Chapter 9,10,11

Example 9-1

Problem Statement:

An air-standard thermodynamic cycle is outlined below. The specific heats of air can be assumed constant throughout the cycle. The four processes that make up the cycle are:

Process 1-2: Air at standard conditions (state 1) (1 atm, 298 K) is heated at constant volume to state 2, where P2 = P1.

Process 2-3: The air at state 2 is expanded isobarically to state 3, where v3 = 3v2.Process 3-4: The air is expanded isentropically until it reaches P1 at state 4.Process 4-1: The air is cooled isobarically to return it to state 1.

(a) Sketch the P-v and T-s diagrams of this cycle. Pay special attention to the relative slopes of the process lines.

(b) Find the numerical value of the thermodynamic efficiency of this heat engine cycle.

Solution:

The diagrams are shown below: Note that

so the slopes of the isobars « slopes of the dv = 0 lines on the T-s diagram.

The temperatures of each state can be found in terms of T1, and the thermo efficiency can also be put in terms of T only, so the problem is probably most easily worked that way.

The efficiency is defined by

The net work done by the cycle is given by

The heat transfer to the cycle occurs in both processes 1-2 and 2-3 so

Now,

Example 9-2

Problem Statement:

An engine operates on an air-standard Otto cycle. The pressure and temperature at the beginning of the compression stroke are 110 kPa and 40°C, respectively. The temperature at the end of the compression stroke is 500°C. If the heat added during

the heat addition process is 1500 kJ/kg, calculate the maximum pressure that the walls must withstand. Assume temperature independent properties for air.

Solution:

Solving for P2:

Solving for P3:

Problem Statement:

A "four-stroke" internal combustion engine operates on an air-standard Otto cycle. The engine is a 454-in3 displacement (7.44 x 10-3 m3) V-8 engine, and it has a compression ratio of 8.5:1.

(a) What is the air-standard efficiency of the engine (percentage)?

(b) If the pressure at the end of the power stroke is 200 kPa just before the exhaust valve opens, how much power (kW) is produced by the engine at 3600 rpm?

Assume temperature independent properties for air

SOLUTION:

Problem Statement:

A hot-rod mechanic has an engine from a 1966 British sports car. The engine displaces 4.2 liters, has an original compression ratio of 10.2:1, and has six cylinders. Assume that the pressure at the end of the power stroke is 300 kPa just before the exhaust valve opens. The pistons in the original engine are 12 cm in diameter.

(a) If the mechanic bores the cylinders and replaces the pistons with new pistons that are 0.5 cm larger in diameter than the originals, by what percentage will the engine efficiency change (assuming the 300 kPa pressure is unchanged)?

(b) If the mechanic replaces the engine crankshaft and piston connecting rods, so that the original pistons have their stroke increased by 0.5 cm, by what percentage will the engine

efficiency change?(c) If the cylinder head is milled so that the original clearance is reduced to 0.4 cm, by what

percentage will the engine efficiency change if the engine retains the original stroke?

Assume in all cases that the ideal air-standard efficiency is desired.

SOLUTION:

Suppose the mechanic replaces the crankshaft and connecting rods as in part (b), and mills the head as in part (c). What will be the resulting dimensions and efficiency?

Example 9-5

Problem Statement:

An engine operates on the air-standard diesel cycle with a compression ratio of 18. The air pressure and temperature at the beginning of the compression are 0.12 MPa and 43°C. If the maximum temperature is 1992 K and the heat addition is 1274 kJ/kg, find the maximum pressure for which the cylinder must be designed. Also find T2.

Solution:

or

Example 9-6

Problem Statement:

The combustion process in an air-standard diesel cycle adds 800 kJ/kg to the medium. The minimum temperature and pressure are 20°C and 0.1 MPa while the maximum temperature is 1000°C. Find the thermal efficiency.

