4F-5 : Performance of an Ideal Gas Cycle 10 pts

An ideal gas, initially at 30C and 100 kPa, undergoes an internally reversible, cyclic process

in a closed system. The gas is first compressed adiabatically to 500 kPa, then cooled at a

constant pressure of 500 kPa to 30C, and finally expanded isothermally to its original state.

a.) Carefully draw this process in a traditional piston-and-cylinder schematic.

b.) Sketch the process path for this cycle on a PV Diagram. Put a point on the diagram

for each state. Be sure to include and label all the important features for a complete

PV diagram for this system.

c.) Calculate Q, W, U and H, in J/mole, for each step in the process and for the entire

cycle. Assume that CP = (7/2) R.

d.) Is this cycle a power cycle or a refrigeration cycle ? Explain. Calculate the thermal

efficiency or COP of the cycle, whichever is appropriate.

Read : Sketch carefully. Understanding what is going on in the problem is half the

battle. Apply the 1st Law, the definitions of boundary work, CP and CV to a cycle

on an ideal gas with constant heat capacities. Take advantage of the fact that step 1-2

is both adiabatic and reversible, so it is isentropic. Power cycles produce a net

amount of work and proceed in a clock-wise direction on a PV Diagram.

Given : T1 303.15 K

P1 100 kPa

T2 ??? K

P2 500 kPa

T3 303.15 K

P3 500 kPa

Q12 0 J/mole

R 8.314 J/mole-K

CP 29.099 J/mole-K

Find : For each of the three steps and for the entire cycle: U ??? J/mole

H ??? J/mole

Q ??? J/mole

W ??? J/mole

Diagrams :

Part a.)

Part b.)

Assumptions: 1. Step 1-2 is adaibatic, Step 2-3 is isobaric, Step 3-1 is isothermal.

2. The entire cycle and all of the steps in the cycle are internally reversible.

3. Changes in kinetic and potential energies are negligible.Changes in kinetic and potential energies are negligible.

4. Boundary work is the only form of work interaction during the cycle. 5. The PVT behavior of the system is accurately described by the ideal

gas EOS.

Equations / Data / Solve :

Part c.) Let's begin by analyzing step 1-2, the adiabatic compression.

Begin by applying the 1st law for closed systems to each step in the Carnot

Cycle. Assume that changes in kinetic and potential energies are negligible.

Eqn 1

Because internal energy is not a function of pressure for an ideal gas, we can

determine DU by integrating the equation which defines the constant volume heat

capacity. The integration is simplified by the fact that the heat capacity for the gas in

this problem has a constant value.

Eqn 2

Eqn 3

Combining Eqns 1 & 3 yields:

Eqn 4

The problem is that we do not know T2. So, our next task is to determine T2.

Since the entire cycle is reversible and this step is also adiabatic, this step is

isentropic.

The fastest way to determine T2 is to use one of the PVT relationships for isentropic

processes.

Eqn 5

Solve Eqn 5 for T2:

Eqn 6

Now, we need to evaluate :

Eqn 7

But for ideal gases :

Eqn 8

Solving Eqn 8 for CV yields :

Eqn 9

Plugging values into Eqn 9 and then Eqn 7 yields : CV 20.785

J/mole

-K

1.4

Now, plug values into Eqn 5 to get T2

and plug that into Eqn 4 to get W12 :

T2 480.13 K

W12 -3678.6 J/mole

Plugging values into Eqn 1 yields :

U12 3678.6 J/mole

Now, we can get H

from its definition :

Eqn 10

But, the gas is an ideal gas:

Eqn 11

Combining Eqns 10 & 11

gives us :

Eqn 12

Now, we can plug values into Eqn 12 :

H12 5150.1 J/mole

Next, let's analyze step 2-3, isobaric

cooling.

T2 480.1349 K

P2 500 kPa

T3 303.15 K

P3 500 kPa

The appropriate form of the 1st Law is:

Eqn 13

Because we assumed that boundary work is the only form of work that crosses the

system boundary, we can determine work from its definition.

