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Question 1
ప్రశ్న 1
For steady inviscid compressible flow along a streamline under gravity, which Bernoulli-type relation is appropriate when density can be expressed as a function of pressure along the process?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady inviscid flow, Euler's equation can be integrated along a streamline. When pressure and density are related through the specified thermodynamic process, the pressure contribution appears as the integral ∫dp/ρ. The resulting relation is V²/2 + ∫dp/ρ + gz = constant. Each term has dimensions of energy per unit mass. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Why is knowledge of the pressure-density relation important when applying a Bernoulli-type equation to compressible flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In compressible flow, density generally changes as pressure changes. The pressure-work term therefore cannot usually be replaced simply by p/ρ as in constant-density flow. A thermodynamic relation between pressure and density is needed to evaluate ∫dp/ρ. Isothermal and isentropic ideal-gas processes provide two important examples of such relations. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For an ideal gas undergoing an isothermal process at absolute temperature T, which relation between pressure and density is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ideal-gas equation per unit mass is p = ρRT. During an isothermal process, T remains constant for a given gas. Therefore p/ρ = RT remains constant. Pressure is consequently directly proportional to density under these conditions. Hence Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
For isothermal flow of an ideal gas at constant temperature T, what is the value of the pressure integral between states 1 and 2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isothermal ideal gas, ρ = p/(RT). Therefore 1/ρ = RT/p. Integrating dp/ρ from p1 to p2 gives RT times the integral of dp/p. The result is RT ln(p2/p1), where the logarithm contains a dimensionless pressure ratio. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
Air undergoes steady, horizontal, inviscid isothermal acceleration at T = 300 K from a section where p1 = 200 kPa and V1 is negligible to a section where p2 = 100 kPa. Taking R = 287 J/(kg·K), what is V2 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For horizontal isothermal ideal-gas flow, V1²/2 + RT ln(p1/pref) = V2²/2 + RT ln(p2/pref), where pref is any common reference pressure. The reference pressure cancels, giving V2²/2 - V1²/2 = RT ln(p1/p2). With negligible V1, V2 = sqrt[2RT ln(p1/p2)]. Substitution gives sqrt[2 × 287 × 300 × ln 2] = approximately 345.5 m/s. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
Air flows steadily, horizontally and isothermally at 300 K with R = 287 J/(kg·K). If its velocity increases from 100 m/s to 300 m/s in an inviscid passage, what is p2/p1 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The isothermal Bernoulli relation gives RT ln(p2/p1) = (V1² - V2²)/2. Substituting the velocities gives an exponent of (100² - 300²)/(2 × 287 × 300). This is approximately -0.4646. Taking the exponential gives p2/p1 approximately 0.6284. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
An ideal gas undergoes an isothermal pressure decrease from 240 kPa to 120 kPa. What is the density ratio ρ2/ρ1?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an ideal gas at constant temperature, p/ρ = RT is constant. Therefore density is directly proportional to absolute pressure. The density ratio is consequently ρ2/ρ1 = p2/p1. Substitution gives 120/240 = 0.50. Hence Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
A calorically perfect ideal gas accelerates steadily and isothermally in a horizontal passage with no shaft work. What additional energy interaction is generally required to maintain constant temperature while its velocity increases?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a calorically perfect ideal gas, enthalpy depends only on temperature. Isothermal flow therefore has no change in enthalpy. If velocity increases at unchanged elevation and there is no shaft work, the steady-flow energy equation requires an energy input equal to the increase in kinetic energy. That energy can be supplied as heat under the stated conditions. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
Which statement correctly distinguishes an adiabatic process from an isentropic process?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An adiabatic process has zero heat transfer across the system boundary. An isentropic process has constant entropy. An adiabatic process is isentropic only when it is also internally reversible, so entropy generation is absent. Friction, shocks or other irreversibilities can make an adiabatic process non-isentropic. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
