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Question 1
ప్రశ్న 1
In a uniform supersonic flow, what is meant by a Mach wave?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A Mach wave represents the limiting form of an infinitesimally weak compressive or expansive disturbance in a supersonic flow. Relative to the local fluid, the disturbance propagates at the local acoustic speed. Because the surrounding flow itself is moving faster than sound, the disturbance is swept downstream. A Mach wave should not automatically be identified with a finite-strength shock wave, which produces finite property changes and entropy generation. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
For a point disturbance moving supersonically through a uniform gas, how is the Mach cone formed?
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Explanation:
A disturbance continuously emits small acoustic waves as it moves through the gas. In supersonic motion, the source travels farther than an individual sound wave can propagate upstream during the same time. The successive spherical wave fronts therefore possess a common downstream envelope. In three dimensions this envelope forms a cone whose half-angle is the Mach angle. Hence Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
If μ is the Mach angle measured between the flow direction and a Mach line or generator of the Mach cone, which relation is correct for M > 1?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
During a time interval, an acoustic disturbance travels a distance proportional to the sound speed a while the supersonic source travels a distance proportional to V. The resulting right-triangle geometry gives sin μ = a/V. Since Mach number is M = V/a, this becomes sin μ = 1/M. The relation requires M greater than or equal to one for a real Mach angle in this geometric interpretation. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A uniform supersonic stream has Mach number M = 2.0. What is its Mach angle?
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Explanation:
The Mach-angle relation is sin μ = 1/M. For M = 2, sin μ = 0.5. The inverse sine of 0.5 is 30 degrees. This is the half-angle of the Mach cone relative to the flow direction. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
What is the Mach angle for a uniform flow at Mach 3.0?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For M = 3, the Mach-angle relation gives sin μ = 1/3. Taking the inverse sine gives μ approximately equal to 19.47 degrees. The result is smaller than the 30-degree angle at Mach 2. This illustrates that the Mach cone narrows as Mach number increases. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
A Mach cone has a half-angle of 20 degrees in a uniform supersonic flow. What is the Mach number approximately?
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Explanation:
The governing relation is M = 1/sin μ. Substituting μ = 20 degrees gives M = 1/sin20 degrees. Since sin20 degrees is approximately 0.3420, the Mach number is approximately 2.924. The flow is therefore clearly supersonic. Hence Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
How does the Mach angle change as the Mach number of a uniform supersonic flow increases?
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Explanation:
Mach angle satisfies sin μ = 1/M. Increasing M decreases the value of 1/M. Over the relevant range from 0 to 90 degrees, a smaller sine corresponds to a smaller angle. Thus increasingly supersonic motion produces a narrower Mach cone. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
Which statement correctly compares disturbance propagation in subsonic and supersonic uniform flow?
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Explanation:
In subsonic flow, the local acoustic speed exceeds the flow speed, so acoustic information can propagate upstream relative to the fixed frame. In supersonic flow, the flow convects disturbances downstream faster than they can propagate upstream acoustically. Small disturbances are consequently confined to the downstream Mach-cone region. As M approaches one from above, sin μ approaches one and μ approaches 90 degrees rather than zero. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
What is meant by the stagnation state associated with a local flowing state in an isentropic compressible-flow analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Stagnation or total properties are reference properties associated with bringing a flowing fluid to rest. For stagnation pressure and the standard isentropic property relations, the deceleration is imagined to occur adiabatically and reversibly. The final velocity at the stagnation state is zero. The kinetic energy of the local flow is thereby converted into thermodynamic energy without an isentropic loss. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
For a calorically perfect ideal gas with specific-heat ratio γ, which relation gives the isentropic stagnation-to-static temperature ratio?
