Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 17.7, Problem 85P

Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma = 2.5. If the pressure and temperature of air are 10.0 psia and 440.5 R, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions.

Expert Solution & Answer
Check Mark
To determine

The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock and for helium undergoing a normal shock under the same conditions.

Answer to Problem 85P

The actual temperature of air after the normal shock is 942R.

The actual pressure of air after the normal shock is 71.25psia.

The stagnation pressure of air after the normal shock is 85.3psia.

The Mach number value of air after the normal shock is 0.513.

The velocity of air after the normal shock is 772ft/s.

The Mach number of helium gas after the normal shock is 0.553.

The actual temperature of helium after the normal shock is 1233R.

The actual pressure of helium after the normal shock is 75.6psia.

The stagnation pressure of air after the normal shock is 115.46psia.

The velocity of air after the normal shock is 2794ft/s.

Explanation of Solution

Refer Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with k=1.4”, write the following expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.5.

T2T1=2.1375 (I)

P2P1=7.125 (II)

P02P01=8.5262 (III)

Ma2=0.513

Here, actual temperature after the shock is T2, actual temperature before the shock is T1, actual pressure after the shock is P2, actual pressure before the shock is P1, stagnation pressure  after the shock is P02, stagnation pressure before the shock is P01, and Mach number after the shock is Ma2.

Write the expression to calculate the velocity of sound after the normal shock.

c2=kRT2 (IV)

Here, velocity of sound after the shock is c2, and gas constant of air is R.

Write the expression to calculate the velocity of air after the normal shock.

V2=Ma2c2 (V)

The value from the table is not considered for Mach number (Ma2), because the value for ratio of specific heats k is not 1.4 for helium.

Write the expression to calculate the Mach number for helium after the normal shock.

Ma2=(Ma12+2/(k1)2Ma12k/(k1)1)1/2 (VI)

Here, Mach number of helium before the normal shock is Ma1, Mach number of helium before the normal shock is Ma2, and ratio of specific heats for helium is k.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

P2P1=1+kMa121+kMa22 (VII)

Here, actual pressure of helium after the shock is P2, and actual pressure of helium before the shock is P1.

Write the expression to calculate the actual temperature of helium gas after the normal shock.

T2T1=1+Ma12(k1)/21+Ma22(k1)/2 (VIII)

Here, actual temperature of helium after the shock is T2, and actual temperature of helium before the shock is T1.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

P02P01=1+kMa121+kMa22(1+(k1)Ma222)k/(k1) (IX)

Here, stagnation pressure of helium after the shock is P02, and stagnation pressure of helium before the shock is P01.

Write the expression to calculate the velocity of sound after the normal shock for helium.

c2,helium=kRT2 (X)

Here, velocity of sound after the shock for helium is c2,helium, and gas constant of helium is R.

Write the expression to calculate the velocity of helium after the normal shock V2,helium.

V2,helium=Ma2c2,helium (XI)

Conclusion:

For air:

Refer Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following properties for air at room temperature.

R=0.06855Btu/lbmRk=1.4

Substitute 440.5R for T1 in Equation (I).

T2440.5R=2.1375

T2=440.5R×2.1375=941.56R=942R

Thus, the actual temperature of air after the normal shock is 942R.

Substitute 10psia for P1 in Equation (II).

P210psia=7.125P2=10psia×7.125P2=71.25psia

Thus, the actual pressure of air after the normal shock is 71.25psia.

The actual pressure before the normal shock (P1) is the equal as the stagnation pressure before the normal shock (P01), since the flow through the nozzle is isentropic,

Substitute 10psia for P01 in Equation (III).

P0210psia=8.5262

P2=10psia×8.5262=85.262psia=85.3psia

Thus, the stagnation pressure of air after the normal shock is 85.3psia.

From Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with k=1.4”, write the following expressions of Mach number after the shock for a Mach number of 2.5.

Ma2=0.513

Thus, the Mach number value of air after the normal shock is 0.513.

Substitute 1.4 for k, 0.06855Btu/lbmR for R, and 942R for T2 in Equation (IV).

c2=1.4×0.06855Btu/lbmR×942R=1.4×0.06855Btu/lbmR×942R(25,037ft2/s21Btu/lbm)=1504.47ft/s

Substitute 0.513 for Ma2, and 1504.47ft/s for c2 in Equation (V).

V2=0.513×1504.47ft/s=771.7ft/s=772ft/s

Thus, the velocity of air after the normal shock is 772ft/s.

For helium:

Refer Table A-E, “Ideal-gas specific heats of various common gases”, obtain the following properties for helium.

R=0.4961Btu/lbmRk=1.667

Substitute 2.5 for Ma1, and 1.667 for k in Equation (VI).

Ma2=(2.52+2/(1.6671)2×2.52×1.667/(1.6671)1)1/2=(9.24830.24)1/2=0.553

Thus, the Mach number of helium gas after the normal shock is 0.553.

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (VII).