Solution:

Example 9-7

Problem Statement:

An air-standard diesel cycle engine operates as follows: At the end of the power stroke, the pressure is 0.24 MPa and the temperature is 550°C, and at the end of the compression stroke the pressure and temperature are 4.5 MPa and 700°C, respectively. Determine the

(a) compression ratio,

(b) cut-off ratio,

(c) heat transfer to the engine per cycle (kJ/kg),

(d) net work output per cycle (kJ/kg), and

(e) thermal efficiency.

Solution:

Example 9-8

Problem Statement:

An ideal air-standard closed Brayton cycle operates between the pressures of 1 atm and P2. Intake air to the compressor is at 27°C and 1 atm. The turbine is limited to a maximum temperature of 1227°C. Heat transfer from the cycle is qout = 100 kJ/kg.

(a) What is the value of the inlet pressure to the turbine?

(b) What is the cycle efficiency?

(c) What is the compressor outlet temperature?

Solution:

State P (atm) T (°C) s (kJ/kg·K)

1 1 27 s1

2 P2   s1

3 P2 1227 s3

4 P1   s3

a) For air as an ideal gas:

Since the process (3) to (4) is an isentropic process,

b) 

c) Since the process (1) to (2) is an isentropic process,

Example 9-9

Problem Statement:

An ideal gas turbine (Brayton cycle) has a net power output of 100 kW. The working medium is air with T1 = 30°C, T3 = 750°C, and T4 = 300 °C. Assume that properties are temperature independent. Determine

(a) the compressor pressure ratio,

(b) the compressor work (kJ/kg),

(c) the mass flow rate of air (kg/s), and

(d) the thermal efficiency.

Solution:

a) 

b) 

c) 

d) 

Example 9-10

Problem Statement:

A closed Brayton cycle has a two-stage turbine. The high-pressure (first) stage has inlet T = 1100 K at 10 atm and an efficiency of 90 percent. Argon is the working fluid. After leaving the first stage at 5 atm, the argon is reheated to 1100 K and enters the second stage, which also has an efficiency of 90 percent. The exhaust from the turbine is at 1 atm. The compressor in the cycle has an efficiency of 95 percent, and the compressor inlet temperature is 27°C.

(a) Draw a T-s diagram for the cycle.

(b) Find the cycle efficiency (percentage).

(c) Find the cycle efficiency without reheat (percentage).

Solution:

Argon behaves as an ideal gas. From Table C.2,

State (1) - (2)

State (2) - (3)

State (3) - (4)

State (4) - (5)

State (5) - (6)

State T (K) s (kJ/kg·K)

1 300 s1 = s1

2 778 s2 = s1 + 0.017

3 1100 s3 = s2 + 0.18 = s1 + 0.197

4 860 s4 = s3 + 0.016 = s1 + 0.213

5 1100 s5 = s4 + 0.128 = s1 + 0.341

6 630 s6 = s5 + 0.045 = s1 + 0.386

b) 

c) Without reheat

Comments:

Reheating normally increases overall cycle efficiency. Usually, a two-stage turbine with reheat will have a slightly lower efficiency in the low-pressure (second) stage than will a two-stage turbine with no reheat, because with reheat the mass passing through the second stage will have higher specific volume. This causes higher velocities and greater frictional losses. We have ignored this effect in the problem by

using ideal relationships in the cycle analysis, and by assuming all turbine stages have the same efficiency in all cases.

Example 9-11

Problem Statement:

A closed Brayton cycle has a two-stage compressor, with each stage having an efficiency (compared to an isentropic compressor) of 95 percent. The working fluid is argon. The compressor inlet is at 1 atm and 27°C. The compressor outlet is at 10 atm. Between the compressor stages, the argon is cooled at P = 5 atm to 100°C and then enters the second compressor stage. The single-stage turbine has an efficiency of 90 percent and an inlet temperature of 1100 K.

(a) Draw a T-s diagram for the cycle.

(b) Find the cycle efficiency (percentage).

(c) Find the cycle efficiency without intercooling (percentage).