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Explanation:
The steady adiabatic energy relation gives h0 = h + V²/2. For a calorically perfect ideal gas, h = cpT and V = Ma. Using a² = γRT and the ideal-gas specific-heat relations gives T0/T = 1 + [(γ - 1)/2]M². This expression relates local static temperature to the corresponding stagnation temperature. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
For a calorically perfect ideal gas brought to rest isentropically, which relation gives the stagnation-to-static pressure ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Isentropic ideal-gas relations connect pressure and temperature through p0/p = (T0/T)^[γ/(γ - 1)]. The stagnation-temperature ratio is 1 + [(γ - 1)/2]M². Substitution gives the expression shown in Option A. The exponent γ/(γ - 1) distinguishes the pressure ratio from the density ratio. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a calorically perfect ideal gas brought to rest isentropically, which relation gives the stagnation-to-static density ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isentropic ideal gas, density and temperature satisfy ρ0/ρ = (T0/T)^[1/(γ - 1)]. The stagnation-temperature ratio is 1 + [(γ - 1)/2]M². Combining the two relations gives the expression in Option A. The exponent is smaller than the corresponding pressure-ratio exponent. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
For air modeled with γ = 1.4 at Mach number M = 1.0, what is T0/T?
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Explanation:
For γ = 1.4, the coefficient (γ - 1)/2 equals 0.2. At M = 1, the stagnation-temperature ratio is therefore 1 + 0.2(1²). This gives T0/T = 1.20. Thus the stagnation temperature is 20 percent higher than the local static temperature at Mach one under the stated ideal-gas assumptions. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
For air with γ = 1.4 at Mach number M = 1.0, what is the isentropic stagnation-to-static pressure ratio p0/p approximately?
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Explanation:
At Mach one and γ = 1.4, the temperature factor is 1 + 0.2 = 1.2. The pressure ratio is p0/p = 1.2^[1.4/0.4]. The exponent is 3.5. Evaluating the expression gives approximately 1.893. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
Air with γ = 1.4 has static temperature T = 280 K and Mach number M = 2.0. What is the stagnation temperature?
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Explanation:
For γ = 1.4, T0/T = 1 + 0.2M². At M = 2, the factor is 1 + 0.2(4) = 1.8. Thus T0 = 1.8 × 280 K. This gives 504 K. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Air with γ = 1.4 has static pressure p = 80 kPa and Mach number M = 2.0. What is its isentropic stagnation pressure approximately?
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Explanation:
For M = 2 and γ = 1.4, the basic factor is 1 + 0.2M² = 1.8. The stagnation-pressure ratio is 1.8^3.5. This evaluates to approximately 7.824. Multiplying 80 kPa by 7.824 gives approximately 626 kPa. Therefore Option D is correct.
వివరణ:
Question 17
ప్రశ్న 17
Air with γ = 1.4 has static density ρ = 0.80 kg/m³ at Mach number M = 2.0. What is the corresponding isentropic stagnation density approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At M = 2 and γ = 1.4, the temperature factor is 1.8. The stagnation-density ratio is 1.8^[1/(0.4)], which is 1.8^2.5. This gives approximately 4.347. Multiplying by the static density 0.80 kg/m³ gives approximately 3.48 kg/m³. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
For air with γ = 1.4, the measured stagnation-to-static temperature ratio is T0/T = 1.50. What is the Mach number approximately?
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Explanation:
For γ = 1.4, the temperature relation is T0/T = 1 + 0.2M². Setting this equal to 1.50 gives 0.2M² = 0.50. Thus M² = 2.5. Taking the positive square root gives M approximately 1.581. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
A calorically perfect ideal gas has static temperature 290 K and velocity 300 m/s. If cp = 1005 J/(kg·K), what is its stagnation temperature when elevation effects are negligible?
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Explanation:
The stagnation-energy relation is cpT0 = cpT + V²/2. Therefore T0 = T + V²/(2cp). Substituting the values gives T0 = 290 + 300²/(2 × 1005). The kinetic-energy contribution is approximately 44.78 K, giving T0 approximately 334.78 K. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
For an isentropic calorically perfect ideal-gas stagnation process, how can the pressure ratio be written in terms of the stagnation-to-static temperature ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isentropic ideal gas, pressure and temperature satisfy p/ρ^γ = constant together with the ideal-gas equation. The resulting pressure-temperature relation is p0/p = (T0/T)^[γ/(γ - 1)]. This relation is consistent with the standard Mach-number stagnation formula. It applies to the isentropically related static and stagnation states of the same gas. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
Which statement about stagnation pressure in a compressible gas is generally correct?
