PRINCIPLES OF TURBOMACHINERY SOLUTIONS MANUAL PDF

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PRINCIPLES OF TURBOMACHINERY SOLUTIONS MANUAL by Seppo A. Korpela Department of Mechanical and Aerospace Engineering January 2012 Copyright ⃝2011-2012, c Seppo A. Korpela Chapter 2 Exercise 2.1 Steam flows through a bank of nozzles shown in the figure below, with wall thickness t2 = 2 mm, spacing s ...

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PRINCIPLES OF TURBOMACHINERY SOLUTIONS MANUAL

by

Seppo A. Korpela Department of Mechanical and Aerospace Engineering January 2012

c Copyright ⃝2011-2012, Seppo A. Korpela

Chapter 2 Exercise 2.1 Steam flows through a bank of nozzles shown in the figure below, with wall thickness t2 = 2 mm, spacing s = 4 cm, blade height b = 2.5 cm, and exit angle α2 = 68◦ . The exit velocity V2 = 400 m/s, pressure is p2 = 1.5 bar, and temperature is T2 = 200 C. Find the mass flow rate. Given: b = 2.5 C

s = 4 cm

T2 = 200 C = 473.15 K

t2 = 0.2 cm

α2 = 68◦

V2 = 400 m/s v2 = 1.4443 m3 /kg

p1 = 1.50 bar = 150 kPa

Find: Mass flow rate. Solution: A2 = b(2 cos α2 − t2 ) = 2.5(4 cos(68◦ ) − 0.2) = 3.25 cm 3.25 · 400 A2 V2 = 4 = 0.09 kg/s ⇐ v2 10 · 1.4443 Note that the specific volume could also be approximated as m ˙ =

v2 =

¯ 2 RT 8.314 · 473.15 = = 1.457 m3 /kg Mp2 18 · 150

V1

t1

s 1

t2

2

α2

V2

2

Exercise 2.2 Air enters a compressor from atmosphere at pressure 102 kPa and temperature 42 C. Assuming that its density remains constant determine the specific compression work required to raise its pressure to 140 kPa in a reversible adiabatic process, if the exit velocity is 50 m/s. Given: Since the air is stagnant in the atmosphere, its conditions are the stagnation conditions. The flow is isentropic and T2 = 42 C = 315.15 K

p1 = 102 kPa

p2 = 140 kPa

V2 = 50 m/s

Find: Specific work done. Solution: For isentropic flow T ds = dh − vdp leads to dh = dp/ρ. In addition h0 = h + V 2 /2 is constant. The flow is assumed incompressible because the pressure changes only slightly and the exit velocity is small. p 2 − p1 1 2 w= + V2 ρ 2 The density is p1 102 ρ= = = 1.128 kg/m2 RT1 0.287 · 315.156 so that 502 140 − 102 + = 43.95 kJ/kg ⇐ w= 1.128 2 · 1000 Exercise 2.3 Steam flows through a turbine at the rate of m ˙ = 9000 kg/h. The ˙ rate at which power is delivered by the turbine is W = 440 hp. The inlet total pressure is p01 = 70 bar and total temperature is T01 = 420 C. For a reversible and adiabatic process find the total pressure and temperature. leaving the turbine. Given: The flow is isentropic and ˙ = 440 hp T01 = 420 C p01 = 70 bar m ˙ = 9000 kg/h W Find: T02 and p02 . Solution:

˙ W 440 · 0.7457 · 3600 = = 131.2 kJ/kg m ˙ 9000 From steam tables h)1 = 3209.8 kJ/kg, s1 = 6.5270 kJ/kg K. At the exit w=

h02 = h01 − w = 3209.8 − 131.2 = 3078.6 kJ/kg

s2 = 6.5270 kJ/kg K

From the superheated steam tables, or using EES, p02 = 43.58 bar

T02 = 348.5 C 3

⇐

Exercise 2.4 Water enters a pump as saturated liquid at total pressure of p01 = 0.08 bar and leaves it at p02 = 30 bar. If the mass flow rate is m ˙ = 10, 000 kg/h and the process can be assumed to take place reversibly and adiabatically, determine the power required. Given: The flow is isentropic and p01 = 0.08 bar

p02 = 30 bar

m ˙ = 10.000 kg/h

˙. Find: W Solution: For isentropic flow w= Therefore

30 − 0.08) · 105 p02 − p01 = = 2.998 kJ/kg ρ 998

10, 000 · 2.998 ˙ = mw = 8.33 kW W ˙ = 3600

⇐

Exercise 2.5 Liquid water at 700 kPa and temperature 20 C flows at velocity 15 m/s. Find the stagnation temperature and stagnation pressure. Given: The flow is isentropic and p = 700 kPa

