4A-1 : Work for a Cycle Carried Out in a Closed System 6 pts

Air undergoes a three-process cycle.  Find the net work done for 2 kg of air if the processes are :Process 1-2: constant pressure expansionProcess 2-3: constant volume    Process 3-1: constant temperature compression                     Sketch the cycle path on a PV Diagram.               Data : T1 100oC          T2 600oC          P1 200kPa                       

Given: T1100oC Find:

T2 600oC Sketch cycle on a PV Diagram.

P1 200kPa Wcycle = ??? kJm 2.0kg

Assumptions: - The gas is a closed system- Boundary work is the only form of work interaction- Changes in kinetic and potential energies are negligible.- Air behaves as an ideal gas.  This must be verified at all three states.

Part a.)

Part b.) Since Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way.

Step 1-2 is isobaric, therefore, the definition of boundary work becomes:

Eqn 1

We can simplify Eqn 1 using the fact that P2 = P1 and the Ideal Gas EOS :

Eqn 2

Eqn 3

We can determine the number of moles of air in the system from the given mass of air and its molecular weight.

Eqn 4

MWair 29g/mole n 68.97mole

Plug values into Eqn 3 : R 8.314J/mole-KW12 286.69 kJ

Because the volume is constant in step 2-3: W23 0kJ

Step 3-1 is isothermal, therefore, the definition of boundary work becomes:

Eqn 5

The problem is that we don't know either P3 or V3.  Either one would be useful in evaluating W31 because we know P1 and we can determine V1 from the Ideal Gas EOS, Eqn 2.

We can evaluate V3 using the fact that V3 = V2.  Apply the the Ideal Gas EOS to state 2.

Eqn 6

V3 2.503m3/mole

Next, we can apply Eqn 6 to state 1 : V1 1.070m3/mole

Now, we can plug values into Eqn 4 to evaluate W13 :

W31 -181.89kJ

Sum the work terms for the three steps to get Wcycle:

Wcycle = 104.8 kJ

4A-2 : Quasi-Equilibrium Expansion of a Gas 4 pts

Warm air is contained in a piston-and-cylinder device oriented horizontally, as shown below. The air cools slowly from an initial volume of 0.003 m3 to a final volume of 0.002 m3. During the process, the spring exerts a force that varies linearly with position from an initial value of 900 N to a final value of zero Newtons. The atmospheric pressure is 100 kPa and the area of the piston face is 0.018 m2. Friction between the piston and the cylinder wall can be neglected. For the air, determine the initial and final pressures, in kPa, and work, in kJ.

         

Read : The key to solving this problem is to determine the slope and intercept for the linear relationship between the force exerted by the spring on the piston and the volume that the gas occupies.  This relationship is linear because, for a cylinder of uniform diameter, gas volume varies linearly with respect to the position of the piston.

Given: V1 = 0.003 m3 Patm 100 kPaV2 = 0.002 m3 Apiston 0.018 m2

F1 = 900 NF2 = 0 N

Find: P1 = ??? kPa W = ??? kJP2 = ??? kPa

Assumptions: - The air in the cylinder is a closed system.

- The process occurs slowly enough that it is a quasi-equilibrium process.- There is no friction between the piston and the cylinder wall.- The spring force varies linearly with position.

Solution :

In the initial and final states, the piston is not accelerating.  In fact, it is not moving.  Therefore, there is no unbalanced force acting on it.  This means that the vector sum of all the forces acting on the piston must be zero.

 

Initial State: Eqn 1 P1 = 150 kPa

Final State: Eqn 2 P1 = 100 kPa

For a quasi-equilibrium process, boundary or PV-work is defined by:

Eqn 3

Because Fspring varies linearly with the position of the piston AND volume also varies linearly with the position of the piston, we can conclude that Fspring must vary linearly with respect to the volume ! 

Eqn 4

m = 900000 N/m3 b = -1800 N

Eqn 5

Answer: W = -125 J

4A-3 : Quasi-Equilibrium Compression of Ammonia 4

Ammonia vapor is compressed inside a cylinder by an external force acting on the piston. The ammonia is initially at 30°C and 500 kPa. The final pressure is 1400 kPa.

   

a.) Determine the work done by the ammonia in the process.

b.) What is the final temperature of the ammonia ?

                 

The following data have been measured for the process.    

                  

  P (kPa) 500 653 802 945 1100 1248 1400  

  V (L) 1.25 1.08 0.96 0.84 0.72 0.60 0.50  

Read : The key concept here is that boundary or PV work is represented by the area under the process path curve on a PV Diagram.  So, once we plot the given data on a PV Diagram, all we need to do is numerically integrate to determine the area under the curve and we will have know the work !We can use the Ammonia Tables on the NIST WebBook to determine the final temperature because we know both Pfinal and Vfinal.  But we don't know the number of moles in the system.  Fortunately, we can use the initial state P1, V1, and T1 to determine the number of moles in this closed system.