P2P1=1+1.667×(2.5)21+1.667×(0.553)2

P2P1=7.5631 (XII)

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (VIII).

T2T1=1+(2.5)2(1.6671)/21+(0.553)2(1.6671)/2

T2T1=2.7989 (XIII)

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (IX).

P02P01=1+1.667×(2.5)21+1.667×(0.553)2(1+(1.6671)(0.553)22)1.667/(1.6671)

P02P01=11.546 (XIV)

Substitute 10psia for P1 in Equation (XII).

P210psia=7.5632P2=10psia×7.5632P2=75.6psia

Thus, the actual pressure of helium after the normal shock is 75.6psia.

Substitute 440.5R for T1 in Equation (XIII).

T2440.5R=2.7989

T2=440.5R×2.7989=1232.91R=1233R

Thus, the actual temperature of helium after the normal shock is 1233R.

Since the flow through the nozzle is isentropic (P01=P1).

Substitute 10psia for P01 in Equation (XIV).

P0210psia=11.546P02=10psia×11.546P02=115.46psia

Thus, the stagnation pressure of air after the normal shock is 115.46psia.

Substitute 1.667 for k, 0.4961Btu/lbmR for R, and 1233R for T2 in Equation (X).

c2,helium=1.667×0.4961Btu/lbmR×1233R=1.667×0.4961Btu/lbmR×1233R(25,037ft2/s21Btu/lbm)=5052.72ft/s

Substitute 0.553 for Ma2, and 5052.72ft/s for c2 in Equation (XI).

V2,helium=0.553×5052.72ft/s=2794.15ft/s=2794ft/s

Thus, the velocity of air after the normal shock is 2794ft/s.

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
-6- 8 من 8 Mechanical vibration HW-prob-1 lecture 8 By: Lecturer Mohammed O. attea The 8-lb body is released from rest a distance xo to the right of the equilibrium position. Determine the displacement x as a function of time t, where t = 0 is the time of release. c=2.5 lb-sec/ft wwwww k-3 lb/in. 8 lb Prob. -2 Find the value of (c) if the system is critically damping. Prob-3 Find Meq and Ceq at point B, Drive eq. of motion for the system below. Ш H -7~ + 目 T T & T тт +
Q For the following plan of building foundation, Determine immediate settlement at points (A) and (B) knowing that: E,-25MPa, u=0.3, Depth of foundation (D) =1m, Depth of layer below base level of foundation (H)=10m. 3m 2m 100kPa A 2m 150kPa 5m 200kPa B
W PE 2 43 R² 80 + 10 + kr³ Ø8=0 +0 R²+J+ kr200 R² + J-) + k r² = 0 kr20 kr20 8+ W₁ = = 0 R²+1) R²+J+) 4 lec 8.pdf Mechanical vibration lecture 6 By: Lecturer Mohammed C. Attea HW1 (Energy method) Find equation of motion and natural frequency for the system shown in fig. by energy method. m. Jo 000 HW2// For the system Fig below find 1-F.B.D 2Eq.of motion 8 wn 4-0 (1) -5- m