Solution:

Assume Argon as an ideal gas. From Table C.2,

State (1) - (2)

State (2) - (3)

State (3) - (4)

State (4) - (5)

State (5) - (6)

a)

State T (K) s (kJ/kg·K)

1 300 s1 = s1

2 585 s2 = s1 + 0.013

3 373 s3 = s2 - 0.234 = s1 - 0.221

4 498 s4 = s3 + 0.006 = s1 - 0.215

5 1100 s5 = s4 + 0.413 = s1 + 0.198

6 504 s6 = s5 + 0.073 = s1 + 0.271

b) 

Without intercooling,

Comments:

Why is the efficiency without intercooling (36.3 percent) higher than that with intercooling (31.0 percent)? Note that the net work output (|wnet|, intercooling = 97 kJ/kg) is higher than (|wnet|, without intercooling = 61 kJ/kg). However, the net heat added to the cycle is also larger when intercooling is added (you can see this by comparing the T-s diagram for this problem with that without intercooling), and the ratio of |wnet|/qin is lower with intercooling. That is indeed the result of this problem. In practice, however, intercooling provides other benefits that we have not considered. The presence of intercooling reduces the temperature and therefore the specific volume of the Argon entering the second compressor stage. That stage can therefore be designed to be smaller and more efficient than a stage operating at higher inlet T, although the problem statement does not allow this to be considered. This increase in compressor efficiency and resulting reduced compressor work often outweighs the increase in heat addition necessary in the cycle. Also, the compressor can be physically smaller.

Example 9-12

Problem Statement:

A closed Brayton cycle uses argon as the working fluid. The engine has a two-stage compressor with intercooling, and each stage has an isentropic efficiency of 95 percent. The turbine is two-stage type with reheat, and each stage has an efficiency of 90 percent. Reheat and intercooling occur at 5 atm, and the cycle operates between 1

and 10 atm. The turbine inlet temperature is 1100 K, and the compressor inlet is at 27°C. Reheating brings the second-stage inlet temperature to 1100 K, and intercooling brings the second-stage compressor inlet temperature to 100°C.

(a) Draw the T-s diagram for the cycle.

(b) Find the cycle efficiency (percentage).

(c) Find the cycle efficiency without intercooling (percentage).

(d) Find the cycle efficiency without reheat (percentage).

(e) Find the cycle efficiency with neither reheat nor intercooling (percentage).

Solution:

From Table C.2, for Argon

State (1) - (2)

State (2) - (3)

State (3) - (4)

State (4) - (5)

State (5) - (6)

State (6) - (7)

State (7) - (8)

a)

State T (K) s (kJ/kg·K)

1 300 s1 = s1

2 585 s2 = s1 + 0.013

3 373 s3 = s2 - 0.234 = s1 - 0.221

4 498 s4 = s3 + 0.006 = s1 - 0.215

5 1100 s5 = s4 + 0.413 = s1 + 0.198

6 860 s6 = s5 + 0.016 = s1 + 0.214

7 1100 s7 = s6 + 0.128 = s1 + 0.342

8 630 s8 = s7 + 0.045 = s1 + 0.387

b) 

c) Without intercooling,

d) 

e) Without intercooling or reheat

Comments:

Why does the cycle efficiency DECREASE as intercooling is used? For two reasons: First, intercooling is used in practice because it reduces the specific volume of the Argon in the second stage of the compressor, and the overall compressor work should be less. By using the constant property relations in our ideal analysis, we cannot see that advantage. Second, the intercooled second stage efficiency would normally be greater than for a single turbine or for a two-stage adiabatic turbine; we ignored this factor in the problem statement.

Example 9-13

Problem Statement:

Consider the actual jet propulsion cycle shown. The compressor is adiabatic. Find

(a) the compressor efficiency, ηc;

(b) the specific work input to the compressor, win (kJ/kg); and

(c) the nozzle exit velocity, v5 (m/s).