T = 20 K

V = 15 m/s

Find: T0 and p0 . Solution: Since water is incompressible T0 = T +

V2 152 = 20 + = 20.027 = 20.027 C 2cp 2 · 4187

⇐

1 998 · 152 p0 = p + ρV 2 = 700 + = 700 + 122.75 = 812.3 kPa 2 2 · 1000

⇐

Exercise 2.6 Water at temperature T1 = 20 C flows through a turbine with inlet velocity V1 = 3 m/s, static pressure p1 = 780 kPa and elevation z1 = 2 m. At the exit the conditions are V2 = 6 m/s, p2 = 100 kPa and z2 = 1.2 m. Find the specific work delivered by the turbine. Given: Assuming that the process is isentropic and given that T1 = 20 C, and p1 = 780 kPa

z1 = 2 m 4

V1 = 3 m/s

p2 = 100 kPa

z1 = 1.2 m

V1 = 6 m/s

Find: w Solution: Since water is incompressible w = h01 − h02s w=

p01 − p02 V12 − V22 = + + g(z1 − z2 ) ρ 2

780 − 100) 103 32 − 62 + + 9.81 (2 − 2.12) = 681.36 − 13.50 + 7.85 998 2 w = 675.71 J/kg ⇐

Note the small contributions from potential and kinetic energy. Exercise 2.7 Air at static pressure of 2 bar and static temperature of 300 K flows with velocity 60 m/s. Find total temperature and pressure. Given: p1 = 2 bar T1 = 300 K V1 = 60 m/s Find: T01 and p01 Solution: The total temperature is T01 = T1 +

V12 602 = 300 + = 301.8 K 2 cp 2 · 1004.5

⇐

and since the velocity is quite small, air may be taken to be incompressible with density p1 2 · 105 ρ= = = 2.323 kg/m3 RT1 287 · 300 and the stagnation pressure is 2.323 · 602 1 = 204.2 kPa p01 = p1 + ρV12 = 2 · 105 + 2 2

⇐

Exercise 2.8 Air at static temperature of 300 K and static pressure of 140 kPa flow with velocity 60 m/s. Evaluate the total temperature and total pressure of air. Repeat the calculation if the air speed is 300 m/s. Given: p1 = 140 kPa T1 = 300 K V1 = 60 m/s 5

Find: T01 and p01 Solution: T01 = T1 +

602 V12 = 300 + = 301.8 K 2 cp 2 · 1004.5

⇐

and since the velocity is quite small, air may be taken to be incompressible with density p1 2 · 105 ρ= = = 2.323 kg/m3 RT1 287 · 300 and the stagnation pressure is 2.323 · 602 1 = 142.93 kPa p01 = p1 + ρV12 = 140 · 103 + 2 2

⇐

and if V1 = 300 m/s then T01 = T1 + and p01 = p1

(

T01 T1

3002 V12 = 300 + = 344.8 K 2 cp 2 · 1004.5

)γ/(γ−1)

= 140

(

344.8 300

)3/5

⇐

= 227.9 kPa

⇐

Exercise 2.9 Air undergoes an increase of 1.75 kPa in total pressure through a blower. The inlet total pressure is one atmosphere and the inlet total temperature is 21 C. Evaluate the exit total temperature if the process is reversible adiabatic. Evaluate the energy added to the air per unit mass flow. Given: p01 = 101.325 kPa

T01 = 294.15 K

∆p0 = 1.74 kPa

Find: T02 Solution: The stagnation density is ρ01 =

101.325 · 103 p01 = = 1.200 kg/m3 RT01 287 · 294.15

Since the process is reversible and adiabatic and the pressure increase is quite small, the work done by the blower may be calculated from w=

1.75 p02 − p01 = = 1.458 kJ/kg ρ01 1.200 6

⇐

and the stagnation temperature leaving the blower is T02 = T01 +

1458 w = 294.15 + = 295.6 K = 22.4 C cp 1004.5

⇐

Exercise 2.10 Air enters a blower from the atmosphere where pressure is 101.3 kPa and temperature is 27 C. Its velocity at the inlet is and with velocity 46 m/s. At the exit the total temperature is 28 C and the velocity is 123 m/s. Assuming that the flow is reversible and adiabatic, determine (a) the change in total pressure in millimeters of water and (b) the change in static pressure, also in millimeters of water. Given: The inlet stagnation pressure is p01 = 101.3 kPa and stagnation temperature and velocity are T01 = 300.15 K

V1 = 46 m/s

T02 = 301.15 K

V2 = 123 m/s

At the exit Find: ∆p0 in millimeters of water. Solution: Work done by the blower is w = cp (T02 − T01 ) = 1004.5(28 − 27) = 1004.5 kJ/kg The stagnation density at the inlet is ρ01 =

p01 101.3 · 103 = = 1.176 kg/m3 RT01 287 · 300.15

The flow is isentropic and the pressure rise is small. Therefore ws =

∆p0 ρ01

∆p0 = ρ01 w = 1.176 · 1.0045 = 1.181 kPa

In millimeters of water this is ∆H =

∆p0 1182 = = 0.121 mm ρw g 998 · 9.81

⇐ (a)

To calculate the difference in the static pressure, first calculate 1 462 p1 = p01 − ρ01 V12 = 300.15 − = 100.06 kPa 2 2 · 1004.5 7

The static temperatures are T1 = T01 −

462 V12 = 300.15 − = 299.10 K 2cp 2 · 1004.5

T2 = T02 −

1232 V22 = 301.15 − = 293.62 K 2cp 2 · 1004.5

The exit stagnation pressure is p02 = 101.300 + 1.182 = 102.482 kPa. and the static pressure is p2 = p02

(

T2 T02

)γ/(γ−1)

= 102.48

(

293.62 301.15

)3.5

= 93.789 kPa

and thus ∆Hs =

p2 − p1 93.789 − 100.06 = = −641 mm ρw g 998 · 9.81

⇐

(b)

Exercise 2.11 The total pressure, static pressure and the total temperature of air at a certain point in a flow are 700 kPa, 350 kPa, and 450 K, respectively. Find the velocity at that point. Given: The properties are T0 = 450 K

p = 350 kPa

p0 = 700 kPa

Find: The velocity Solution: T = T0

(

p p0

)(γ−1)/γ

= 450

(

350 700

)1/3.5

= 369.15 K

and √ √ V = 2cp (T0 − T ) = 2 · 1004.5(450 − 369.15) = 403.0 m/s

Exercise 2.12 Air has a static pressure of 2 bar and static temperature 300 K while it flows at speed 1000 m/s. (a) Assuming air obeys the ideal gas law with constant specific heats, determine its stagnation temperature and stagnation pressure. (b) Repeat part (a) using the air tables. 8

Given: The properties are T1 = 300 K

p1 = 2 bar

V1 = 1000 m/s

Find: p01 and T01 Solution: 10002 1 2c = 300 + = 797.8 K ⇐ (a) p V12 2 · 1004.5 ( ( )γ/(γ−1) )3/5 T01 797.8 = p1 = 200 = 61.33 bar ⇐ (a) T1 300

T01 = T1 + p01

Using air tables: At the inlet pr1 = 1.386 h01

V12 10002 =h+ = 300.19 + = 800.19 kJ/kg 2 2cdot1000

⇐

(b)

Thus with the same entropy and this stagnation enthalpy pr0 = 43.38, and T01 = 780.15 K. Therefore p01 = p1

pr0 43.38 =2 = 6.26 bar pr 1.386

⇐

(b)

Exercise 2.13 At a certain location the velocity of air flowing in a duct is 321.5 m/s. At that location the stagnation pressure is 700 kPa and stagnation temperature is 450 K. What is the static density at this location. Given: The properties are T01 = 450 K

p1 = 700 kPa

V1 = 321.5 m/s

Find: ρ1 . Solution: 321.52 V12 = 450 − = 398.55 kPa T1 = T01 − 2cp 2 · 1004.5 ( )3.5 )γ/(γ−1) ( 389.55 T1 = 457.66 kPa = 700 p1 = p01 T01 430

The density is then

ρ1 =

p1 398550 = = 4.00 kg/m3 RT1 287 · 457.66 9

⇐

Exercise 2.14 Air flows in a circular duct of diameter 4 cm at the rate of 0.5 kg/s. The flow is adiabatic with stagnation temperature 288 K. At certain location the static pressure is 110 kPa. Find the velocity at this location. Given: The properties are T0 = 288 K

p = 110 kPa

m ˙ = 0.5 kg/s

D = 0.04 m

Find: The velocity Solution: Solving m ˙ = ρAV = pAV /RT for temperature gives T =

pAV mR ˙

Since T = T0 − V 2 /2cp a quadratic equation for velocity if found V2+

2cp pA V − 2cp T0 = 0 mR ˙

the solution of which is cp pA V =− mR ˙

(

1−

√

2T0 m ˙ 2 R2 1+ cp p2 A2

)

The terms are 2 · 288 · 0.52 · 2872 2cp m ˙ 2 R2 = = 0.618 cp p2 A2 1004.5 · 1100002 (π · 0.022 )2

Therefore

cp RA 10004.5 · 110000 · π · 0.02 = = 967.61 mR ˙ 0.5 · 287 V = −967.61(1 −

√

1 + 0.6180) = 263.2 m/s

⇐

Exercise 2.15 Saturated steam enters a nozzle at static pressure 14 bar at velocity 52 m/s. It expands isentropically to pressure 8.2 bar. Mass flow rate is m ˙ = 0.7 kg/s. Find the exit area if, (a) steam is assumed to behave as an ideal gas with γ = 1.135, and cp = 2731 J/kg K; (b) the end state is calculated with properties obtained from the steam tables.

10

Given: The properties are T1 = 468.2 K

p1 = 14 bar

V1 = 52 m/s

m ˙ = 0.7 kg/s

γ = 1.135

Find: A2 Solution: From steam tables the value of enthalpy is h1 = 2789.4 kJ/kg and the specific volume is V1 = 0.1408 m3 /kg. Then assuming ideal gas behavior with constant specific heats, gives T01 = T1 +

522 V12 = 468.2 + = 468.7 K 2cp 2 · 2731

and the stagnation pressure is 522 1 = 14.096 bar p01 = p1 + ρ1 V12 = 14 · 105 + 2 2 · 0.1408 State 2 has p2 = 820 bar, so that ( )(γ−1)/γ )0.135 ( p2 820 T2 = T 1 1.135 = 439.3 K = 468.2 p1 1400 Since no work is done T02 = T01 and √ √ V2 = 2cp (T02 − T2 ) = 2 · 2731(468.7 − 439.3) = 400.5 m/s The specific volume is v2 = v1

(

p2 p1

)1/γ

= 0.1408

(

1400 820

)

1/1.315 = 0.2256 m3 /kg

Hence A = mv ˙ 2 /V2 = 0.7 · 0.2256/400.5 = 3.944 cm2

⇐

Using steam tables x2 =

s2 − sf 6.4684 − 2.0559 = = 0.9596 sg − Sf 6.6546 − 2.0559

h2 = hf + x2 hf g = 725.6 + 0.9596(2770.0 − 725.36) = 2687.0 kJ/kg

v2 = vf + x2 vf g = 1.116 · 10−3 + 0.9596(0.2354 − 1.116 · 10−3) = 0.2259 m3 /kg 11

Therefore V2 so that

√ √ 2(h02 − h2 ) = 2(2791.0 − 2687.0 = 454.9 m2 /s

A = mv ˙ 2 /V2 = 0.7 · 0.2259/54509 = 0.3469 cm2

⇐

Exercise 2.16 A fluid enters a turbine with total temperature of 330 K and total pressure of 700 kPa. The outlet total pressure is 100 kPa. If the expansion process through the turbine is isentropic, evaluate (a) the work per unit mass flow if the fluid is incompressible and having a density equal to 1000 kg/m3 , (b) the work per unit mass flow if the fluid is air Given: The properties are T01 = 330 K

p01 = 700 kPa

p02 = 100 kPa

Find: w for water and air. Solution: Since the process is isentropic, for water ws =

700 − 100 p01 − p02 = = 0.6 kJ/kg ρ 998

For air T02 = T01

(

p02 p01

)(γ−1)/γ

= 330

(

100 700

)1/3/5

⇐

= 189.3 K

so that ws = h01 − h02 = cp (T01 − T02 ) = 1.0045(330 − 189.3) = 141.4 kJ/kg

⇐

Note that the air temperature leaving the turbine is very low, and operation such as this would be unusual. It might happen in an expander in cryogenic applications. Exercise 2.17 Air flows through a turbine which has a total pressure ratio 5 to 1. The total-to-total efficiency is 80% and the flow rate is 1.5 kg/s. The desired output power is to be 250 hp(186.4 kW). Determine: (a) The inlet total temperature; (b) The outlet total temperature; (c) the outlet static temperature if the exit velocity is 90 m/s; (d) draw the process on a T S-diagram and determine the total-to-static efficiency of the turbine.