Eqn 14

Isobaric process:

Eqn 15

Because the system contains and ideal gas:

Eqn 16

W23 -1471.5 J/mole

Next we can calculate U by

applying Eqn 3 to step 2-3:

Eqn 17

U23 -3678.6 J/mole

Now, solve Eqn 13 to

determine Q23 :

Eqn 18

Q23 -5150.1 J/mole

Now, we apply Eqn 12 to

step 2-3 to determine H :

Eqn 19

H23 -5150.1 J/mole

Next, we analyze step 3-1, isothermal expansion.

For ideal gases, U and H are

functions of T only. Therefore : U31 0.0 J/mole

H31 0.0 J/mole

The appropriate form

of the 1st Law is:

Eqn 20

Again, because we assumed that boundary work is the only form of work that crosses

the system boundary, we can determine work from its definition.

Eqn 21

Ideal Gas EOS :

Eqn 22

Solve Eqn 22 for P and

substitute the result into

Eqn 21 to get :

Eqn 23

Eqn 24

Integrating Eqn 24 yields

:

Eqn 25

We can use the Ideal Gas EOS to avoid calculating V1 and V3 as follows:

Apply Eqn 22 to both states 3 and 1 :

Eqn 26

Cancelling terms and rearranging leaves :

Eqn 27

Use Eqn 27 to eliminate the V's from Eqn 25 :

Eqn 28

Now, plug values into Eqn 28 and then Eqn 20 : W31 4056.4 J/mole

Q31 4056.4 J/mole

Finally, we can calculate Qcycle and Wcycle from :

Eqn 29

Eqn 30

Plugging values into Eqns 29 & 30 yields :

Wcycle -1093.7 J/mole

Qcycle -1093.7 J/mole

This result confirms what an application of the 1st Law to the entire cycle tells us:

Qcycle = Wcycle.

Part d.) The cycle is a refrigeration cycle because both Wcycle and Qcycle are

negative.

The coefficient of performance of a

refrigeration cycle is defined as :

Eqn 31

QC is the heat absorbed by the system during the cycle. In this case, QC =Q31.

W is the work input to the system during the cycle. In this case, W = - Wcycle.

Therefore :

QC 4056.4 J/mole

W 1093.7 J/mole

Plug values into Eqn 31 to get

: COPR 3.709

Verify : The ideal gas assumption needs to be verified.

We need to determine the specific

volume at each state and check if :

Eqn 32

V1 25.20 L/mol

Eqn 33

V2 7.98 L/mol

V3 5.04 L/mol

The specific volume at each state is greater than 5 L/mol for all states and Air can be

considered to be a diatomic gas, so the ideal gas assumption is valid.

Answer

s: a.) See diagram above. b.) See diagram above.

c.) Step U H Q W

1 - 2 3679 5150 0 -3679

All values in

this table are

in J/mole.

2 - 3 -3679

-

5150 -5150 -1471

3 - 1 0 0 4056 4056

Cycle 0 0 -1094 -1094

d.) Refrigeration or Heat Pump Cycle.

COPR 3.7

COPHP 4.7

5C-4 : Expansion of Steam Through a Throttling Valve 5 pts

Steam at 9000 kPa and 600C passes through a throttling process so that the pressure is

suddenly reduced to 400 kPa.

a.) What is the expected temperature after the throttle ?

b.) What area ratio is necessary for the kinetic energy change to be zero ?

Read : We know the values of two intensive variables for the inlet steam, so we can

determine the values of all of its other properties, including the specific enthalpy,

from the steam tables. If changes in kinetic and potential energy are negligible and

the throttling device is adiabatic, then the throttling device is isenthalpic. In this

case, we then know the specific enthalpy of the outlet stream. The pressure of the

outlet stream is given, so we now know the values of two intensive properties of

the outlet stream and we can determine the values of any other property using the

steam tables. Part (b) is an application of the 1st Law. The area must be greater at

the outlet in order to keep the velocity the same because the steam expands as the

pressure drops across the throttling device.

Diagram

:

Given : P1 9000 kPa

Find : T1 ??? C

T1 600 C

A2 /

A1 ???

