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

Answer to Problem 90P
The Mach number value of air after the normal shock through the nozzle is
The actual temperature of air after the normal shock through the nozzle
is
The actual pressure of air after the normal shock through the nozzle is
The stagnation pressure of air after the normal shock though the nozzle
is
The velocity of air after the normal shock through the nozzle is
Thus, the Mach number of helium gas after the normal shock through the nozzle
is
Thus, the actual temperature of helium after the normal shock through the nozzle is
Thus, the actual pressure of helium after the normal shock through the nozzle
is
Thus, the stagnation pressure of helium after the normal shock though the nozzle
is
Thus, the velocity of helium after the normal shock through the nozzle is
Comparison between the obtained results of air and helium is shown in below Table:
Parameters/ Conditions | Air | Helium |
Mach number value | ||
Actual temperature of air | ||
Actual pressure of air | ||
Stagnation pressure | ||
Velocity |
Explanation of Solution
Write the expression for the velocity of sound after the normal shock.
Here, velocity of sound after the shock is
Write the expression for the velocity of airafter the normal shock.
Write the expression for the Mach number for helium after the normal shock.
Here, Mach number of helium before the normal shock is
before the normal shock is
Write the expression for the actual pressure of helium gas after the normal shock.
Here, actual pressure of helium after the shock is
Write the expression for the actual temperature of helium gas after the normal shock.
Here, actual temperature of helium after the shock is
Write the expression for the actual pressure of helium gas after the normal shock.
Here, stagnation pressure of helium after the shock is
Write the expression for the velocity of sound after the normal shock.
Here, velocity of sound after the shock is
Write the expression for the velocity of helium after the normal shock.
Conclusion:
Refer to Table A-33, “One-dimensional normal-shock functions for an ideal gas with k 5 1.4”, obtain the expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.6 before the shock.
Thus, the Mach number value of air after the normal shock through the nozzle is
Here, actual pressure after the shock is
Substitute
Thus, the actual temperature of air after the normal shock through the nozzle
is
Substitute
Thus, the actual pressure of air after the normal shock through the nozzle is
The actual pressure before the normal shock is the same as the stagnation pressure before the normal shock
Substitute
Thus, the stagnation pressure of air after the normal shock though the nozzle
is
Refer to thermodynamics properties table and interpret the value of k, and R for a temperature of
Substitute 1.4 for k,
Substitute 0.5039for
Thus, the velocity of air after the normal shock through the nozzle is
Substitute 2.6for
Thus, the Mach number of helium gas after the normal shock through the nozzle
is
Substitute 1.667 for k, 2.6for
Substitute 1.667 for k, 2.6for
Substitute 1.667 for k, 2.6for
Substitute
Thus, the actual temperature of helium after the normal shock through the nozzle is
Substitute
Thus, the actual pressure of helium after the normal shock through the nozzle
is
Since,
Substitute
Thus, the stagnation pressure of air after the normal shock though the nozzle
is
Refer Table A–1, “Molar mass, gas constant, and critical2point properties”, obtain
the value of k, and R for a temperature of
Substitute 1.667 for k,
Substitute 0.5455 for
Thus, the velocity of air after the normal shock through the nozzle is
Want to see more full solutions like this?
