Fundamentals of Chemical Engineering Thermodynamics (MindTap Course List)
Fundamentals of Chemical Engineering Thermodynamics (MindTap Course List)
1st Edition
ISBN: 9781111580704
Author: Kevin D. Dahm, Donald P. Visco
Publisher: Cengage Learning
Question
Book Icon
Chapter 4.8, Problem 29P

(A)

Interpretation Introduction

Interpretation:

The mass flow rate (m˙in) entering the turbine

Concept Introduction:

Write the expression to calculate the mass flow rate (m˙in) entering the turbine.

m˙in=VinAV^=V(πD2/4)V^

Here, initial velocity of turbine is Vin, area of turbine is A, specific volume is V^, and diameter of pipe is D.

(A)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Turbine: Pressure of 14bar and temperature of 300°C.

Inlet velocity is 2.5m/s.

Pipe diameter is 5 cm.

Refer the Appendix table A-3, “Superheated steam”, obtain the value of specific volume corresponding to pressure of 14bar and temperature of 300°C.

V^=0.182m3/kg

Calculate the mass flow rate (m˙in) entering the turbine.

m˙in=V(πD2/4)V^        (1)

Substitute 2.5m/s for Vin, 0.182m3/kg for V^, and 0.05m for D in Equation (1).

m˙in=Vin(πD2/4)V^=(2.5m/s)[π(0.05m)2/4]0.182m3/kg=0.270kg/s

Thus, the mass flow rate (m˙in) entering the turbine is 0.270kg/s.

(B)

Interpretation Introduction

Interpretation:

The power output (W˙S) from the turbine

Concept Introduction:

Write the steady state energy balance equation around the turbine.

ddt{M(U^+V22+gh)}=[m˙in(H^in+Vin22+ghin)m˙out(H^out+Vout22+ghout)+W˙S+W˙EC+Q˙]

Here, time taken is t, total mass is M, specific internal energy is U^, velocity is V, acceleration due to gravity is g, height is h, initial mass flow rate  is m˙in, initial specific enthalpy is H^in, rate of heat is added or removed from the system is Q˙, initial velocity is Vin, initial height of the gas is hin, final mass flow rate is m˙out, final height of the gas is hout, rate of shaft work is added to the system is W˙S, and rate of work is added to the system through expansion or contraction of the system is W˙EC.

(B)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Outlet datum is 1.5 m.

Exhaust steam pressure is 0.9 bar.

Calculate the work output using steady state energy balance equation around the turbine.

ddt{M(U^+V22+gh)}=[m˙in(H^in+Vin22+ghin)m˙out(H^out+Vout22+ghout)+W˙S+W˙EC+Q˙]        (2)

Rewrite the steady state Equation (2) by neglecting the heat transfer and work done by expansion or contraction process.

0=m˙in(H^in+Vin22+ghin)m˙out(H^out+Vout22+ghout)+W˙SW˙S=m˙out(H^out+Vout22+ghout)m˙in(H^in+Vin22+ghin)        (3)

Here, initial specific enthalpy is H^in, initial volume is Vin, initial enthalpy is hin, final mass flow rate m˙out, final specific enthalpy H^out, final volume is Vout, and final enthalpy is hout.

Refer the Appendix table A-3, “Superheated steam”, obtain the value of initial specific enthalpy corresponding to pressure of 14bar and temperature of 300°C.

H^in=3,040.9kJ/kg

Refer the Appendix table A-1, “Saturated steam-pressure increments”, obtain the value of final specific enthalpy and specific volume corresponding to pressure of 0.9bar

H^out(@H^V)=2,670.3kJ/kg

V^out(@V^V)=1.869m3/kg

Calculate the velocity while assuming that the mass flow rate is constant.

m˙in=m˙outVinV^in=VoutV^outVinV^in(Ainlet)=VoutV^out(Aoutlet)

VinV^in(πDinlet2/4)=VoutV^out(πDoutet2/4)Vout=(Vin)(πDinlet2/4)(V^out)(V^in)(πDoutet2/4)Vout=(Vin)(Dinlet2)(V^out)(V^in)(Doutet2)        (4)

Here, initial mass is Min, final mass is Mout, initial specific volume is V^in, final velocity is Vout, final specific volume is V^out, area of inlet is Ainlet, area of outlet is Aoutlet, diameter of inlet is Dinlet, and diameter of outlet is Doutlet.

Substitute 2.5m/s for vin, 5cm for Dinlet, 1.869m3/kg for V^out, 0.182m3/kg for V^in, and 20cm for Doutlet in Equation (4).

Vout=(vin)(Dinlet2)(V^out)(V^in)(Doutet2)=(2.5m/s)(5cm)2(1.869m3/kg)(0.182m3/kg)(20cm)2=(2.5m/s)[((5cm)0.01m/1cm)]2(1.869m3/kg)(0.182m3/kg)[(20cm)(0.01m/1cm)]2

=(2.5m/s)(0.05m)2(1.869m3/kg)(0.182m3/kg)(0.2m)2Vout=1.605m/s

Substitute 0.270kg/s for m˙out, 2,670.3kJ/kg for H^out, 1.605m/s for Vout, 9.8m/s2 for g, 1.5m for hout. 0 for ghin, 0.270kg/s for m˙in, 3,040.9kJ/kg for H^in, and 2.5m/s for Vin in Equation (3).

W˙S=(0.270kgs){[(2,670.3kJkg)+(1.605ms)22+(9.8ms2×1.4 m)][(3,040.9kJkg)+(2.5ms)22+0]}=(0.270kgs){[(2,670.3kJkg)+1.30m2s2+(14.7m2s2)][(3,040.9kJkg)+3.125m2s2]}=(0.270kgs){{(2,670.3kJkg)+[1.30m2s2(Jkgm2/s2)(1kJ1,000J)]+[(14.7m2s2(Jkgm2/s2)(1kJ1,000J))]}[(3,040.9kJkg)+[(3.125m2s2(Jkgm2/s2)(1kJ1,000J))]]}=(0.270kgs){[(2,670.3kJkg)+0.0013kJkg+0.0147kJkg][(3,040.9kJkg)+0.00313kJkg]}

W˙S=100.05kJ/s

Thus, the power output (W˙S) from the turbine is 100.05kJ/s.

(C)

Interpretation Introduction

Interpretation:

The efficiency of the turbine

Concept Introduction:

Write the expression to calculate the efficiency of the turbine (η).

η=W˙SW˙S,rev

Here, rate of actual shaft work is W˙S, and rate of reversible shaft work is W˙S,rev.

Write the entropy balance equation around the boiler.

d(MS^)dt=m˙inS^inm˙outS^out+Q˙T+S˙gen

Here, total mass is M, specific entropy is S^, initial specific entropy is S^in, final specific entropy is S^out, and rate at which entropy is generated within the boundaries of the system is S˙gen.

(C)

Expert Solution
Check Mark

Explanation of Solution

Calculate the reversible entropy using entropy balance equation around the boiler.

d(MS^)dt=m˙inS^inm˙outS^out+Q˙T+S˙gen        (5)

Rewrite the Equation (5) by neglecting the surrounding, entropy generation, and reversible entropy term.

0=m˙inS^inm˙outS^out

Here, initial specific entropy is S^in, and final specific entropy is S^out.

Refer the Appendix table A-3, “superheated steam”, write the value of initial specific entropy corresponding to pressure of 14bar and temperature of 300°C.

S^in=6.955kJ/kgK

As the initial and final specific entropy is same,

S^in=S^out=6.955kJ/kgK

Calculate the work output using steady state energy balance equation around the boiler. ddt{M(U^+V22+gh)}=[m˙in(H^in+Vin22+ghin)m˙out(H^out+Vout22+ghout)+W˙S,rev+W˙EC+Q˙]        (6)

Rewrite the steady state Equation (6) by neglecting the heat transfer and work done by expansion or contraction process.

W˙S,rev=m˙out(H^out+Vout22+ghout)m˙in(H^in+Vin22+ghin)        (7)

Refer the Appendix table A-3, “Superheated steam”, obtain the value of final specific enthalpy corresponding to pressure of 1bar and an initial specific entropy of 6.955kJ/kgK.

H^out=2,524.4kJ/kg

Substitute 0.270kg/s for m˙out, 2,524.4kJ/kg for H^out, 1.605m/s for Vout, 14.7m2/s2 for ghout, 0 for ghin, 0.270kg/s for m˙in, 3,040.9kJ/kg for H^in, and 2.5m/s for Vin in Equation (7).

W˙S,rev=(0.270kgs){[(2,524.4kJ/kg)+(1.605ms)22+(14.7m2s2)][(3,040.9kJkg)+(2.5ms)22+0]}=(0.270kgs){[(2,524.4kJ/kg)+1.30m2s2+(14.7m2s2)][(3,040.9kJkg)+3.125m2s2]}=(0.270kgs){{(2,524.4kJ/kg)+[1.30m2s2(Jkgm2/s2)(1kJ1,000J)]+[(14.7m2s2)(Jkgm2/s2)(1kJ1,000J)]}[(3,040.9kJkg)+[(3.125m2s2)(Jkgm2/s2)(1kJ1,000J)]]}=(0.270kgs){[(2,524.4kJ/kg)+0.0013kJkg+0.0147kJkg][(3,040.9kJkg)+0.00313kJkg]}

W˙S,rev=139.45kJ/s

Calculate the efficiency of the turbine (η).

η=W˙SW˙S,rev        (8)

Substitute 100kJ/s for W˙S and 139.4kJ/s for W˙S,rev in Equation (8).

η=W˙SW˙S,rev=100kJ/s139.4kJ/s=0.717=71.7%

Thus, the efficiency of the turbine (η) is 71.7%.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A piston-fitted cylinder with a 5-cm diameter contains 1.65 g of nitrogen. The mass of the piston is 6.50 kg, and a 35.0-kg weight rests on the piston. The gas temperature is 35.0°C, and the pressure outside the cylinder is 3.50 atm. d=5 m₁g N₂ m3kg dcm m2=6.50 m1 1.65 m3 = 35.0 Calculate the pressure in the cylinder. i x 105 Pa eTextbook and Media Save for Later Volume What is the volume of the gas in the cylinder? V = i L m2 kg If the 35.0 kg weight is removed from the piston, how far does the piston move up? i Save for Later m Heat Transferred A How much heat is transferred to (positive) or from (negative) the gas during the expansion? J
chemical engineering The answer for the specific molar volume of nitrogen gas is 12.089x10^(-5) m^3/mol.  How was this answer determined?  You need to use the ideal gas law to determine the specific molar volume.  Do not determine the third specific enthalpy.
Using Raoult's law for water and Henry's law for nitrogen, calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.500 mole% N2 and 99.50 mole% water in equilibrium with nitrogen gas and water vapor at 70.0 °C. The Henry's law constant for nitrogen in water is recommended by NIST to be well represented by KH = 0.000625 exp[1300 (1/T - 1/298.15)] mol N2/(kg H2O bar), where T is measured in Kelvin. Physical Property Tables Unit Conversion Check the unit conversions and examine the definition of H. Estimate the Henry's law constant H [atm/(mole fraction N2)] for nitrogen in water at T = 70.0 °C. i ! x 104 atm/(mole fraction N2)
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The