P2 400 kPa

Assumptions : 1 - The throttling device is adiabatic.

2 - Changes in potential energy are negligible.

3 - Changes in kinetic energy are negligible because the cross-sectional

area for flow in the feed and effluent lines have been chosen to

make the fluid velocity the same at the inlet and the outlet.

Equations / Data / Solve

:

Part a.) Begin by looking up the specific enthalpy of the feed in the steam

tables.

At a pressure of 9000 kPa, the saturation temperature is : Tsat 303.4 C

Because T1 > Tsat, we conclude that the feed is superheated steam and we must

consult the Superheated Steam Tables. Because 9000 kPa is not listed in the table,

interpolation is required.

At 600oC : P (kPa) H (kJ/kg) V (m3/kg)

8000 3642.0 0.04845

9000 3633.7 0.04341

H1 3633.7 kJ/kg

10000 3625.3 0.03837

V1 0.04341 m3/kg

The 1st Law for a throttling device that is adiabatic and causes negligible changes

in kinetic and potential energies is :

Eqn 1

Because the pressure drops in the throttling device and the feed is a superheated

vapor, the effluent must also be a superheated vapor. So, to answer part (a), we

must use the Superheated Steam Tables to determine the temperature of 400 kPa

steam that has a specific enthalpy equal to H2.

At 400 kPa : H (kJ/kg) T (oC) V (m3/kg)

3592.9 550 0.9475

3633.7 568.6 0.9693

T2 568.6 C

3702.4 600 1.006

V2 0.96927 m3/kg

Part b.) We need to use the definition of kinetic energy to determine how much the area of

the outlet pipe must be greater than the area of the inlet pipe in order to keep the

kinetic energy (and therefore the velocity) constant.

Eqn 2

Eqn 3

Because the mass flow rate at the inlet

and outlet is the same, Eqn 3 simplifies

to :

Eqn 4

Next, we need to consider the relationship between velocity, specific volume and

cross-sectional area.

Eqn 5

Now, substitute Eqn 5 into Eqn 4 to

get :

Eqn 6

Solve for the area ratio, A2 / A1 :

Eqn 7

A2 /

A1 22.328

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

verified.

Answers

: T2 569 C

A2 /

A1 22.3

5C-5 : Open Feedwater

Heater 6 pts

An open feedwater heater in a steam power plant operates at steady-state with liquid entering

at T1 = 40oC and P1 = 7 bar. Water vapor at T2 = 200

oC and P2 = 7 bar enters in a second

feed stream. The effluent is saturated liquid water at P3 = 7 bar. If heat exchange with the

surroundings and changes in potential and kinetic energies are negligible, determine the ratio

of the mass flow rates of the two feed streams, mdot1 / mdot,2.

Read :

The feedwater heater is just a fancy mixer. When we write the MIMO form of the

1st Law at steady-state, there are three unknowns: the three mass flow rates. The

states of all three streams are fixed, so we can determine the specific enthalpy of

each of them.

Mass conservation tells us that m3 = m1 + m2. We can use this to eliminate m3

from the 1st Law. Then we can solve the 1st Law for m1 / m2 !

Given : T1 40

oC

Find :

P1 700 kPa

mdot1 / mdot,2 = ???

T2 200

oC

P2 700 kPa

P3 700 kPa

Q 0 kW

Diagram:

Assumptions:

1 - The feedwater heater operates at steady-state.

2 - Changes in potential and kinetic energies are negligible.

3 - Heat transfer is negligible.

4 - No shaft work crosses the system boundary in this process.

Equations / Data / Solve :

An open feedwater heater is essentially a mixer in which superheated vapor is used

to raise the temperature of a subcooled liquid. We can begin our analysis with the

steady-state form of the 1st Law.

Eqn 1

The assumptions in the list above allow us to simplify the 1st Law considerably:

Eqn 2

Conservation of mass on the feedwater heater operating at steady-state tells us that :

Eqn 3

We can solve Eqn 3 for mdot,3 and use the result to eliminate mdot,3 from Eqn 2. The

result is:

Eqn 4

The easiest way to determine mdot,1 / mdot,2 is to divide Eqn 4 by mdot,2.

