Fundamentals Of Engineering Thermodynamics, 9e
9th Edition
ISBN: 9781119391432
Author: MORAN
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 4, Problem 4.7CU
To determine
The condition for a flow to be idealized as a throttling process.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Note:-
• Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism.
• Answer completely.
• You will get up vote for sure.
a
(a)
Air
in
(b)
(c)
Diffuser
Compressor
Combustors
2
State
State 1 80
State 2
3300
State 3 3200
State 4
400
State 5
80
wwww
Pressure (kPa)
3
Turbine
4
Figure 1: Figure for Problem 2.
In a modern jet engine, air passes through the following states from the inlet to the outlet,
as shown in Figure 1:
Product
gases out
260
780
1500
900
640
K
5
>
Nozzle
justify them) to find the compressor specific work.
Temperature (K)
For this problem, you may neglect any heat transfer, as well as neglect kinetic energy except
at the outlet (state 5). Use the tables for obtaining properties (do not assume constant
specific heat). Assume air is an ideal gas with ideal gas constant of R = 0.287 kJ/kg-K.
Use the appropriate conservation equations and make approximations (and
In a similar manner, find the turbine specific work.
And finally, using similar arguments, find the nozzle exit velocity.
substance wWhose properties are uniform throughout a sample is referred to as
a. A solid
b. An ideal substance
c. A pure substance
d. A standard substance
6. Which of the following statements is true for an isentropic steady-flow process?
a.
P1
b. Work = -V (P2 – P1)
C.
V2
d. 4= 2
T1
T2
Chapter 4 Solutions
Fundamentals Of Engineering Thermodynamics, 9e
Ch. 4 - Prob. 4.1ECh. 4 - Prob. 4.2ECh. 4 - Prob. 4.3ECh. 4 - Prob. 4.4ECh. 4 - Prob. 4.5ECh. 4 - Prob. 4.6ECh. 4 - Prob. 4.7ECh. 4 - Prob. 4.8ECh. 4 - Prob. 4.9ECh. 4 - Prob. 4.10E
Ch. 4 - Prob. 4.11ECh. 4 - Prob. 4.12ECh. 4 - Prob. 4.13ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.15ECh. 4 - Prob. 4.1CUCh. 4 - Prob. 4.2CUCh. 4 - Prob. 4.3CUCh. 4 - Prob. 4.4CUCh. 4 - Prob. 4.5CUCh. 4 - Prob. 4.6CUCh. 4 - Prob. 4.7CUCh. 4 - Prob. 4.8CUCh. 4 - Prob. 4.9CUCh. 4 - Prob. 4.10CUCh. 4 - Prob. 4.11CUCh. 4 - Prob. 4.12CUCh. 4 - Prob. 4.13CUCh. 4 - Prob. 4.14CUCh. 4 - Prob. 4.15CUCh. 4 - Prob. 4.16CUCh. 4 - Prob. 4.17CUCh. 4 - Prob. 4.18CUCh. 4 - Prob. 4.19CUCh. 4 - Prob. 4.20CUCh. 4 - Prob. 4.21CUCh. 4 - Prob. 4.22CUCh. 4 - Prob. 4.23CUCh. 4 - Prob. 4.24CUCh. 4 - Prob. 4.25CUCh. 4 - Prob. 4.26CUCh. 4 - Prob. 4.27CUCh. 4 - Prob. 4.28CUCh. 4 - Prob. 4.29CUCh. 4 - Prob. 4.30CUCh. 4 - Prob. 4.31CUCh. 4 - Prob. 4.32CUCh. 4 - Prob. 4.33CUCh. 4 - Prob. 4.34CUCh. 4 - Prob. 4.35CUCh. 4 - Prob. 4.36CUCh. 4 - Prob. 4.37CUCh. 4 - Prob. 4.38CUCh. 4 - Prob. 4.39CUCh. 4 - Prob. 4.40CUCh. 4 - Prob. 4.41CUCh. 4 - Prob. 4.42CUCh. 4 - Prob. 4.43CUCh. 4 - Prob. 4.44CUCh. 4 - Prob. 4.45CUCh. 4 - Prob. 4.46CUCh. 4 - Prob. 4.47CUCh. 4 - Prob. 4.48CUCh. 4 - Prob. 4.49CUCh. 4 - Prob. 4.50CUCh. 4 - Prob. 4.51CUCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.31PCh. 4 - Prob. 4.32PCh. 4 - Prob. 4.33PCh. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Prob. 4.37PCh. 4 - Prob. 4.38PCh. 4 - Prob. 4.39PCh. 4 - Prob. 4.40PCh. 4 - Prob. 4.41PCh. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - Prob. 4.48PCh. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - Prob. 4.51PCh. 4 - Prob. 4.52PCh. 4 - Prob. 4.53PCh. 4 - Prob. 4.54PCh. 4 - Prob. 4.55PCh. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Prob. 4.58PCh. 4 - Prob. 4.59PCh. 4 - Prob. 4.60PCh. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - Prob. 4.63PCh. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Prob. 4.66PCh. 4 - Prob. 4.67PCh. 4 - Prob. 4.68PCh. 4 - Prob. 4.69PCh. 4 - Prob. 4.70PCh. 4 - Prob. 4.71PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - Prob. 4.77PCh. 4 - Prob. 4.78PCh. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - Prob. 4.81PCh. 4 - Prob. 4.82PCh. 4 - Prob. 4.83PCh. 4 - Prob. 4.84PCh. 4 - Prob. 4.85PCh. 4 - Prob. 4.86PCh. 4 - Prob. 4.87PCh. 4 - Prob. 4.88P