Solution:

a) 

b) 

c) 

Example 10-1

Problem Statement:

A steam power plant has a mass flow rate of steam of 20,000 kg/h. Water enters the boiler (state 1) at 10 MPa and leaves the boiler (state 2) as saturated vapor. The working fluid leaves the condenser (state 4) at 10 kPa as saturated liquid. Cooling water to the condenser circulates at 1×106 kg/hr, and rises in temperature by 8.0°C in passing through the condenser. Find:

(a) The thermal efficiency of the cycle.

(b) The turbine efficiency.

(c) The power output of the turbine (kW).

Construct a state table to aid in your solution.

Solution:

State P (kPa) T (°C) ha (kJ/kg) ss (kJ/kg·K) hs (kJ/kg)

1 10000   (201.9)    

2 10000 (311.0) (2724.5) (5.6139)  

3 10   (1877.0)   (1776.32)

4 10 (45.8) (191.8)    

First Law on condenser:

a) Thermal efficiency

b) Turbine efficiency

c) Power output

Example 10-2

Problem Statement:

Saturated steam leaves a boiler at P = 200 bars. Heat transfer occurs in a superheater at constant P until the temperature reaches 600°C. The steam then enters a steam

turbine, where it does work on the turbine, and leaves the turbine at a pressure of P = 7.0 bars. The turbine has an efficiency of 75 percent.

(a) How much work is done on the turbine?

(b) How much heat transfer occurs in the superheater?

(c) If liquid water enters the boiler at 165°C and 200 bars, what is the Rankine cycle efficiency? (Neglect pump work.)

Solution:

State P (bars) T (°C) h (kJ/kg) s (kJ/kg·K)

1 200 165 (708.6)  

2 200   (2413.6)  

3 200 600 (3536.7) (6.5039)

4,s 7.0   (2673.9) 6.5039

a) 

b) 

c) 

Example 10-3

Problem Statement:

An actual steam power plant operates as shown. The boiler and condenser operate at constant pressure. The turbine and pump are adiabatic. Find wt, wp, and ηth.

T1 = 30 °C T3 = 600 °C

P1 = 0.01 MPa hp = 0.80

P2 = 3.0 MPa ht = 0.90

Solution:

Example 10-4

Problem Statement:

Steam enters the high-pressure turbine in an ideal reheat cycle with a mass flow rate of 1000 g/s. Find wt, wp, and ηth.

T3 = 600°C P4 = 0.8 MPa

P3 = 15 MPa T5 = 600°C

  P6 = 6 kPa

Solution:

Example 10-5

Problem Statement:

A Rankine cycle with superheat has a condenser pressure of 8 kPa, a boiler pressure of 6 MPa, and a turbine efficiency of 88 percent and superheats the steam leaving the boiler to 400°C. Compare the efficiency of this cycle to a reheat cycle operating with the same boiler, superheater, and condenser conditions in which steam leaves the high-pressure stage of a two-stage turbine at 300 kPa, is reheated to 400°C, and is

then fed to the low-pressure turbine stage. Assume that both turbine stages have efficiencies of 88 percent.

Solution:

The solution for the cycle with superheat but no reheat is first discussed.

The state table for the two cases is given below:

Property Cycle with Superheat

P2 (kPa) 6000

s2 (kJ/kg·K) 5.8886

s3 (kJ/kg·K) 6.5404

h2 (kJ/kg) 2783.9

h3 (kJ/kg) 3177.0

h5 (kJ/kg) 173.85

h4,s (kJ/kg) 2045.5

h1 (kJ/kg) 179.1

x4,a 0.836

ηR (%) 33.2

For the pump,

The cycle efficiency is found from

When reheat is added the cycle is as shown. To determine the cycle efficiency, the relation is

The state table becomes:

Property Cycle with Reheat and Superheat

P3 (kPa) 6000

s3 (kJ/kg·K) 6.5404

h3 (kJ/kg) 3177.0

h4,s (RH) (kJ/kg) 2541.6

h4,a (RH) (kJ/kg) 2617.8

h5 (RH) (kJ/kg) 3274.9

s5 (RH) (kJ/kg·K) 8032.7

h5,s (RH) (kJ/kg) 2515.0

h5,a (kJ/kg) 2606.2

h2 (kJ/kg) 179.1

ηR (%) 33.6

Adding a reheat loop to the cycle has improved the cycle efficiency from 33.2 to 33.6 percent. A simple cycle (no superheat or reheat) had a cycle efficiency of 31.9 percent when operating between the same boiler and condenser pressures.