Eqn 5

Now, we can solve Eqn 5 for mdot,1 / mdot,2 :

Eqn 6

Now, all we need to do is to determine the specific enthalpy of all three streams and

plug these values into Eqn 6 to complete the problem.

First we must determine the phase(s) present in each stream.

Tsat(700kPa)= 164.95

oC

Therefore:

Stream 1 is a subcooled liquid because T1 < Tsat

Stream 2 is a superheated vapor because T2 > Tsat

T3 = Tsat because it is a saturated liquid.

Data from the Steam Tables of the NIST Webbook (using the default reference

state) :

H1 168.15 kJ/kg

H2 2845.3 kJ/kg

H3 697 kJ/kg

Now, plug these values into Eqn 6 to obtain : mdot1 / mdot,2 = 4.062

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

Answers

: mdot1 / mdot,2 = 4.06

5C-6

: Analysis of a Steam Power Cycle 8 pts

A simple steam power plant operates on 8 kg/s

of steam. Losses in the connecting pipes and

through the various components are to be

neglected. Calculate

a.) The power output of the turbine

b.) The power needed to operate the pump

c.) The velocity in the pump exit pipe

d.) The heat transfer rate necessary in the

boiler

e.) The heat transfer rate realized in the

condenser

f.) The mass flow rate of cooling water required

g.) The thermal efficiency of the cycle

Data

: P1 20 kPa

T1 50 C

P2 8000 kPa P4 20 kPa

T2 50 C x4 0.92 kg vap/kg total

D2 0.05 m

P3 8000 kPa Tcw,in 20 C

T3 600 C Tcw,out 50 C

Read

: Cycle problems of this type usually require you to work your way around the cycle,

process by process until you have determined the values of all of the unknowns. This is

a good approach here because the problem statement asks us to determine the values of

unknowns in every process in the cycle. The only decision is where to begin. We can

begin with the turbine because that is the 1st question and also because we have enough

information to answer part (a). We know T3 and P3, so we can determine H3. Stream 4

is saturated mixture with known P4 and x4, so we can also determine H4. With the

usual assumtions about kinetic and potential energy, we can determine Wturb. In fact,

because we know the T and P of streams 1 and 2 as well, we can analyze the processes

in this cycle in any convenient order. So, we will let the questions posed in the problem

determine the order in which we analyze the processes. We will apply the 1st law to the

pump, the boiler and the condenser, in that order. Use the Steam Tables in the NIST

Webbook.

Give

n : m 8 kg/s

P3 8000 kPa

P1 20 kPa

T3 600 C

T1 50 C

P4 20 kPa

P2 8000 kPa

x4 0.92 kg vap/kg total

T2 50 C

Tcw,in 20 C

D2 0.05 m

Tcw,out 50 C

Find

: Wturb ??? MW

Qboil ??? MW

Wpump ??? kW

Qcond ??? MW

v2 ??? m/s

mcw ??? kg/s

th ???

Assumptions : 1 - Changes in kinetic and potential energy are negliqible in all the

processes in the cycle

2 - The pump and turbine are adiabatic.

3 - All of the heat that leaves the working fluid in the condenser is

transferred to the cooling water. No heat is lost to the

surroundings.

Equations / Data / Solve :

Part

a.)

Begin by writing the 1st Law for the turbine, assuming that changes in kinetic and

potential energy are negligible. This makes sense because we have no elevation,

velocity or pipe diameter information to use.

Eqn 1

If we assume that the turbine is adiabatic, we can solve Eqn 1 for the shaft work of the

turbine :

Eqn 2

Now, we must use the steam tables to determine H3 and H4. Let's begin with stream 3.

