What-if scenario: How would the answers change if the intial temperature were 0 ^o C? 

Step 2: Calculate the b and f States
 
 
 
 
 
 
 
 
 
 
 
 
 

Step 3: Load the States on the Process panel, enter process variables, Calculate and Super-Calculate.
 
 
 
 
 
 
 

Step 4: Change a parameter, Calculate and Super-Calculate.
 

Answering the six questions described in the Approach   section leads you to the appropriate daemon page:  HOME. Daemons. Systems. Closed. Process. Generic. Uniform.  IdealGas .

In this problem the uniform closed system, which can be represented by a single state at any given instant, executes a closed process as it goes from a clearly defined b-State to an unique f-State. We first evaluate the two states (State-1 and State-2 ) from the given data, and then embark on a process and exergy analysis.

State-1: Enter m1, T1, p1, and   Calculate.

State-2: Enter p2 ('=p1'), Vol2('=Vol1*4'), and   Calculate.

On the Analysis panel, load State-1 as the b-State and State-2 as the f-State. Enter the only known process variable Q (=0) and T_B to 25 ^o C.  A Calculate and Super-Calculate produce W_O (Other or electrical work) as -977.5 kJ  and  S_gen=1.5135 kJ/K   as well as all other process variables. Go to the States panel to find the final temperature T2 as  899^o C . 

For the exergy (availability) analysis, go back to the States panel, choose State-0 as the dead state, enter pressure (100 kPa) and temperature (25 ^o C), and Calculate . Go to the Exergy panel. Load State-0 as the dead state. A Calculate produce the reversible work W_rev as -526 kJ and irreversibility I as 451 kJ .

For the what-if study, go to States panel,  choose State-1, change the temperature, press the Enter key, and   Super-Calculate . All the answers are updated. Another way is to adjust the TEST-Code on the I/O panel, Load and Super-Calculate . 

What-if scenario:   (a) How would the answer change if the amount of heat transfer were 400 kJ?

Step 2: Calculate the States
 
 
 
 
 
 
 
 

Step 3: Calculate the Process
 
 

Step 4: Super-Calculate 
to obtain p2
 
 
 
 

Step 5: For the what-if study, change Q to new value, Calculate and Super-Calculate
 

Answering the six questions described in the Approach   section leads you to the appropriate daemon page:  HOME. Daemons. Systems. Closed. Process. Generic. Uniform. PhaseChange . 

In this problem the uniform closed system, which can be represented by a single state at any given instant, executes a closed process as it goes from a clearly defined b-State to an unique f-State. We calculate the two states ( State-1 and State-2 ) from the given data, do a process analysis, and Super-Calculate   the pressure in the tank at the end of the process. 

State-1: Enter Vol1, p1, x1 and   Calculate.

State-2: Enter Vol2 ('=Vol1'),  and   Calculate.

On the Analysis panel, load State-1 as the b-State and State-2 as the f-State. Enter the known process variables W_O(=0) and Q (=200 kJ).  Calculate and then Super-Calculate to produce the final pressure p2 as 135 kPa.

To evaluate the effect of heat transfer on the final pressure in the tank, change Q to the new value, Calculate and Super-Calculate  to produce the final pressure p2 as 193 kPa . 

What-if scenario:   (a) How would the answer change if the gas were pure N2?

Step 2: Calculate State-1 and State-2
 
 

Step 3: On the Analysis panel, enter the known process variables. A Calculate and Super-Calculate produce everything you want to know about the process and its anchor states
 
 

Step 4: Change a parameter, Calculate and Super-Calculate the new answer
 
 
 
 
 

Answering the six questions described in the Approach   section leads you to the appropriate daemon page:  HOME. Daemons. Systems. Closed. Process. Generic. Uniform.  IdealGasBinaryMixture . 

We have a uniform mixture, identifiable by a single State at any instant, undergoing a closed process. Let us represent the b-State and the f-State with State-1 and State-2 respectively. 

State-1: Choose N2 and H2 as Gas-A and Gas-B. Enter y_A1=0.5 (volume fraction is same as mole fraction), m1 (=1 kg), p1 and T1.  Calculate.

State-2: Calculate (with a calculator) Vol2 from the polytropic equation p1*Vol1^1.2=p2*Vol2^1.2 as 0.439 m³ . Enter Vol2, o r type in the expression '=Vol1*(p1/p2)^(1/1.2)' . Also enter m2 ('=m1'),  p2 and y_A2 ('=y_A1').  Calculate.

On the process panel, load State-1 and  State-2 as the b- and f-States.  Notice how W_B is automatically calculated to produce  Q = -129 kJ.

To obtain a solution for a different mixture, change y_A1 to the new value, 100% in this case, Calculate and Super-Calculate produce Q =-69 kJ and  and W_B= -138 kJ. Notice how severely the energy transfers change with the mixture composition. 

What-if scenario: If the large chamber were 100 times larger, how would that affect the result?

Solution

Answering the six questions, described in the Approach   section, leads you to the appropriate daemon page:  TEST. Daemons. Systems. Closed. Process. Generic. NonUniformMixing. PhaseChange . H2O is the default fluid. 
 

Step 2: Calculate the composite begin States and the single f-State.
 
 
 
 
 

Step 3: Calculate and Super-Calculate the process.
 

State-1: Enter  m1=4 kg, p1=200 kPa, T1=90 deg^oC and Calculate.

State-2: Enter m2 (=0), p1=0 and Vol2 ('=10*Vol1').   Calculate.  Variables e, j and s are set to zero as the mass is  zero.

State-3: Enter Vol3 as '=Vol1+Vol2', and  Calculate. At this point the State cannot be fully evaluated.

On the Analysis panel, load State-1 as the bA-State , State-2 as the bB-State and State-3 as the f-State. Enter the known process variables W (=0) and Q (=0).  Calculate to produce  T3=87.94 ^oC , p3=64.77 kPa , and S_gen=0.007223 kJ/K (Look for the exact value on the Message Panel as you move the pointer over S_gen widget) . Therefore, I=298*0.007223 =2.15 kJ.

In the States panel, change the Vol2 to '=100*Vol1'.A Calculate and Super-Calculate yield the entropy generation as 0.05897 kJ/K. Therefore, I=298*0.05897 =17.57 kJ. Notice the ten fold increase in the irreversibility.

Also determine the heat transfer necessary to bring the mixture back to the ambient temperature of 20 ^oC. 

What-if scenario:   How would the answers change if both the gases were CO ₂?

Step 2: Calculate State-1 and State-2
 
 

Step 3: On the Analysis panel, enter the known process variables. A Calculate and Super-Calculate produce everything you want to know about the process and its anchor states.
 
 
 
 
 

Step 4: Change a parameter, Calculate and Super-Calculate the new answer.
 
 
 
 
 
 
 
 
 
 
 
 
 

Step 5: User of Super-Iterate button is necessary in rare occasions. Why is Process-B almost reversible?
 
 
 
 
 
 

Step 6: More food for thought

Answering the six questions described in the Approach   section leads you to the appropriate daemon page:  HOME. Daemons. Systems. Process. Generic. NonUniformMixing. PerfectGases . 

In this problem the non-uniform closed system, which can be represented by two states ( State-1 and State-2 ) at the beginning and a single state ( State-3 ) at the end of the process. We calculate the three states from the given data, do a process analysis, and Super-Calculate   the answers. 

State-1: Choose H2 and CO2 as Gas-A and Gas-B, enter x_A1=1,  Vol1=10 m3, p1=200 kPa, T1=20 deg^oC and Calculate. The mass is calculated as 1.64 kg.

State-2: Enter x_A2=0,  Vol2=0.2 m3, p2=600 kPa, T2=100 deg^o C and Calculate. The mass is calculated as 1.7 kg.