 

Given: P (kPa) V (L)

500 1.25

653 1.08

802 0.96

945 0.84

1100 0.72

1248 0.60

1400 0.50

T1 = 30 oC

Find: T2 = ??? oC

Assumptions: - Each state in the data table is an equilibrium state.

- The process is a quasi-equilibrium process.

- The system is a closed system.

- The trapezoidal rule gives an acceptable estimate of the area under the process path in the PV Diagram

Part a.) The area of each trapezoid under the process path in the PV Diagram is the product of the average pressure for that trapezoid and the change in volume across the trapezoid.

Trapezoid

Pavg

(kPa)

ΔV (L) W (J)

A 1324 0.10 132

B 1174 0.12 141

C 1023 0.12 123

D 874 0.12 105

E 728 0.12 87

F 577 0.17 98 Wtotal = 686 J

Part b.) If we knew the specific volume of the ammonia in the final state, we could use the ammonia tables to determine the temperature.  But, at this point all we know is the total volume in the final state.  Therefore, we need to determine the mass of ammonia in  the system is the same in the initial and final state.  So, we can use the information we have for the initial state to determine the mass of ammonia in the system.

First, look up the specific volume of ammonia in the initial state in the isothermal thermodynamic tables: V1 = 0.28099

m3

/kg

 

Then, to calculate mNH3, use:

Eqn 1 mNH3 = 0.00445k

g

Finally, calculate V2,hat using:

Eqn 2 V2,hat = 0.112396

m3

/kg

Now, use the isobaric thermodynamic tables at a pressure of 1400 kPa.  Specify a temperature range that you are sure brackets V2.  Here is a table that I cut-and-pasted from the NIST WebBook.

T(C)

P(MPa)

V(

m3/kg)Phase  

70 1.4 0.10881 vapor  

72 1.4 0.1097 vapor  

74 1.4 0.11059 vapor  

76 1.4 0.11148 vapor  

78 1.4 0.11236 vapor   V2 falls between 78oC and

80oC80 1.4 0.113

23 vapor   So, now, we must interpolate.

Answer: T2 = 78.1 oC

4A-4 : Expansion of a Gas in a Cylinder Against a Spring 5

An unstretched spring is attached to a horizontal, frictionless piston.  Energy is added to the gas inside the cylinder until the presure in the cylinder is 400 kPa.  Determine the work done by the gas on the piston.  Use Patm = 75 kPa.

                   

Read : The key to solving this problem is to determine the slope and intercept for the linear relationship between the force exerted by the spring on the piston and the pressure within the gas.  This relationship is linear because the pressure within the cylinder is atmospheric pressure plus the spring force divided by the cross-sectional area of the piston.

Given: P2 400 kPa Dpiston

0.050 m

Patm 75 kPa k 1 kN/m

Find: W = ??? kJ

Assumptions: - The air in the cylinder is a closed system.

- The process occurs slowly enough that it is a quasi-equilibrium process.

- There is no friction between the piston and the cylinder wall.- The spring force varies linearly with position.

Solution:For a quasi-equilibrium process,boundary or PV-work is defined by:

Eqn 1

It is critical to note that the gas must overcome the force due to atmospheric pressure AND the force of the spring during this expansion process.  Because the force exerted by the linear spring on the piston is linear, we can write the following equation relating the force exerted by the gas on the piston to the displacement of the piston from its original, unstretched position.

Eqn 2 Where x is the displacement of the piston from its initial position.

Plug Eqn 2 into Eqn 1and integrate to get :

Eqn 3

Where x2 is the displacement of the spring in the final state.So, our next objective is to determine how far the piston moved during this process.

In the initial and final states, the piston is not accelerating.  In fact, it is not moving.  Therefore, there is no unbalanced force acting on it.  This means that the vector sum of all the forces acting on the piston must be zero.

Initial State: Eqn 4

P1 75 kP

a

The relationship between force and pressure is:

  Eqn 5

Where : Eqn 6 Apiston

1.96E-03

m2

Fatm 0.1473 kN

Final State:

Eqn 7 or :

Eqn 8

Now, plug numbers into Eqn 8 : F2 0.6381k

N

Because the spring is linear :

Eqn 9 or :

Eqn 10

x2 0.6381M

Finally, substitute back into Eqn 3 to evaluate the work done by the gas in the cylinder on its surroundings during this process :

Answer: W = 0.29758kJ W = 0.298 kJ

4A-5 : Quasi-Equilibrium Expansion of a Gas4 pts

Estimate the work done by a gas during an unknown process.  Data obtained that relates pressure and volume are :

                   

  P (kPa) 200 250 300 350 400 450 500  

  V (cm3) 800 650 550 475 415 365 360  

                   

Read : The key concept here is that boundary or PV work is represented by the area under the process path curve on a PV Diagram.  So, once we plot the given data on a PV Diagram, all we need to do is numerically integrate to determine the area under the curve and we will have the work !