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Explanation:
The incompressible Bernoulli result p0 = p + ρV²/2 is an appropriate low-compressibility approximation. At appreciable Mach number, both density and other thermodynamic properties change during deceleration. The isentropic compressible-flow pressure relation must then be used instead. For a nonzero Mach number, isentropic stagnation pressure exceeds the corresponding static pressure. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
For an isentropic deceleration of a calorically perfect ideal gas from a nonzero Mach number to rest, how do stagnation temperature, pressure and density compare with their local static values?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Isentropic deceleration converts the local kinetic energy into thermodynamic energy. The stagnation-temperature relation gives T0/T greater than one whenever M is nonzero. The corresponding isentropic pressure and density ratios are also greater than one. At zero Mach number, all three stagnation quantities reduce to their static counterparts. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
A calorically perfect gas flows through an adiabatic device with no shaft work and negligible elevation change, but the process is irreversible because of friction. Which statement about stagnation properties is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady adiabatic flow with no shaft work and negligible potential-energy change, total enthalpy is conserved. For a calorically perfect gas, this means stagnation temperature remains constant. Friction generates entropy, however, so the process is not isentropic. Entropy generation causes a loss of stagnation pressure even though the stagnation temperature can remain unchanged. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement most accurately distinguishes a static property from a stagnation property in compressible flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Static pressure, temperature and density are local thermodynamic properties of the gas while it is moving. Their existence does not require the gas itself to be motionless. Stagnation properties describe a related reference state obtained by bringing that local flow to zero velocity according to the specified thermodynamic process. Static and stagnation properties coincide when the local velocity, and therefore Mach number, is zero. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements: I. For a uniform supersonic flow, sin μ = 1/M. II. Increasing supersonic Mach number decreases the Mach angle. III. For a calorically perfect ideal gas, T0/T = 1 + [(γ - 1)/2]M². IV. In an adiabatic irreversible flow with no shaft work, stagnation temperature may remain constant while stagnation pressure decreases. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is the standard geometric Mach-angle relation for a small disturbance in uniform supersonic flow. Statement II follows directly because 1/M decreases as supersonic Mach number increases. Statement III is the calorically perfect ideal-gas stagnation-temperature relation. Statement IV correctly reflects conservation of total enthalpy together with stagnation-pressure loss caused by entropy generation. Therefore all four statements are correct and Option D is the valid answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. An infinitesimally weak pressure disturbance propagating at the local speed of sound relative to the fluid సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. It is the envelope of successive acoustic wave fronts emitted by the moving disturbance సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. sin μ = 1/M సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. 30 degrees సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. 19.47 degrees సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 2.92 సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. The Mach angle decreases సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. In subsonic flow small disturbances can propagate upstream, whereas in supersonic flow they are confined downstream within the Mach-cone region సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. The state obtained by bringing the fluid locally to zero velocity isentropically సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. T0/T = 1 + [(γ - 1)/2]M² సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. p0/p = [1 + ((γ - 1)/2)M²]^[γ/(γ - 1)] సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. ρ0/ρ = [1 + ((γ - 1)/2)M²]^[1/(γ - 1)] సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. 1.20 సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 1.89 సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 504 K సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: D. 626 kPa సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: C. 3.48 kg/m³ సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. 1.58 సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: C. 334.8 K సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. p0/p = (T0/T)^[γ/(γ - 1)] సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. It cannot in general be found by simply adding the incompressible dynamic pressure ρV²/2 to the static pressure at appreciable Mach number సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. T0 > T, p0 > p and ρ0 > ρ సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. Stagnation temperature can remain constant while stagnation pressure decreases సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. A static property describes the local moving state, while a stagnation property refers to the corresponding state obtained when the flow is brought to rest under the specified stagnation process సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.