For an isentropic process of a calorically perfect ideal gas with specific-heat ratio γ, which pressure-density relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A calorically perfect ideal gas undergoing an isentropic process satisfies p/ρ^γ = constant. This relation couples pressure and density changes during reversible adiabatic compression or expansion. It differs from the isothermal relation p/ρ = constant. The exponent γ is the ratio of specific heats cp/cv. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
For an isentropic ideal-gas process, which relation between temperature and density is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The isentropic relation is p/ρ^γ = constant. Combining this with p = ρRT gives T/ρ^(γ - 1) = constant. Therefore T2/T1 = (ρ2/ρ1)^(γ - 1). An isentropic ideal-gas expansion generally involves both density and temperature changes. Hence Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For steady inviscid isentropic flow of a calorically perfect ideal gas along a streamline, which Bernoulli-type relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isentropic ideal gas, integrating dp/ρ gives γp/[(γ - 1)ρ]. Therefore the steady inviscid Bernoulli-type equation becomes V²/2 + γp/[(γ - 1)ρ] + gz = constant. The thermodynamic term is also equal to the specific enthalpy h for a calorically perfect ideal gas. Thus the same relation may be written h + V²/2 + gz = constant. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
For a calorically perfect ideal gas, why may the isentropic compressible-flow energy equation be written as cpT + V²/2 + gz = constant?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a calorically perfect ideal gas, specific enthalpy is h = cpT. Also p/ρ = RT. Using cp = γR/(γ - 1), the term γp/[(γ - 1)ρ] becomes cpT. The isentropic Bernoulli relation can therefore be expressed directly in terms of temperature, velocity and elevation. Hence Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
For an ideal gas with γ = 1.4 and R = 287 J/(kg·K), what is cp according to cp = γR/(γ - 1)?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an ideal gas, cp - cv = R and γ = cp/cv. These relations give cp = γR/(γ - 1). Substitution gives cp = 1.4 × 287/0.4. This evaluates to 1004.5 J/(kg·K). Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
Air flows steadily, adiabatically and isentropically through a horizontal passage with cp = 1005 J/(kg·K). At section 1, T1 = 320 K and V1 = 100 m/s. If T2 = 300 K, what is V2 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the stated horizontal isentropic flow, cpT1 + V1²/2 = cpT2 + V2²/2. Rearranging gives V2² = V1² + 2cp(T1 - T2). Substitution gives V2² = 10000 + 2 × 1005 × 20. Taking the square root gives approximately 224.1 m/s. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A calorically perfect ideal gas expands isentropically so that p2/p1 = 0.50. If γ = 1.4, what is ρ2/ρ1 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isentropic ideal gas, p/ρ^γ is constant. Hence ρ2/ρ1 = (p2/p1)^(1/γ). Substitution gives 0.50^(1/1.4). This evaluates to approximately 0.6095. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the same isentropic ideal-gas expansion with p2/p1 = 0.50 and γ = 1.4, what is T2/T1 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The isentropic ideal-gas temperature-pressure relation is T2/T1 = (p2/p1)^[(γ - 1)/γ]. Here the exponent is 0.4/1.4. Substitution gives 0.50^(0.2857). The result is approximately 0.8203. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
An ideal gas at T1 = 300 K expands isentropically to half its initial absolute pressure. If γ = 1.4, what is T2 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isentropic ideal-gas expansion, T2/T1 = (p2/p1)^[(γ - 1)/γ]. For a pressure ratio of 0.50 and γ = 1.4, this ratio is approximately 0.8203. Multiplying by 300 K gives about 246.1 K. The static temperature decreases during the specified isentropic expansion. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
Air flows steadily and adiabatically through a horizontal frictionless passage with no shaft work. If cp = 1005 J/(kg·K), T1 = 330 K, V1 = 100 m/s and V2 = 250 m/s, what is T2 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The steady adiabatic frictionless energy relation is cpT1 + V1²/2 = cpT2 + V2²/2. Solving gives T2 = T1 + (V1² - V2²)/(2cp). Substituting the values gives 330 + (10000 - 62500)/(2010). The result is approximately 303.9 K. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
For a calorically perfect ideal gas in steady adiabatic flow with no shaft work and negligible elevation change, the stagnation temperature is defined by T0 = T + V²/(2cp). If T = 300 K, V = 200 m/s and cp = 1005 J/(kg·K), what is T0 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The kinetic-energy contribution expressed as an equivalent temperature is V²/(2cp). With V = 200 m/s, this contribution is 40000/(2 × 1005). It equals approximately 19.90 K. Adding it to the static temperature of 300 K gives about 319.9 K. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