 

Given: P (kPa) V (L)

200 0.800

250 0.650

300 0.550

350 0.475

400 0.415

450 0.365

500 0.360

Find: W ??? J

Assumptions: - Each state in the data table is an equilibrium state.- The process is a quasi-equilibrium process.- The system is a closed system.

- The trapezoidal rule gives an acceptable estimate of the area under the process path in the PV Diagram.

Solution : The area of each trapezoid under the process path in the PV Diagram is the product of the average pressure for that trapezoid and the change in volume across the trapezoid.

Trapezoid

Pavg

(kPa)

ΔV(L) W (J)

A 475-

0.005

-2

B 425-

0.050

-21

C 375-

0.060

-23

D 325-

0.075

-24

E 275-

0.100

-28

F 225-

0.150

-34      

     

Answer: Wtotal = -132 J

5C-1 : Cross-Sectional Area Requirement for an Adiabatic Nozzle 6 pts

Steam enters a nozzle operating at steady-state with P1 = 40 bar, T1 = 400oC and a velocity of 10 m/s.  The steam flows through the nozzle with negligible heat transfer and no significant change in

potential energy.  At the exit, P2 = 15 bar and the velocity is 665 m/s.  The mass flow rate is 2 kg/s.  Determine the exit area of the nozzle in m2.

Read :The key here is that we know both the mass flow rate and velocity of the effluent stream.  If we can determine the specific volume of the effluent, we can determine the cross-sectional area for flow at the effluent, A2.  We are given the value of one intensive variable for the effluent, P2, but we need to know another in order to completely determine the state of the effluent.  Once know the sate of the effluent, we can use the steam tables to determine the specific volume and then the cross-sectional area.  We must apply the steady-state form of 1st Law for open systems to this process.If we assume that heat transfer and changes in potential energy are negligible and that no shaft work occurs, we can solve for the specific enthalpy of the effluent and thereby fix the state of the system.  This allows us to complete the problem.

Given: P1 4000kPa Find:

T1 400oC A2 ??? m2

v1 10m/s

P2 1500kPa

v2 665m/s

mdot 2kg/s

Diagrams :

Assumptions:

1 - The nozzle operates at steady-state.

2 - Heat transfer is negligible.

3 - No shaft work crosses the system boundary.

4 - The change in the potential energy of the fluid from the inlet to the outlet is negligible.

Equations / Data / Solve :

Let's begin by writing the steady-state form of the 1st Law for open systems.

Eqn 1

Based on the assumptions listed above, we can simplify Eqn 1 as follows :

Eqn 2

The only unknown in Eqn 2 is H2 because we can look up H1 and the velocities are both given.

So, let's look up H1 and solve Eqn 2 for H2 :

Eqn 3 H13214.

5kJ/kg

H22993.

4kJ/kg

We could use H2 and P2 to determine T2 using the Steam Tables, but we are more interested in V2 because :

Eqn 4

or : Eqn 5

Once we know the specific volume at state 2, we can use Eqn 5 to determine the cross-sectional area of the effluent pipe.

Interpolating on the Steam Tables at 1.5 MPa :

T (oC)H (kJ/kg)

V (m3/kg)

250 2923.9 0.15201

T22993.4 V2 T2 280oC

300 3038.2 0.16971 V2

0.16278m3/kg

Now, plug V2 into Eqn 5 : A2

4.896E-04 m2

Verify : None of the assumptions made in this problem solution can be verified.

Answers : A24.90E-04 m2

5C-2 : Heat Losses form a Steam Turbine 5 pts

Steam enters a turbine operating at steady-state with a mass flow rate of 4600 kg/h.  The turbine develops a power output of 1000 kW.  In the feed, the pressure is 60 bar, the temperature is 400oC and the velocity is 10 m/s. For the effluent, the pressure is 0.1 bar, the quality is 90% and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and the surroundings in kW.               

Read :Apply the steady-state form of the 1st Law for open systems and solve for Q.  Assume changes in potential energy are negligible.  We know the values of two intensive variables for state 1, so we can look up H1.  We know the pressure and quality for state 2, so we can also determine H2.  Then, just plug back into the 1st Law to get Q !

Given : m 4600kg/h v1 10m/s

1.278kg/s P2 10kPa

Ws 1000kW x2 0.90

P1 6000kPa v2 50m/s

T1 400oC

Find : Q ??? kW

Diagrams :

Assumptions:

1 - The turbine operates at steady-state.

2 - The change in the potential energy of the fluid from the inlet to the outlet is negligible.

Equations / Data / Solve :

Let's begin by writing the steady-state form of the 1st Law for open systems.

Eqn 1

Solve Eqn 1 for Q : Eqn 2

We must use the Steam Tables to determine H2 and H1 :

Eqn 3

H1 3178.2kJ/kg

At P2 = 10 kPa : Hsat liq 191.81kJ/kg

Hsat vap 2583.9kJ/kg

H2 2344.7kJ/kg

Now, we can plug values into Eqn 2 to evaluate Q :

Q -63.51kW

Verify : None of the assumptions made in this problem solution can be verified.

Answers : Q -63.5kW