Comments:

The enthalpy of the exit steam from the low pressure turbine stage is above the enthalpy of saturated steam at 8 kPa, so that the exit steam is superheated for this case. A turbine redesign or slight change in operating conditions could be made to extract more enthalpy from the steam before it enters the condenser.

Example 11-1

Problem Statement:

In the ideal vapor-compression refrigerator using refrigerant 134a, the evaporator temperature is -20°C, and the inlet temperature to the condenser is 30°C. Saturated vapor enters the compressor. Calculate

(a) The work of the compressor

(b) The heat transfer from the condenser

(c) The heat transfer to the evaporator

(d) The coefficient of performance

Solution

Compressor work:

T2 is specified in the problem statement. T2 = 30°C. s2 is determined by using the ideal compressor assumption. s2 = s1

We can see in Table <?> that T2 = 30°C and s2 = 0.9932 kJ/kg·K is a state between P = 6.0 bars and 7.0 bars.

a) 

b) 

c) 

d) 

Example 11-2

Problem Statement:

A refrigeration system using refrigerant 134a is to have a capacity of 25 tons. The cycle is the ideal vapor compression cycle in which the evaporator pressure is 0.22 MPa and he condenser pressure is 1.0 MPa. Determine the horsepower required to drive the compressor.

P4 = P1 = 0.22 MPa , P2 = P3 = 1.0 MPa

Find Wcomp

Solution:

Example 11-3

Problem Statement:

An ideal vapor-compression refrigeration system using refrigerant 134a operates with an evaporator temperature of -28°C and a condenser exit temperature of 40°C and requires a 100-hp motor to drive the compressor. Calculate the refrigerator's capacity in tons.

Find the capacity in tons. 

Solution:

REQUIRES DOUBLE INTERPOLATION

Example 11-4

Problem Statement:

Refrigerant 134a enters the adiabatic compressor in an air conditioner with P1 = 0.32 MPa and x1 = 1.0. If P2 = 1.0 MPa and the compressor efficiency is 70 percent, determine the actual work (kJ/kg).

Solution:

Example 11-5

Problem Statement:

A 10-ton vapor-compression refrigeration device uses ammonia as the working fluid. The saturated liquid entering the throttling valve has T = 30°C, and the saturated vapor entering the compressor is at -25°C. The compressor is ideal with an efficiency of 100 percent.

(a) What is the ratio of the COP of this cycle to the COP of a Carnot cycle operating between the same saturation temperatures?

(b) How much ammonia must be circulated through the device?

Solution:

a) 

[or if the highest cycle temperature (state 2a) is used then (COP)Carnot = 1.16]

State T (°C) P (kPa) v (m3/kg) h (kJ/kg) s (kJ/kg·K) x

1 -25 (151.5) (0.7722) (1430.87) (5.9844) 1.0

2s (122) 1167.1 (0.1582) (1733) s2s = s1  

3 30 (1167.1) (0.00168) (342.4) (1.4906) 0.0

4 -25 (151.5) (0.1485) 342.4 (1.5977)  

qc = h1 - h4 = 1430.87 - 342.4 = 1088.47 kJ/kg

w= h2s - h1 = 1733 - 1430.87 = 302.13 kJ/kg

COP = 1088.47/ 302.13 = 3.6

Thus

COPactual/COPCarnot = 3.6/4.51 = 0.8

b) 

Thus,