State-3: Enter x_A3 as '=m1/(m1+m2)' and Vol3 as '=Vol1+Vol2', and  Calculate. At this point the State cannot be fully evaluated. Notice that the mixture is heavily biased towards hydrogen (look at the molecular mass, for instance).

On the Analysis panel, load State-1 as the bA-State and State-2 as the bB-State and State-3 as the f-State.  Enter the known process variable W (=0).  A Calculate and Super-Calculate produce  T3=25 ^o C , p3=209 kPa , and S_gen=1.423 kJ/K . Entropy is generated in a mixing process.

State-4: Enter x_A4 as '=x_A3',  Vol4 as '=Vol3', T4= 20 deg^o C and Calculate. 

On the Analysis panel, choose Process-B . Load State-3 as the bA-State , State-4 as the bA-State and State-3 as the f-State. Enter the known process variable W (=0) and Q (=0).  A Calculate and Super-Calculate produce  Q=-89.5 kJ. Notice that S_gen is not determined for Process-B . It is because, this is a rare instance where a Super-Calculate does not finish up the job. More iterations can be performed through Super-Iterate to complete the solution. 

Suppose  Process-A did not take place at all, and the tanks were allowed to cool  without ever having the valve opened. Would Q be the same in that situation. In this new process, Process-C , State-2 will cool down to State-5 , a new State with Vol5=Vol2, x_A5=x_A2, m5=m2. The first tank, being at the same temperature as the ambient, will not participate in any heat transfer with the surroundings. A process between State-2 and State-5 with W=0 produces Q=-89.5 kJ , identical to the answer in Process-B. 

What-if scenario: How would the answers change if the block had a mass of 2 lbm, instead?

Step 2: Calculate the composite begin States and the finish States.
 
 
 
 
 
 
 
 
 
 
 
 

Step 3: A process analysis produces Q
 
 
 

Step 4: A trial-and-error procedure to obtain the 
desired final temperature 
 

Answering the six questions, described in the Approach   section, leads you to the appropriate daemon page:  HOME. Daemons. Systems. Closed. Process. Generic. NonUniformUnmixed. Solid/Liquid . In this problem the non-uniform closed system, the copper block (part-A) in water (part-B), involves four states with the copper block going from a begin-state ( State-1 or bA-State ) to a finish-state ( State-2 or fA-State ) and the water going from a different begin-state ( State-3 or bB-State ) to a different finish-state ( State-4 or fB-State ). We calculate the two begin states from the given data. We guess a final temperature, do a process analysis to find Q , and improve our guess of the final temperature until a zero Q is achieved. 

Select English system.

State-1: Choose copper as the working substance from the left  menu (Model-1). Enter  m1 (1 lbm) and T1 (300^oF), and Calculate.

State-2: Choose water as the working substance from the right menu (Model-2). Enter m2 (5 lbm) and  T2 (70 ^oF), and Calculate.
 
State-3:   Choose copper as the working substance from the left  menu (Model-1). Enter  m3 (=m1) and a guessed value for T3, say, 100 deg-F.  Calculate.  

State-4:  Choose water as the working substance from the right menu (Model-2). Enter T4 (=T3), and Calculate.

On the Analysis panel, load State-1 as the bA-State , State-2 State-3 as the fA-State, and State-4 as the fB-State.Enter the known process variable W (=0).  A Calculate   produces Q=131 Btu , which must be added to the tank to obtain the final temperature of 100 ^oF.  Go back to State-2 and change T2 to a lower value, press the Enter key and Super-Calculate. Repeat until Q is close to zero. T3 is approximately 74.16 ^oF . 

For the what-if study, change m1 to 2 lbm at State-1. A Calculate registers the changes made in State-1.  A Super-Calculate yields  Q=-20.9 Btu . The trial-and-error procedure outlined above leads to a finial temperature of 78^oF .

Note that for solids or liquids temperature alone determines properties u and s, and, therefore, information on pressure is not necessary to solve this problem.