Knowledge Booster
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.Similar questions
- Phase change occurs atarrow_forwardSolve for mas flow rate, and compressor power in horsepower. Step by step solution please thank youarrow_forwardGIVEN: Gas HAS CONSTANT MASS IDEAL GAS SPECIFIC HEAT RATIO OF GAS Y= 1.6 GAS CYCLE P^29 At 1: P1 = 2 BAR At 2 + 2 = 300° K : At 3: ASK: A) B) c) P2 V 2 P3 T2 = 12: 23: 31: T3 ISENTRO PIC REVERSIBLE VOLUME IS ISOTHERMAL EXPANSION CONSTANT T1 = 149.8 x 150⁰ K V₁ = √3 3 V₁ = 0.0005 marrow_forward
- For a certain gas with constant, R = 320 J/(kg-K) and constant volume specific heat, Cv = 184 J/(kg-K), what is the value of constant pressure specific heat, Cp if the system undergoes a reversible non flow constant pressurearrow_forwardAssume 5.08 lb/sec of fluid enter a steady- state, steady-flow system with P1 = 97.71 psia, p1 = 0.309 lb/ft, v1 = 94.19 fps, u1 = 793.23 BTU/lb, and leave with P, = 15.54 %3D %3D %3! %3D %3D psia, p2 = 0.148 Ib/ft°, v2 = 470.44 fps, and u2 = 772.14 BTU/lb. During the passage %3D %3D through the open system, each pound rejects 12.4 BTU of heat. Determine the work in hp.arrow_forwardFinding exit temperature in F, and volumetric flow rate. Step by step solution please thank youuuarrow_forward
- a. Find an appropriate expression for the change in entropy in the following two cases: 1) S=S(TV) 2) S=S(TP) Where: S is entropy. T is temperature. V is volume, P is pressure b. Prove the following two thermodynamic property relationships ac Where: T. P. V are temperature, pressure and volume, respectively. C. and C, are specific heats at constant volume and constant pressure, respectively.arrow_forwardExactly 10 kg of steam entered a heat exchanger at 100 bar and 560o C. At the exit the Steam was at 30 bar and a quality of 0.47. Please determine a. The exit temperature in oC. b. The mass of the liquid water in kg.arrow_forwardCHR 6. An inventor comes to you with a device that produces work. She tells you helium gas at 10 atm and 650 K enters the device and exits at 2 atm and 260 K. Assume an ideal gas with constant specific heat and the surroundings are at 25 C, Cp = 5.2 kJ/kg-K. Is this process possible assuming steady-state operation?arrow_forward
- I need some help in how to solve this problem. Thank!!!arrow_forwardGas supply line- Gas A 2.1 m3 tank initially contains a gas at 170 kPa, 370 K. The tank is connected a line carrying same gas at 1700 kPa and 370 K. What should be heat transfer during this process, in kJ, so that the final pressure and temperature are 1700 kPa and 370 K, respectively. (cp=D0.92 kJ/kg.K, c=0.64 kJ/kg.K).arrow_forwardCompressed natural gas, which can be modelled as methane, (CHa, Molecular weight-16, C,=2254 J/kgK; C=1738 J/kgK) is filled into a rigid 100 litre tank of a vehicle. The initial state of gas in the tank is 1 bar, 27°C, and the final pressure is 150 bar. The filling is done from a large reservoir at 300 bar, 27°C. Heat ioss during filling can be neglected, and the gas can be treated as ideal. Compute the final mass and temperature of the gas in the tank.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University Press
Mechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSON
Thermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEY
Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Engineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Fluid Mechanics - Viscosity and Shear Strain Rate in 9 Minutes!; Author: Less Boring Lectures;https://www.youtube.com/watch?v=_0aaRDAdPTY;License: Standard youtube license