Chapter 17 Solutions

Thermodynamics: An Engineering Approach

Ch. 17.7 - Prob. 11PCh. 17.7 - Prob. 12PCh. 17.7 - Prob. 13PCh. 17.7 - Prob. 14PCh. 17.7 - Prob. 15PCh. 17.7 - Prob. 16PCh. 17.7 - Prob. 17PCh. 17.7 - Prob. 18PCh. 17.7 - Prob. 19PCh. 17.7 - Prob. 20PCh. 17.7 - Prob. 21PCh. 17.7 - Prob. 22PCh. 17.7 - Prob. 23PCh. 17.7 - Prob. 24PCh. 17.7 - Prob. 25PCh. 17.7 - Prob. 26PCh. 17.7 - Prob. 27PCh. 17.7 - The isentropic process for an ideal gas is...Ch. 17.7 - Is it possible to accelerate a gas to a supersonic...Ch. 17.7 - Prob. 30PCh. 17.7 - Prob. 31PCh. 17.7 - A gas initially at a supersonic velocity enters an...Ch. 17.7 - Prob. 33PCh. 17.7 - Prob. 34PCh. 17.7 - Prob. 35PCh. 17.7 - Prob. 36PCh. 17.7 - Prob. 37PCh. 17.7 - Prob. 38PCh. 17.7 - Air at 25 psia, 320F, and Mach number Ma = 0.7...Ch. 17.7 - Prob. 40PCh. 17.7 - Prob. 41PCh. 17.7 - Prob. 42PCh. 17.7 - Prob. 43PCh. 17.7 - Prob. 44PCh. 17.7 - Prob. 45PCh. 17.7 - Prob. 46PCh. 17.7 - Is it possible to accelerate a fluid to supersonic...Ch. 17.7 - Prob. 48PCh. 17.7 - Prob. 49PCh. 17.7 - Consider subsonic flow in a converging nozzle with...Ch. 17.7 - Consider a converging nozzle and a...Ch. 17.7 - Prob. 52PCh. 17.7 - Prob. 53PCh. 17.7 - Prob. 54PCh. 17.7 - Prob. 55PCh. 17.7 - Prob. 56PCh. 17.7 - Prob. 57PCh. 17.7 - Prob. 58PCh. 17.7 - Prob. 59PCh. 17.7 - Prob. 62PCh. 17.7 - Prob. 63PCh. 17.7 - Prob. 64PCh. 17.7 - Prob. 65PCh. 17.7 - Air enters a nozzle at 0.5 MPa, 420 K, and a...Ch. 17.7 - Prob. 67PCh. 17.7 - Are the isentropic relations of ideal gases...Ch. 17.7 - What do the states on the Fanno line and the...Ch. 17.7 - It is claimed that an oblique shock can be...Ch. 17.7 - Prob. 73PCh. 17.7 - Prob. 74PCh. 17.7 - For an oblique shock to occur, does the upstream...Ch. 17.7 - Prob. 76PCh. 17.7 - Prob. 77PCh. 17.7 - Prob. 78PCh. 17.7 - Prob. 79PCh. 17.7 - Prob. 80PCh. 17.7 - Prob. 81PCh. 17.7 - Prob. 82PCh. 17.7 - Prob. 83PCh. 17.7 - Prob. 84PCh. 17.7 - Air flowing steadily in a nozzle experiences a...Ch. 17.7 - Air enters a convergingdiverging nozzle of a...Ch. 17.7 - Prob. 89PCh. 17.7 - Prob. 90PCh. 17.7 - Consider the supersonic flow of air at upstream...Ch. 17.7 - Prob. 92PCh. 17.7 - Prob. 93PCh. 17.7 - Prob. 96PCh. 17.7 - Prob. 97PCh. 17.7 - Prob. 98PCh. 17.7 - Prob. 99PCh. 17.7 - What is the effect of heat gain and heat loss on...Ch. 17.7 - Consider subsonic Rayleigh flow of air with a Mach...Ch. 17.7 - What is the characteristic aspect of Rayleigh...Ch. 17.7 - Prob. 103PCh. 17.7 - Prob. 104PCh. 17.7 - Air is heated as it flows subsonically through a...Ch. 17.7 - Prob. 106PCh. 17.7 - Prob. 107PCh. 17.7 - Prob. 108PCh. 17.7 - Air is heated as it flows through a 6 in 6 in...Ch. 17.7 - Air enters a rectangular duct at T1 = 300 K, P1 =...Ch. 17.7 - Prob. 112PCh. 17.7 - Prob. 113PCh. 17.7 - Prob. 114PCh. 17.7 - What is supersaturation? Under what conditions...Ch. 17.7 - Prob. 116PCh. 17.7 - Prob. 117PCh. 17.7 - Steam enters a convergingdiverging nozzle at 1 MPa...Ch. 17.7 - Prob. 119PCh. 17.7 - Prob. 120RPCh. 17.7 - Prob. 121RPCh. 17.7 - Prob. 122RPCh. 17.7 - Prob. 124RPCh. 17.7 - Prob. 125RPCh. 17.7 - Using Eqs. 174, 1713, and 1714, verify that for...Ch. 17.7 - Prob. 127RPCh. 17.7 - Prob. 128RPCh. 17.7 - 17–129 Helium enters a nozzle at 0.6 MPa, 560...Ch. 17.7 - Prob. 130RPCh. 17.7 - Prob. 132RPCh. 17.7 - Prob. 133RPCh. 17.7 - Nitrogen enters a convergingdiverging nozzle at...Ch. 17.7 - An aircraft flies with a Mach number Ma1 = 0.9 at...Ch. 17.7 - Prob. 136RPCh. 17.7 - Helium expands in a nozzle from 220 psia, 740 R,...Ch. 17.7 - 17–140 Helium expands in a nozzle from 1 MPa,...Ch. 17.7 - Air is heated as it flows subsonically through a...Ch. 17.7 - Air is heated as it flows subsonically through a...Ch. 17.7 - Prob. 145RPCh. 17.7 - Prob. 146RPCh. 17.7 - Air is cooled as it flows through a 30-cm-diameter...Ch. 17.7 - Saturated steam enters a convergingdiverging...Ch. 17.7 - Prob. 151RPCh. 17.7 - Prob. 154FEPCh. 17.7 - Prob. 155FEPCh. 17.7 - Prob. 156FEPCh. 17.7 - Prob. 157FEPCh. 17.7 - Prob. 158FEPCh. 17.7 - Prob. 159FEPCh. 17.7 - Prob. 160FEPCh. 17.7 - Prob. 161FEPCh. 17.7 - Consider gas flow through a convergingdiverging...Ch. 17.7 - Combustion gases with k = 1.33 enter a converging...
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.
Similar questions
SEE MORE QUESTIONS
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
What is entropy? - Jeff Phillips; Author: TED-Ed;https://www.youtube.com/watch?v=YM-uykVfq_E;License: Standard youtube license