At a pressure of 8000 kPa, the saturation temperature is : Tsat 295.01 C

Because T3 > Tsat, we conclude that stream 3 is superheated steam and we must consult

the Superheated Steam Tables. Fortunately, there is an entry in the table for 8000 kPa

and 600oC.

http://webbook.nist.gov/chemistry/fluid/http://webbook.nist.gov/chemistry/fluid/

H3 3642.4 kJ/kg

Stream 4 is a saturated mixture at 20 kPa, so we need to use the properties of saturated

liquid and saturated vapor at 20 kPa in the following equation to determine H4 :

At 20 kPa :

Hsat liq 251.42 kJ/kg

Eqn 3

Hsat vap 2608.9 kJ/kg

H4 2420.3 kJ/kg

Now, we can plug H3 and H4 back into

Eqn 2 to answer part (a) : Wturb 9.777 MW

Part

b.) Write the 1st Law for the pump, assuming that changes in kinetic and potential energy

are negligible. This makes sense because we have no elevation or velocity data and we

are given only the outlet pipe diameter. Also, assume the pump is adiabatic, Qpump =

0.

Eqn 4

Eqn 5

Now, we must determine H1 and H2. We know the T and P for both of these streams,

so we should have no difficulty determining the H values.

Tsat(P1) 60.058 C T1 < Tsat, therefore we must consult the Subcooled

Water Tables.

Tsat(P2) 295.01 C T2 < Tsat, therefore we must consult the Subcooled

Water Tables.

H1 209.35 kJ/kg

H2 216.22 kJ/kg

Now, we can plug H1 and H2 back into

Eqn 5 to answer part (b) : Wpump -54.960 kW

Part

c.)

Here, we need to consider the relationship between velocity, specific volume and cross-

sectional area.

Eqn 6 where :

Eqn 7

A2 0.001963 m2

From the NIST Webbook :

V2 0.0010086 m3/kg

Now, we can plug values into Eqn 6 to answer part (c) : v2 4.109 m/s

Part

d.) Write the 1st Law for the boiler, assuming that changes in kinetic and potential energy

are negligible. This makes sense because we have no elevation, velocity or pipe

diameter data. There is no shaft work in a boiler.

http://webbook.nist.gov/chemistry/fluid/

Eqn 8

Eqn 9

We determined H2 in part (b) and H3 in part (a), so all we need to do is plug numbers

into Eqn 9.

Qboil 27.409 MW

Part

e.) Write the 1st Law for the condenser assuming that changes in kinetic and potential

energy are negligible. This makes sense because we have no elevation, velocity or pipe

diameter data. Use the working fluid as the system so that Qcond is the amount of heat

transferred to the cooling water. There is no shaft work in a condenser.

Eqn 10

Eqn 11

We determined H1 in part (b) and H4 in part (a), so all we need to do is plug numbers

into Eqn 11.

Qcond -17.688 MW

Part

f.) In order to determine the mass flow rate of the cooling water, we must write the 1st Law

using the cooling water as our system. For this system, Qcw = - Qcond because heat

leaving the working fluid for the cycle enters the cooling water.

Qcw 17.688 MW

Assume that changes in kinetic and potential energy are negligible. This makes sense

because we have no elevation, velocity or pipe diameter data. There is no shaft work for

the cooling water system.

Eqn 12

We cannot use the Steam Tables to determine the enthalpy of the cooling water because

we do not know the pressure in either stream. The next best thing we can do is to use

the specific heat of the cooling water to determine Hcw using:

Eqn 13

If we further assume that the specific heat of liquid water is constant over the

temperature range 20C - 50C, than Eqn 13 simplifies to:

Eqn 14

We can then combine Eqn 14 with Eqn 12 to obtain

:

Eqn 15

Finally, we can solve Eqn 15 for mcw :

Eqn 16

All we need to do is look up the average heat capacity of water between 20C and 50C.

NIST : CP,cw(50C) 4.1813 kJ/kg-K

CP,cw(20C) 4.1841 kJ/kg-K

CP,cw 4.1827

kJ/kg-

K

Let's use : CP,cw 4.18 kJ/kg-K

Then :

mcw 141.05 kg/s

Part

g.)

The thermal efficiency of this power cycle can be determined directly from its

definition.

Eqn 17

th 0.3547