Psychrometric Software for Moist Air and Moist Gas Properties - Web-Based Analysis of HVAC systems

Ex 12-1 Calculate the mass of water vapor in a room of size 30m x 10m x 5m for the following conditions: (a) DBT=30^oC, RH=50%; (b) DBT=30^o C, RH=5%; (c) WBT=15^oC, DPT=-30^oC; (d) h=42.4 kJ/kg, omega=0.01081 kg/kg of dry air. Assume the total pressure to be 100 kPa. Draw the states on a psychrometric plot. (e) What-if scenario How would these answers change if the total pressure was doubled?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>States>System>MoistAir (MA) Model

States    {
    State-1: MoistAir;
    Given:       { p1= 100.0 kPa;   T1= 30.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 1500.0 m^3;   RH1= 50.0 %;   }

    State-2: MoistAir;
    Given:       { p2= "p1" kPa;   T2= 30.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Vol2= "Vol1" m^3;   RH2= 5.0 %;   }

    State-3: MoistAir;
    Given:       { p3= "p1" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   Vol3= "Vol1" m^3;   T_dp3= -30.0 deg-C;   T_wb3= 15.0 deg-C;   }

    State-4: MoistAir;
    Given:       { p4= "p1" kPa;   h4= 42.2 kJ/kg;   Vel4= 0.0 m/s;   z4= 0.0 m;   Vol4= "Vol1" m^3;   omega4= 0.01081 kg/kg;   }
    }

Solution Moisture in air is handled by the moist air model in TEST. Since the problem involves evauation of state properties, the system state daemon with the MA (moist air) model, located at Daemons. States. System. MoistAir page, is the right choice although the state panel of the closed-process daemon can also be used for this purpose.

Fig. 1.1 Image of the Daemons.States.Mixture.MoistAir page. Note that the dew point
temperature can be less than the freezing point of water.

Moist air is the default working fluid.

State-1: Choose State-1, enter p1=100 kPa, Vol1=50*10*3 m3, T1 (dry bulb temperature, 30^oC) and RH1 (50%), and click Calculate . The mass of dry air is calculated as m_v1=22.77 kg.

State-2: Choose State-2, enter p2 ('=p1'), Vol2 ('=Vol1'), T2 and RH2, and Calculate . The mass of dry air is calculated as m_v2=2.277 kg.

State-3 : Choose State-3, enter p3 ('=p1') , Vol3 ('=Vol1'), T_wb3 and T_dp3, and Calculate. The mass of dry air is calculated as m_v3=0.402 kg .

State-4 Choose State-4, enter p4 ('=p1'), Vol4 ('=Vol1'), h4 and omega4, and Calculate. The mass of dry air is calculated as m_v4=19.286 kg .

Choose Psychrometric Plot from the diagram selector. The plot appears on a pop-up window.

To recalculate the set of states, simply change p1 to 200 kPa, Calculate and Super-Calculate . The new values are 22.77 kg, 2.277 kg, 0.42 kg and 38.57 kg respectively.

Ex 12-2 Consider 100 m³ of moist air at 100 kPa, 35^oC and 20% R.H. Determine (a) the amount of water vapor condensed and (b) the heat transfer if the air is cooled down to 5^oC in a constant pressure process. (c) What-if scenario: How would the answers change if the air was saturated to start with?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Specific>HVAC

   States    {
    State-1: MoistAir;
    Given:       { p1= 100.0 kPa;   T1= 35.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 100.0 m^3;   RH1= 20.0 %;   }

    State-2: MoistAir;
    Given:       { p2= 100.0 kPa;   T2= 5.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   RH2= 100.0 %;   }

    State-3: H2O;
    Given:       { p3= 100.0 kPa;   T3= "T2" deg-C;   Vel3= 0.0 m/s;   z3= 0.0 m;   }
    }

Analysis    {
    Process-A: b-State = State-1; f-State = State-2, State-3;
    Given: { W= 0.0 kJ;   }
    }

Solution Realizing that the problem involves a closed process (closed system transitioning from b-state to f-state), launch the closed-process psychrometric (HVAC) daemon located at Daemons. Systems. Closed. Process. Specific. HVAC page. It is assumed that you have already worked with a few generic closed process daemons. In this case the system goes from a single b-state to a composite final state. We will use bA-state as the initial state (ignoring the bB-state), fA-state for the final state of saturated air, and fB-state for condensed water.

State-1: Choose State-1 as the bA-State. Enter p1, T1, Vol1, RH1, and Calculate . There is no bB-State in this problem as the beginning-state is made up of an uniform working fluid. Note that the final temperature is below the dew point temperature. Therefore, the final relative humidity is 100%.

State-2: Choose State-2, as the fA-State. Enter p2 ('=p1'), T2, Vol2('=Vol1') and RH2 (100%), and Calculate.

State-3 : Select Water as the working fluid. Choose State-3 , as the fB-State. Enter p3 ('=p1'), T3 ('=T2'), and Calculate.

On the Analysis panel, load State-1 as the bA-State , State-2 as the fA-State State-3 as the fB-State. Enter the only known process variable W (=0). A Calculate produces Q=-2864 kJ and m_fB (=m3)=0.18 kg.

Go back to State-1 and change RH1 to 100%. A Calculate Super-Calculate produce Q=-10467 kJ and Evidently it is much costlier to cool humid air than relatively dry air.

Fig. 2.1 Image of the process panel in Ex. 2. The entire beginning state is occupied by the sub-system A.

Ex 12-3 A 50 m³ chamber containing air at 5^oC, 100 kPa, R.H. 100%, is connected to another 50 m³ chamber containing air at 20^o C, 100 kPa, R.H. 100%. The valve is opened and the system is allowed to reach thermal equilibrium. (a) Will there be condensation? (b) If so, how much? Assume negligible change of total pressure. (c) What-if scenario: How would the answers be affected if the R.H. in the first chamber was 10% instead?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Specific>HVAC

States    {
    State-1: MoistAir;
    Given:       { p1= 100.0 kPa;   T1= 5.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 50.0 m^3;   RH1= 100.0 %;   }

    State-2: MoistAir;
    Given:       { p2= "p1" kPa;   T2= 20.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Vol2= 50.0 m^3;   RH2= 100.0 %;   }

    State-3: MoistAir;
    Given:       { p3= "p1" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   m3= "m1+m2" kg;   Vol3= "Vol1+Vol2" m^3;   RH3= 100.0 %;   }

    State-4: H2O;
    Given:       { p4= "p1" kPa;   T4= "T3" deg-C;   Vel4= 0.0 m/s;   z4= 0.0 m;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3, State-4;
    Given: { W= 0.0 kJ;   }
    }

Solution The only difference with the previous example is that the beginning-state is composite, consisting of bA-state for one chamber and bB-state for another. The final state, however, has two possibilities - a single state of moist air (if there is no condensation) or a composite state (fA and fB state) if there is any condensation. Launch the closed process psychrometric (HVAC) daemon.

State-1: Choose State-1 as the bA-State. Enter p1, T1, Vol1 and RH1 and Calculate .

State-2: Choose State-2 as the bB-State. Enter p2 ('=p1'), T2, Vol2 and RH2, and Calculate.

State-3: Choose State-3, as the fA-State. Enter p3 ('=p1'), m3('=m1+m2') and Vol3('=Vol1+Vol2'). Let us assume that the final state is saturated, i.e., RH2= 100% . If this turns out to be a wrong assumption, the solution will produce a negative amount of condensate. Notice that we also ignore any slight change in pressure that may result from this process (possible temperature change is not much and the volume occupied by liquid water, if any, is almost negligible). Calculate .

State-4: Select Water as the working fluid. Choose State-4 , as the fB-State. Enter p4 ('=p1'), T4 ('=T3'), and Calculate .

On the Analysis panel, load State-1 as the bA-State , load State-2 as the bB-State , State-3 as the fA-State and State-4 as the fB-State. Enter the only known process variable W (=0). (The tanks are not insulated). A Calculate and Super-Calculate produce m4=0.093 kg. In case m4 is found to be negative, that would signal that the air is not saturated after mixing and water must be added to make it so.

Note that the process is non-adiabatic and is accompanied by heat rejection. If the tanks were insulated, the final pressure would have changed. In that case the solution would involve trial and error (guess p3, Calculate and Super-Calculate) until Q is close to zero.

Go back to State-1 and change RH1 to 10%. A Calculate Super-Calculate produce m4=-0.15 kg. A negative answer indicates that there is no condensation in this case.

Ex 12-4 Moist air at 12^oC and 80% R.H. enters a duct at a rate of 150 m³/min. The mixture is electrically heated at a rate of 1 kW. The pressure remains constant at 100 kPa. (a) Determine the relative humidity at the exit. (b) What-if scenario: How would the R.H. change if the electrical heating was intensified to 10 kW?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>Specific>HVAC

States {
               State-1: MoistAir;
               Given:       { p1= 100.0 kPa;   T1= 12.0 deg-C;
                          Vel1= 0.0 m/s;   z1= 0.0 m;
                          Voldot1= 150.0 m^3/min;   RH1= 80.0 %;   }

State-2: MoistAir;
Given: { p2= 100.0 kPa; Vel2= 0.0 m/s; z2= 0.0 m; }
}

Analysis {
               Device-A: i-State = State-1; e-State = State-2;
                     CoolingTower: false;
               Given: { Qdot= 0.0 kW;   Wdot_ext= -1.0 kW;   }
}

Solution This is clearly an open device operating at steady state. Launch the open, steady HVAC daemon located at Open. Steady. Specific. page.

Although the daemon can handle up to two inlets and two exits, we will use State-1 for the i1-State and State-2 for the e1-State, ignoring the i2 and e2 states that are supplied to handle more complex problems.

State-1: Choose State-1, enter T1 and phi1 , and Voldot1. The default pressure is 100 kPa. Calculate. The mass flow rate of dry air is calculated as water vapor and dry air are calculated as: mdot1=181 kg/min.

State-2: Choose State-2. The pressure is the only known variable. Enter p2 as '=p1' (or leave it at its default value of 100 kPa). Calculate .

Go to the Analysis panel and load State-1 as the i1-State and State-2 as the e1-State. Generic Device is default. Enter the known device variables Qdot (=0) and Wdot_ext (-1 kW). Calculate and Super-Calculate . Go to the States panel to find the answer: RH2=78.2% .Note that (see Fig. below) a number of state variables calculated in the Analysis panel have been exported back to State-2. This is indicated by a different background color of these variable.

For the parametric study, simply change Wdot_ext to the new value (-10 kW). A Calculate and a Super-Calculate RH2=64.6% .

Fig. 4.1 Image of the I/O panel after a Super-Calculate in the second part of Ex. 4.

Ex 12-5 Moist air at 40^oC and 90% R.H. enters a dehumidifyer at the rate of 300 m³/min. The condensate and the saturated air exit at 10^oC through separate exits. The pressure remains constant at 100 kPa. Determine (a) the mass flow rate of dry air, (b) the water removal rate, and (c) the required refrigeration capacity, in tons. (d) What-if scenario: How would the answers change if the pressure was 1000 kPa instead?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>Specific>HVAC

States {
               State-1: MoistAir;
               Given:       { p1= 100.0 kPa;   T1= 40.0 deg-C;
                          Vel1= 0.0 m/s;   z1= 0.0 m;
                         Voldot1= 300.0 m^3/min;   RH1= 90.0 %;   }

               State-2: MoistAir;
                           Given:       { p2= "p1" kPa;   T2= 10.0 deg-C;
                           Vel2= 0.0 m/s;   z2= 0.0 m;   RH2= 100.0 %;   }

               State-3: H2O;
                            Given:       { p3= "p1" kPa;   T3= "T2" deg-C;
                          Vel3= 0.0 m/s;   z3= 0.0 m;   }
}

Analysis {
                 Device-A: i-State = State-1; e-State =
                          State-2, State-3;
                         CoolingTower: false;
               Given: { Wdot_ext= 0.0 kW;   }
}

Solution This is also an open steady device as in the previous example, except condensation must be handled through a separate exit, requiring the use of two exit states. Launch the open steady HVAC daemon.

We will use State-1 for the i1-State , State-2 for the e1-State and , State-3 for the e2-State.