An ideal gas flows steadily and isentropically with unchanged velocity through a passage that rises by 100 m. If cp = 1005 J/(kg·K) and g = 9.81 m/s², by how much does the static temperature decrease?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With unchanged velocity, the energy relation gives cp(T1 - T2) = g(z2 - z1). The elevation increase is 100 m. Therefore the temperature decrease is 9.81 × 100/1005. This evaluates to approximately 0.976 K. Hence Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
An ideal gas is compressed from pressure p1 to 2p1. Compare the density increase for an isothermal process with that for an isentropic process having γ = 1.4.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isothermal ideal-gas compression, density is directly proportional to absolute pressure, so doubling pressure doubles density. For isentropic compression, ρ2/ρ1 = 2^(1/γ). With γ = 1.4, this is approximately 1.641. The accompanying temperature rise in isentropic compression means the density does not need to double to support the doubled pressure. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
In the relations V²/2 + RT ln(p/pref) + gz = constant for isothermal ideal-gas flow and cpT + V²/2 + gz = constant for isentropic ideal-gas flow, what are the dimensions of every term?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The kinetic-energy term V²/2 has units of m²/s². The gravitational term gz has the same units. Both RT and cpT have units of J/kg, which are equivalent to m²/s². The logarithm ln(p/pref) is dimensionless because it contains a ratio of pressures expressed with the same reference units. Therefore every additive term has dimensions of energy per unit mass and Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
When using an isothermal ideal-gas relation containing ln(p2/p1), which pressures should be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Ideal-gas thermodynamic relations are written in terms of absolute pressure. The logarithmic pressure ratio must therefore use absolute pressures. Gauge pressure is obtained by subtracting atmospheric pressure and does not preserve pressure ratios. Using gauge pressures inside ln(p2/p1) can therefore produce an incorrect result. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about isothermal and isentropic compressible flow of an ideal gas: I. In isothermal flow, p/ρ is constant when T is constant. II. In isentropic flow, p/ρ^γ is constant. III. For a calorically perfect ideal gas, the isentropic Bernoulli thermodynamic term γp/[(γ - 1)ρ] equals cpT. IV. Adiabatic flow is not necessarily isentropic if irreversible entropy generation occurs. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I follows directly from the ideal-gas equation at constant temperature. Statement II is the standard pressure-density relation for an isentropic calorically perfect ideal gas. Statement III follows from p/ρ = RT together with cp = γR/(γ - 1). Statement IV correctly distinguishes the absence of heat transfer from the absence of 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. V²/2 + ∫(dp/ρ) + gz = constant సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. Because the term ∫dp/ρ cannot generally be evaluated unless the relation between p and ρ along the process is known సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. p/ρ = RT = constant సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. ∫(from p1 to p2) dp/ρ = RT ln(p2/p1) సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: C. 345 m/s సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. 0.628 సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. 0.50 సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: A. Heat must generally be supplied to the gas సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. Adiabatic means no heat transfer, while isentropic additionally requires no entropy generation సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. p/ρ^γ = constant సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. T2/T1 = (ρ2/ρ1)^(γ - 1) సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. V²/2 + [γ/(γ - 1)](p/ρ) + gz = constant సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: A. Because h = cpT and γp/[(γ - 1)ρ] = cpT సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 1004.5 J/(kg·K) సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. 224 m/s సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: B. 0.610 సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 0.820 సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. 246 K సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. 303.9 K సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 319.9 K సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 0.976 K సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. Isothermal gives ρ2/ρ1 = 2, while isentropic gives approximately 1.641 సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. Energy per unit mass, equivalent to m²/s² సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. Absolute thermodynamic pressures సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.