12

Given: The properties of air are p01 /p02 = 5

ηtt = 0.8

m ˙ = 1.5 kg/s

˙ = 186.4 kW W

V2 = 90 m/s

Find: T01 , T02 , T2 , the TS-diagram and ηts . Solution: w=

˙ 186.4 W = = 124.27 kJ/kg m ˙ 1.5

ws =

w 124.27 = = 155.33 kJ/kg ηtt 0.8

ws = cp (T01 − T02s )

T01 =

ws 155.33 = = 419.51 K (γ−1)/γ cp [1 − (p02 /p01 ) ] 1.0045(1 − 5−1/3.5 ) T02 = T01 − T2 = T02 −

w 124.27 = 419.51 − = 295.80 K cp 1.0045

Assuming that V2s = V2 T2s = T02s − Thus ηts =

⇐

V22 902 = 295.80 − = 291.76 K 2cp 2 · 1.0045

T02s = T01 −

⇐

⇐

155.33 w = 419.51 − = 264.9 K 2cp 2 · 1.0045 V2s2 902 = 260.8 K = 264.9 − 2cp 2 · 1.0045

419.51 − 295.80 T01 − T02 = = 0.78 T01 − T2s 419.51 − 260.8

⇐

Exercise 2.18 A blower has a change in total enthalpy of 6000 J/kg, an inlet total temperature 288 K, and inlet total pressure 101.3 kPa. Find: (a) the exit total temperature if the working fluid is air; (b) the total pressure ratio across the machine if the total-to-total efficiency is 75%. Given: The properties of air are h02 − h01 = 6000 J/kg

T01 = 288 K

Find: T02 , p02 /p01 .

13

p01 = 101.3 kPa

ηtt = 0.75

Solution: T02 = T01 +

6000 h02 − h01 = 288 + = 294.0 K cp 1004.5

Next ηtt =

⇐

T02s − T01 T02 − T01

T02s = T01 + ηtt (T02 − T01 ) = 288 + 0.75 · 6 = 292.5 K so that p02 = p01

(

T02s T01

)γ/(γ−1)

=

(

292.5 288

)3/5

= 1.0555

⇐

Exercise 2.19 A multi-stage turbine has a total pressure ratio 2.5 across each of four stages. The inlet total temperature is T01 = 1200 K and the total-to-total efficiency of each stage is 0.87. Evaluate the overall total-to-total efficiency of the turbine if steam is flowing through it. Steam can be assumed to behave as a perfect gas with γ = 1.3. Why is the overall efficiency higher than the stage efficiency? Given: The properties of air are p02 = 2.5 p01

T01 = 1200 K

ηtt = 0.87

Find: the total-to-total efficiency for the multistage turbine. Solution: p01 p02 p03 p04 p01 = = 2.54 = 39 p05 p02 p03 p04 p05 Also [ ( )γ/(γ−1) ] p02 T02 = T01 1 − ηtt p01 and since the efficiency of each stage is the same [ ( )γ/(γ−1) ] p02 T03 = T02 1 − ηtt p01 [

T03 = T01 1 − ηtt 14

(

p02 p01

)γ/(γ−1) ]2

and so forth to [

T05 = T01 1 − ηtt

(

p02 p01

)γ/(γ−1) ]4

so that [

(

T05 = 1200 1 − 0.87 1 −

1 2.50.3/1.3

)]4

= 581.1 K

The overall efficiency is therefore η0 =

1 − T05 /T01 T01 − T05 = T01 − T05s 1 − (p05 /p01 )γ/(γ−1)

η0 =

1 − 581.1/1200 = 0.904 1 − (1/2.54 )0.3/1.3

The overall efficiency is larger than the stage efficiency because the internal energy at the exit of each stage is higher because of internal heating and this becomes available during the next expansion process. Exercise 2.20 Gases from a combustion chamber enter a gas turbine at total pressure of 700 kPa and total temperature of 1100 K. The total pressure and total temperature at the exit of the turbine are 140 kPa and 780 K. If γ = 4/3 is used for the mixture of combustion gases, which has a molecular mass of 28.97, find the total-to-total efficiency and the total-to-static efficiency of the turbine, if the exit velocity is 210 m/s. Given: The inlet and exit conditions are and V2 = 210 m/s. p01 = 700 kPa

p02 = 140 kPa

T01 = 1100 K

T02 = 780 K

Combustion gases with γ = 4/3 and R = 287 J/kg. Find: the total-to-total efficiency. Solution: γR cp = = 4 · 287 = 1148 J/kg γ−1 ( )(γ−1)/γ )1/4 ( p02 140 T02s = T01 = 1100 = 735.6 K p01 700 ηtt =

T01 − T02 1100 − 780 = = 0.878 T01 − T02s 1100 − 735.6 15

⇐

T2s = T02s − ηts =

V2s2 2102 =...