State-2: Choose State-2. Enter p2 as '=p1' and T1. Calculate .

State-3: Choose State-3. Choose H2O as the working fluid. Enter p3 as '=p1' and T3 as '=T2'. Calculate.

Go to the Analysis panel and load State-1 as the i1-State, State-2 as the e1-State and State-2 as the e1-State. Enter the known device variable Wdot_ext (=0 kW). Calculate and Super-Calculate . Go to the States panel to find the answer: mdot3=11.39 kg/min.

For the parametric study, simply change p1 to the new value (1000 kPa). A Calculate and a Super-Calculate produce: mdot3=11.31 kg/min .

Fig. 5.1 Image of the device panel with the Generic Device button selected.

Ex 12-6 A wet cooling tower is used to cool 40 kg/s of cooling water from 40^o C to 25^oC at a location where the atmospheric air is at 90 kPa, 20 deg-C and 50% RH. Air, forced through the tower by a fan, cooling tower by a fan, leaves saturated at 35 deg-C. Neglecting the power input to the fan, determine (a) the volume flow rate of air, and (b) the mass flow rate of the make up water. (c) What-if scenario: How would the answers change if the day was a humid one with a RH of 90%?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>Specific>HVAC

States    {
    State-1: MoistAir;
    Given:       { p1= 90.0 kPa;   T1= 20.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   RH1= 50.0 %;   }

    State-2: MoistAir;
    Given:       { p2= "p1" kPa;   T2= 35.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   RH2= 100.0 %;   }

    State-3: H2O;
    Given:       { p3= 90.0 kPa;   T3= 40.0 deg-C;   Vel3= 0.0 m/s;   z3= 0.0 m;   mdot3= 40.0 kg/s;   }

    State-4: H2O;
    Given:       { p4= 90.0 kPa;   T4= 25 deg-C;   Vel4= 0.0 m/s;   z4= 0.0 m;   }
    }

Analysis    {
    Device-A: i-State = State-1, State-3; e-State = State-2, State-4; CoolingTower: true;
    Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   }
    }

Solution The wet cooling tower is a special open, steady HVAC device that cools a flow of water through evaporative cooling while the air flow is humidified. Launch the open steady HVAC daemon.

We will designate State-1 and 2 for the i1 and e1 states of air, and State-3 and 4 for the i2 and e2 states to handle the flow of water.

States: Evaluate the states as described in the TEST-codes above.

Go to the Analysis panel. There, you will find radio buttons to toggle between a cooling tower and a generic device. Select cooling tower. Lload State-1 as the i1-State, State-2 as the e1-State, State-2 as the e1-State, and State-4 as the e2 state. Enter the known device variable Qdot = Wdot_ext (=0 kW). Calculate and Super-Calculate. All the desired answers are found. The difference between m2 and m4 is the amount of evaporation and must be made up through the make up water.

For the parametric study, simply change RH1 to its new value. A Calculate and a Super-Calculate updates all the results.

Fig. 6.1 Image of the device panel with the Cooling Tower button selected.

Ex 12-7 A tank of volume 20 m³ contains a mixture of carbon-di-oxide and water vapor at 18^oC and 100 kPa at a relative humidity of 80 percent. Determine (a) the partial pressure of dry CO₂, (b) the specific humidity of the mixture, (c) the dew point, (d) the amount of vapor, and (e) the mass of CO₂. (f) What-if-scenario: How would the answer in (d) change if the mixture was composed of dry air and water vapor? (e) What-if scenario How would these answers change if the total pressure was doubled?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>States>System>MoistAir

States {
State-1: MoistCO2;
Given: { p1= 100.0 kPa; T1= 18.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; Vol1= 20.0 m^3; RH1= 80.0 %; }
}

Solution The preblem involves evaluation of a state involving moist carbon-di-oxide. The moist air state daemon, located at Daemons. States. System. MoistAir page, can handle not only moist air, but also moist gases, and pure phase-change fluids such as H2O, R-134a, etc.

Launch the daemon. MoistAir is the default working fluid. Select MoistCO2, enter the known properties, and press hte Calculate button. Note that m represents mass of dry gas (CO2 in this case).