Heating, Ventillation, and Air Conditioning

( Visit the Closed-Process and Open-Steady pages first. )

States	Closed Process	Closed Steady	Open Steady
Open Process	Closed Cycle	Open Cycles	States-II
	Combustion	Equilibrium	Gas Dynamics

(Each section above is divided into two sub-sections - Manual and Applications.)
Manual

(This page builds upon the States-II, Closed Process and Open Steady pages.)

Daemons>Heating, Ventillation and AC> Applications

EXAMPLE-1

Calculate the mass of water vapor in a room of size 30m x 10m x 5m for the following conditions: (1) DBT=30^oC, RH=50%; (2) DBT=30^o C, RH=5%; (3) WBT=15^oC, DPT=-30^oC; (4) 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 psychrometricd plot.

What-if scenario: How would these answers change if the total pressure were doubled?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# HOME>Daemons>States>Volume>MoistAir

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;   }
    }

Step 1: Launch
the moist air State Daemon.

Solution:

To arrive at the appropriate daemon on the primary window click on Daemons, States, Volume and MoistAir in sequence after launching TEST on the primary browser. Or use the TEST-Map and jump to the daemon page directly.

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

Step 2: Select a State.

Step 3: Input known
variables and Calculate.

Step 5: Repeat for all other states.

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.

Step 6: Parametric study

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.

EXAMPLE-2

Consider 100 m³ of moist air at 100 kPa, 35^oC and 20% R.H. Determine the amount of water vapor condensed and the heat transfer if the air is cooled down to 5^oC in a constant pressure process.

What-if scenario: (a) How would the answers change if the air were saturated to start with?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# ..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: CondensedWater;
    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;   }
    }

Step 1: Launch the specific process daemon.

Step 2: Calculate the b and f States.

Step 3: Go to the Analysis Panel.

Step 4: Enter the process variable.

Step 5: Calculate.

Step 6: Change a parameter.

Step 7: Calculate and Super-Calculate to update all answers.

Solution

Classify the problem (as a closed process problem) and arrive at the appropriate daemon on the primary window by clicking on Daemons, Systems, Closed, Process, Specific, HVAC in sequence. Or use the TEST-Map to jump to the daemon page directly. It is assumed that you have already visited the generic Closed Process chapter in this tutorial.

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 begin-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=-2872 kJ and m_fB (=m3)=0.183 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.2.1Image of the Analysis Panel in Ex. 2. The entire begin state is occupied by the sub-system A.

EXAMPLE-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. Will there be condensation? If so, how much? Assume negligible change of total pressure.

What-if scenario: (a) How would the answers be affected if the R.H. in the first chamber were 10% instead?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# ..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: CondensedWater;
    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;   }
    }

Step 1: Launch the HVAC Process Daemon.

Step 2: The b-state and f-state are both non-uniform, requiring four states.

Step 3: State-1 and 2 correspond to the moist air in tanks A and B.

Step 4: State 3 and 4 correspond to the two working fluids moist air and condensed water.

Step 5: The only process variable that is known is the work transfer.

Step 6: After changing a variable use the Calculate button to register the change and then use the Super-Calculate button to update all calculations.

Solution

Classify the problem (as a closed process problem) and arrive at the appropriate daemon, Daemons, Systems, Closed, Process, Specific, HVAC, as in the previous problem.

Here, we have a non-uniform begin-State, identified by bA and bB (states 1 and 2) and possibly a non-uniform finish state, identified by fA and fB (states 3 and 4). While A and B refer to two different chambers for the b-states, they signify the two different working fluids, moist air and liquid water, in the f-states.

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 finish 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=0.0927 kg. 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.

EXAMPLE-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. Determine the relative humidity at the exit.

What-if scenario: How would the R.H. change if the electrical heating were intensified to 10 kW?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# ...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;   }
}

Step 1: Launch the Open, Steady HVAC Daemon.

Step 2: Calculate the inlet and exit States.

Step 3: Load these states on the Analysis panel, enter the device variables, and Calculate. To update all the unfinished state calculations Super-Calculate.

Step 4: Change a parameter and click the Calculate and Super-Calculate buttons to update all calculations.

Solution

Classify the system as Open, Steady, Specific, HVAC and navigate to the corresponding page on the primary window page by page or using the TEST-Map.

Although the daemon can handle upto two inlets and two exits, we will use State-1 for the i1-State and State-2 for the e1-State.

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. Note that the TEST
Code is generated as the first part of the detailed output.

EXAMPLE-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.

What-if scenario: How would the answers change if the pressure were 1000 kPa instead?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# ...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: CondensedWater;
                            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;   }
}

Step 1: Launch the Open, Steady HVAC Daemon.

Step 2: Calculate the inlet and exit States.

Step 3: Load these states on the Analysis panel, enter the device variables, and Calculate. To update all the unfinished state calculations Super-Calculate.

Step 4: Change a parameter and click the Calculate and Super-Calculate buttons to update all calculations.

Solution
Classify the system as Open, Steady, Specific, HVAC and navigate to the corresponding page on the primary window page by page or using the TEST-Map.

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 CondesedWater 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=11.44 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.

EXAMPLE-6

A wet cooling tower is 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 intput to the fan, determine (a) the volume flow rate of air, and (b) the mass flow rate of the make up water

What-if scenario: How would the answers change if the day was a humid one with a RH of 90%?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button followed by the Super-Calculate button.

Step 1: Launch the Open, Steady HVAC Daemon.

Step 2: Calculate the 4 states from the known properties.

Step 3: Setup the Device Panel for the cooling tower by clicking the CoolingTower button. Load the anchor states as shown in Fig. 6.1 below. Press the Enter key and Super-Calculate. The answers are calculated as Voldot1=24.3 m^3/s and 0.86 kg/s . (On the I/O Panel type in '=mdot3-mdot4' to calculate the flow rate of the make up water).

Step 4 In State-1, change RH1 to 90%, Calculate and Super-Calculate. The new answers are: Voldot1=29.5 m^3/s and 0.83 kg/s .

# ...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: CondensedWater;
    Given:       { T3= 40.0 deg-C;   Vel3= 0.0 m/s;   z3= 0.0 m;   mdot3= 40.0 kg/s;   }

    State-4: CondensedWater;
    Given:       { T4= 25.0 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;   }
    }

Fig. 6.1 Image of the Device Panel with the Cooling Tower button selected.

Your input!

For more solved examples on this topic, visit the Slide Show and the Home.TEST.Problems pages. If you detect an error or any inconsistent instructions on this page, or would like to see more examples on a particular topic, please write to me using the Home.Comments page. Your input will be appreciated.


Manual