Chapter 17 Solutions
Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684
- Can you explain the algebra steps that aren't shown but stated to be there, on how to get this equationarrow_forwardCorrect answer and complete fbd only. I will upvote. A flanged bolt coupling consists of two concentric rows of bolts. The inner row has 6 nos. of 16mm diameterbolts spaced evenly in a circle of 250mm in diameter. The outer row of has 10 nos. of 25 mm diameter bolts spaced evenly in a circle of 500mm in diameter. If the allowable shear stress on one bolt is 60 MPa, determine the torque capacity of the coupling. The Poisson’s ratio of the inner row of bolts is 0.2 while that of the outer row is 0.25 and the bolts are steel, E =200 GPa.arrow_forwardCorrect answer and complete fbd only. I will upvote. 10: The constant wall thickness of a steel tube with the cross sectionshown is 2 mm. If a 600-N-m torque is applied to the tube. Use G = 80 GPa forsteel.1. Find the shear stress (MPa) in the wall of the tube.2. Find the angle of twist, in degrees per meter of length.arrow_forward
- CORRECT ANSWER WITH COMPLETE FBD ONLY. I WILL UPVOTE. A torque wrench is used to tighten the pipe shown.Dimensions: S1 = 400 mm; S2 = 250 mm; S3 = 100 mmModulus of Rigidity G = 78 GPa1. The diameter of the solid pipe is 20 mm. How much is themaximum force P (N) that can be applied based on theallowable shear stress of 60 MPa?2. For a hollow pipe with 50 mm outside diameter and is 6 mmthick, compute for the maximum force P (kN) that can beapplied such that the angle of twist at A does not exceed 5degrees.3. The torque applied to tighten the hollow pipe is 200 N-m.Given: Pipe outside diameter = 50 mm Pipe thickness = 6 mmSolve for the resulting maximum shear stress (MPa) in the pipe.arrow_forwardCorrect answer and complete fbd only. I will upvote. 6: The shaft carries a total torque T0 that is uniformly distributedover its length L. Determine the angle of twist (degrees) of the shaft in termsif T0 = 1.2 kN-m, L = 2 m, G = 80 GPa, and diameter = 120 mm.arrow_forward2. Calculate the force in all members of the trusses shown using the method of joints. A 5525 lb C 3500 lb BY 14'-0" D 10'- 0" 6250 lb 10'- 0" Earrow_forward
- Correct answer and complete fbd only. I will upvote. 8: The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall at A andjoined together by a pin at B. Because of a slight misalignmentof the pre-drilled holes, the torque T0 = 750 N-m was appliedto the steel rod before the pin could be inserted into theholes. Determine the torque (N-m) in each component afterT0 was removed. Use G = 80 GPa for steel and G = 28 GPa foraluminumarrow_forwardCorrect answer and complete fbd only. I will upvote. 9: The two steel shafts, each with one end builtinto a rigid support, have flanges attached to their freeends. The flanges are to be bolted together. However,initially there is a 6⁰ mismatch in the location of the boltholes as shown in the figure. Determine the maximumshear stress(ksi) in each shaft after the flanges have beenbolted together. The shear modulus of elasticity for steelis 12 x 106 psi. Neglect deformations of the bolts and theflanges.arrow_forwardCorrect answer and complete fbd only. I will upvote. The tapered, wrought iron shaft carriesthe torque T = 2000 lb-in at its free end. Determine theangle of twist (degrees) of the shaft. Use G = 10 x 106psi for wrought ironarrow_forward
- Correct answer and complete fbd only. I will upvote. The compound shaft, consisting of steel and aluminumsegments, carries the two torques shown in the figure. Determine themaximum permissible value of T subject to the following designconditions: τst ≤ 83 MPa, τal ≤ 55 MPa, and θ ≤ 6⁰ (θ is the angle ofrotation of the free end). Use G =83 GPa for steel and G = 28 GPa foraluminum.arrow_forwardThe solid compound shaft, made of threedifferent materials, carries the two torques shown. Theshear moduli are 28 GPa for aluminum, 83 GPa for steel,and 35 GPa for bronze.1. Calculate the maximum shear stress (MPa) in eachmaterial.2. Find the angle of rotation (degrees) of the free endof the shaft.arrow_forwardCorrect answer only please. I will upvote. The velocity of a particle moves along the x-axis and is given by the equation ds/dt = 40 - 3t^2 m/s. Calculate the acceleration at time t=2 s and t=4 s. Calculate also the total displacement at the given interval. Assume at t=0 s=5m.Write the solution using pen and draw the graph if needed.arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY





