time we must enter a range of T values so that we bracket the given H value. I chose to go from 100oC to 500oC by 10oC steps. Here are the two rows in the resulting table that bracket the H value of 3161.7 kJ/kg.

    800 340 3141.1 800 ??? 3162.2 800 350 3162.2

    T 350.00 oC

    TFT

    Hsat liq =TFProp("Water","SI_C","P",800000,"X",0,"H") Eqn 12 Hsat vap =TFProp("Water","SI_C","P",800000,"X",1,"H")

    Hsat liq 720.68 kJ/kg Eqn 13 Hsat vap 2768.67 kJ/kg

    Since :

    T =TFProp("Water","SI_C","P",800000,"H",3161.7,"T") Eqn 14 T 350.23 oC Verify :

    Answers : NIST Webbook

    T (oC) P (kPa) H (kJ/kg) x (kg vap/kg) 120.21 200 2045.75 0.7

    140 361.54 1800 0.565 177.66 950 752.74 0

    80 500 335.37 N/A 350.2 800 3162.2 N/A

    TFT T (oC) P (kPa) H (kJ/kg) x (kg vap/kg) 120.23 200 2045.72 0.7

    140 361.29 1800 0.565 177.69 950 752.68 0

    80 500 334.78 N/A 350.2 800 3162.2 N/A

    Pressure (kPa)

    Temp. (oC)

    H (kJ/kg)

    the system contains a superheated vapor, quality is undefined and we must use the Isobaric properties option to determine T.

    The first step here is to use the given P to obtain data using the Saturation properties — pressure increments option. This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases present in the system at equilibrium.

    None of the assumptions made in the solution of this problem can be verified based on the given information.

    Phase Description Sat'd Mixture (VLE) Sat'd Mixture (VLE) Sat'd Liquid

    the system contains a superheated vapor, quality is undefined and we can directly calculate T using the following formula.

    Pressure (kPa)

    Temp. (oC)

    Now, we can determine H from this table by inspection !

    Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present.

    H (kJ/kg)

    Subcooled Liquid

    Sat'd Liquid Subcooled Liquid Superheated Vapor

    Superheated Vapor

    Phase Description Sat'd Mixture (VLE) Sat'd Mixture (VLE)

    sat vap

    ˆ ˆH H

    sat vap

    ˆ ˆH H

    Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 4/12/2011

  • Baratuci HW #2

    Problem : 3.29E - R-134a NIST/TFT Fundamentals - 2 pts 14-Apr-11

    T (oF)

    P (psia)

    H (Btu/lbm)

    x (lbm vap/lbm)

    80 78 15 0.6 10 70

    180 129.46 110 1.0

    Read :

    It also tests your understanding and ability to use quality, x.

    Given : Two pieces of data for each part, (a) through (e).

    Find : Complete the table.

    Assumptions: The system is in an equilibrium state.

    Solution :

    Part a.) P 80 psia H 78 Btu/lbm NIST :

    Here is the relevant data : Sat. Liq Sat. Vap

    80 65.922 97.162 176.02

    Since :

    This time we must enter a range of T values so that we bracket the given H value. I chose to go from 0oF to 100oF by 10oF steps. Here are the two rows in the resulting table that bracket the H value of 78 Btu/lbm .

    80 0 75.994 80 ??? 78 80 10 79.107

    T 6.44 oF

    TFT :

    Hsat liq =TFProp("R-134a","EE_F","P",80,"X",0,"H") Eqn 2 Hsat vap =TFProp("R-134a","EE_F","P",80,"X",1,"H") Eqn 3

    Hsat liq 97.11 Btu/lbm Hsat vap 175.90 Btu/lbm

    Since :

    T =TFProp("R-134a","EE_F","P",80,"H",78,"T") Eqn 4

    T 6.62 oF

    Part b.) T 15 oF x 0.60 lbm vap/lbm NIST :

    Now, we can determine T from this table by interpolation.

    The quality of the water in the system is not defined because it is a subcooled liquid.

    ENGR 224 - Thermodynamics

    Complete the following table for R-134a using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default reference state for both t