Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 6, Problem 161RQ

(a)

To determine

The mass flow rate of the steam.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The inlet pressure (P1) is 7 MPa.

The inlet temperature (T1) is 600°C.

The inlet velocity (V1) is 60 m/s.

The outlet pressure (P1) is 25 kPa.

The quality of the steam at exit (x2) is 95%.

The heat lost to the surroundings (Q) is 20 kJ/kg.

The area of inlet (Ainlet) is 150 cm2

The area of outlet (Aout) is 1400cm2

The outlet temperature (T1) is 600°C.

Calculation:

At inlet:

The steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

Obtain the inlet enthalpy (h1) and specific volume (v1) corresponding to the pressure of 7MPa and the temperature of 600°C.

h1=3650.6kJ/kgv1=0.055665m3/kg

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Calculate the inlet mass flow rate.

  m˙=A1V1v1

  m˙=(150cm2)(60m/s)0.055665m3/kg=(150cm2×1m2104cm2)(60m/s)0.055665m3/kg=16.1681kg/s16.17kg/s

Thus, the mass flow rate of the steam is 16.17kg/s.

(b)

To determine

The exit velocity of the steam.

(b)

Expert Solution
Check Mark

Explanation of Solution

At exit:

The steam is with the quality of 95%.

Refer Table A-5, “Saturated water—Pressure table”.

Obtain the following corresponding to the pressure of 25kPa.

vf=0.001020m3/kgvg=6.2034m3/kghf=271.96kJ/kghfg=2345.5kJ/kg

Write the formula for exit enthalpy (h2) of the steam with the consideration of its quality.

  h2=hf+xhfg

  h2=271.96kJ/kg+95%(2345.5kJ/kg)=271.96kJ/kg+95100(2345.5kJ/kg)=271.96kJ/kg+2228.225kJ/kg=2500.185kJ/kg

Write the formula for exit specific volume (v2) of the steam with the consideration of its quality.

  v2=vf+x(vgvf)

  v2=0.001020m3/kg+95%(6.2034m3/kg0.001020m3/kg)=0.001020m3/kg+95100(6.20238m3/kg)=0.001020m3/kg+5.892261m3/kg=5.893281m3/kg

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the formula for exit mass flow rate.

  m˙=A2V2v2

Rearrange the above Equation to obtain exit velocity (V2).

  V2=m˙v2A2

  V2=(16.17kg/s)(5.893281m3/kg)1400cm2=(16.17kg/s)(5.893281m3/kg)1400cm2×1m2104cm2=680.6739m/s

Thus, the exit velocity of the steam is 680.6739m/s.

(c)

To determine

The power output of the turbine.

(c)

Expert Solution
Check Mark

Answer to Problem 161RQ

The power output of the turbine is 14562kW.

Explanation of Solution

Consider the steam flows at steady state. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the energy rate balance equation for one inlet and one outlet system.

  [Q˙1+W˙1+m˙(h1+V122+gz1)][Q˙2+W˙2+m˙(h2+V222+gz2)]=ΔE˙system        (VI)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The refrigerant flows at steady state through the compressor. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

Heat loss occurs at the rate of 20kJ/kg. Neglect the potential energy changes. Here, the work done is by the system (turbine) and the work done on the system is zero i.e. W˙1=0.

The Equations (VI) reduced as follows to obtain the work output (W˙2).

  [0+0+m˙(h1+V122+0)][Q˙2+W˙2+m˙(h2+V222+0)]=0m˙(h1+V122)[Q˙2+W˙2+m˙(h2+V222)]=0m˙(h1+V122)=Q˙2+W˙2+m˙(h2+V222)W˙2=m˙(h1+V122)m˙(h2+V222)Q˙2

  W˙2=m˙(h1h2+V12V222)Q˙2W˙2=m˙(h2h1+V22V122)Q˙2        (VII)

Here, Q˙2=m˙Q2.

Rewrite the Equation (VII) as follows.

  W˙2=m˙(h2h1+V22V122)m˙Q2        (VIII)

  W˙2=16.17kg/s[2500.185kJ/kg3650.6kJ/kg+(680.6739m/s)2(60m/s)22]16.17kg/s(20kJ/kg)=[16.17kg/s(1150.415kJ/kg+229858.4791m2/s2×1kJ/kg1000m2/s2)323.4kJ/s]=16.17kg/s(920.5565kJ/kg)323.4kJ/s=14885.3989kJ/s323.4kJ/s

  =14561.9989kJ/s×1kW1kJ/s14562kW

Thus, the power output of the turbine is 14562kW.

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!

Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

Ch. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - A house is maintained at 1 atm and 24°C, and warm...Ch. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - The kinetic energy of a fluid increases as it is...Ch. 6 - Prob. 23PCh. 6 - Air enters a nozzle steadily at 50 psia, 140°F,...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Air at 13 psia and 65°F enters an adiabatic...Ch. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Steam at 4 MPa and 400°C enters a nozzle steadily...Ch. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Helium is to be compressed from 105 kPa and 295 K...Ch. 6 - Carbon dioxide enters an adiabatic compressor at...Ch. 6 - Air is compressed from 14.7 psia and 60°F to a...Ch. 6 - Prob. 47PCh. 6 - An adiabatic gas turbine expands air at 1300 kPa...Ch. 6 - Steam flows steadily into a turbine with a mass...Ch. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Refrigerant-134a is throttled from the saturated...Ch. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - A thin-walled double-pipe counter-flow heat...Ch. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - The components of an electronic system dissipating...Ch. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - A house has an electric heating system that...Ch. 6 - Prob. 95PCh. 6 - Refrigerant-134a enters the condenser of a...Ch. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 100PCh. 6 - Air enters the duct of an air-conditioning system...Ch. 6 - Prob. 102PCh. 6 - A rigid, insulated tank that is initially...Ch. 6 - Prob. 105PCh. 6 - Prob. 106PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - An air-conditioning system is to be filled from a...Ch. 6 - Prob. 111PCh. 6 - A 0.06-m3 rigid tank initially contains...Ch. 6 - A 0.3-m3 rigid tank is filled with saturated...Ch. 6 - Prob. 114PCh. 6 - A 0.3-m3 rigid tank initially contains...Ch. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - The air in a 6-m × 5-m × 4-m hospital room is to...Ch. 6 - Prob. 124RQCh. 6 - Prob. 125RQCh. 6 - Prob. 126RQCh. 6 - Prob. 127RQCh. 6 - Prob. 128RQCh. 6 - Prob. 129RQCh. 6 - Prob. 130RQCh. 6 - Prob. 131RQCh. 6 - Prob. 132RQCh. 6 - Steam enters a nozzle with a low velocity at 150°C...Ch. 6 - Prob. 134RQCh. 6 - Prob. 135RQCh. 6 - Prob. 136RQCh. 6 - In large steam power plants, the feedwater is...Ch. 6 - Prob. 138RQCh. 6 - Prob. 139RQCh. 6 - Prob. 140RQCh. 6 - Prob. 141RQCh. 6 - Prob. 142RQCh. 6 - Prob. 143RQCh. 6 - Prob. 144RQCh. 6 - Prob. 145RQCh. 6 - Prob. 146RQCh. 6 - Repeat Prob. 6–146 for a copper wire ( = 8950...Ch. 6 - Prob. 148RQCh. 6 - Prob. 149RQCh. 6 - Prob. 150RQCh. 6 - Prob. 151RQCh. 6 - Prob. 152RQCh. 6 - Prob. 153RQCh. 6 - An adiabatic air compressor is to be powered by a...Ch. 6 - Prob. 156RQCh. 6 - Prob. 157RQCh. 6 - Prob. 158RQCh. 6 - Prob. 159RQCh. 6 - Prob. 160RQCh. 6 - Prob. 161RQCh. 6 - Prob. 162RQCh. 6 - Prob. 163RQCh. 6 - Prob. 164RQCh. 6 - Prob. 166RQCh. 6 - Prob. 167RQCh. 6 - Prob. 168RQCh. 6 - Prob. 169RQ
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Power Plant Explained | Working Principles; Author: RealPars;https://www.youtube.com/watch?v=HGVDu1z5YQ8;License: Standard YouTube License, CC-BY