Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which statement correctly expresses conservation of mass for a fixed control volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The continuity equation is the mathematical expression of conservation of mass. For a control volume, mass may accumulate within the region or cross its control surface. Using outward mass flux as positive, the rate of mass accumulation plus the net outward mass-flow rate equals zero. In steady operation the accumulation term becomes zero, so total mass inflow equals total mass outflow. Therefore Option A gives the general control-volume statement.
వివరణ:
Question 2
ప్రశ్న 2
For uniform flow crossing a plane area A with velocity component Vn normal to the area and density ρ, which expression gives the mass-flow rate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Mass-flow rate is the rate at which mass crosses a specified surface. The volumetric flow rate through a uniform section is AVn, where Vn is the velocity component normal to the area. Multiplying this volume rate by density gives ṁ = ρAVn. Its SI unit is kg/s because kg/m³ multiplied by m² and m/s gives kg/s. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
If V denotes the average velocity normal to a flow section of area A, which expression gives the volumetric discharge Q?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Volumetric discharge is the volume of fluid crossing a section per unit time. For an average velocity normal to an area A, the volume swept per unit time is AV. Hence Q = AV. Density is required when converting volumetric discharge to mass-flow rate but not for the definition of Q itself. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
For steady incompressible flow through a pipe having one inlet section 1 and one outlet section 2, which relationship is correct when section-average velocities are used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Steady mass conservation requires the mass-flow rate entering the pipe to equal the mass-flow rate leaving it. For an incompressible fluid of the same constant density at both sections, ρA1V1 = ρA2V2. The common density cancels from the equation. This leaves the familiar relation A1V1 = A2V2. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
An incompressible fluid flows steadily through a circular pipe that contracts from diameter 200 mm to diameter 100 mm. What is the ratio V2/V1 of the mean downstream velocity to the mean upstream velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady incompressible flow, A1V1 = A2V2. Circular-pipe area is proportional to the square of diameter. Therefore A1/A2 = (D1/D2)² = (200/100)² = 4. Hence V2/V1 = A1/A2 = 4. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
Water flows steadily through a circular pipe that contracts from 200 mm diameter to 100 mm diameter. If the mean velocity in the 200 mm section is 3.0 m/s, what is the mean velocity in the 100 mm section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady incompressible flow, A1V1 = A2V2. The area ratio is A1/A2 = (0.20/0.10)² = 4. Therefore V2 = 4V1. With V1 = 3.0 m/s, V2 = 12 m/s. Hence Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
Water flows through a 150 mm diameter circular pipe with a mean velocity of 2.0 m/s. What is the volumetric discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The cross-sectional area is A = πD²/4. With D = 0.150 m, the area is approximately 0.01767 m². The discharge is Q = AV = 0.01767 × 2.0. This gives Q approximately equal to 0.03534 m³/s. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
A circular pipe must carry an incompressible discharge of 0.050 m³/s at a mean velocity of 2.5 m/s. What pipe diameter is required approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The required area is A = Q/V = 0.050/2.5 = 0.020 m². For a circular pipe, A = πD²/4. Hence D = sqrt(4A/π). Substitution gives D approximately equal to 0.1596 m. This is about 160 mm, so Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
Water of density 1000 kg/m³ flows through an 80 mm diameter pipe with a mean velocity of 3.0 m/s. What is the mass-flow rate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pipe area is A = π(0.08)²/4 = approximately 0.005027 m². The volumetric discharge is Q = AV = 0.005027 × 3.0 = approximately 0.01508 m³/s. Mass-flow rate is ṁ = ρQ. Thus ṁ = 1000 × 0.01508 = approximately 15.08 kg/s. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
For steady one-dimensional flow of a compressible fluid through a stream tube with one inlet and one outlet, which relationship expresses continuity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Steady conservation of mass requires the same mass-flow rate through every section of the stream tube. The mass-flow rate at a section is ρAV when mean velocity is normal to the section. Therefore ρ1A1V1 = ρ2A2V2. Unlike incompressible flow, the volumetric discharge AV need not remain constant if density changes. Hence Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
A gas enters a steady duct at density 1.2 kg/m³, area 0.040 m² and mean velocity 20 m/s. It leaves at density 0.80 kg/m³ through an area of 0.025 m². What is the outlet mean velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Steady mass conservation requires ρ1A1V1 = ρ2A2V2. The inlet mass-flow rate is 1.2 × 0.040 × 20 = 0.96 kg/s. The outlet coefficient ρ2A2 is 0.80 × 0.025 = 0.020 kg/m. Therefore V2 = 0.96/0.020 = 48 m/s. Hence Option D is correct.
వివరణ:
Question 12
ప్రశ్న 12
A steady compressible flow enters a duct with density 1.5ρ0 and area A0. At a downstream section the density is ρ0 and the area is 0.50A0. What is V2/V1?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady compressible flow, ρ1A1V1 = ρ2A2V2. Therefore V2/V1 = ρ1A1/(ρ2A2). Substituting the ratios gives V2/V1 = 1.5/(1 × 0.50). This evaluates to 3.0. Therefore Option D is correct.
వివరణ:
Question 13
ప్రశ్న 13
For a steady control volume with no mass generation or destruction, what must be true?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Steady operation means that the amount of mass stored inside the control volume does not change with time. Conservation of mass therefore requires the sum of all mass-flow rates entering to equal the sum leaving. For a compressible fluid, volumetric flow rates may differ because density may change. Equal inlet and outlet areas are also unnecessary. Hence Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
An unsteady control volume receives mass at 5 kg/s and discharges mass at 3 kg/s. Assuming these are the only mass flows, at what rate is mass accumulating inside the control volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The mass accumulation rate equals total mass inflow minus total mass outflow. Here the inflow is 5 kg/s and the outflow is 3 kg/s. Therefore the accumulation rate is 5 - 3 = 2 kg/s. The positive sign means that the mass stored within the control volume is increasing. Hence Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
In the integral continuity equation for a fixed control volume, the control-surface normal is conventionally taken outward. What does a positive value of the surface integral of ρV·n represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The control-surface area vector is conventionally directed outward from the control volume. Where velocity has an outward normal component, V·n is positive. Multiplication by density gives a positive outward mass flux. Integration over the entire control surface therefore gives the net outward mass-flow rate. Hence Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
Which equation is the general differential continuity equation in Cartesian coordinates for a compressible, possibly unsteady, three-dimensional flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The differential continuity equation expresses local conservation of mass. In Cartesian coordinates, mass accumulation contributes ∂ρ/∂t. The three mass-flux divergence terms are ∂(ρu)/∂x, ∂(ρv)/∂y and ∂(ρw)/∂z. Their sum with the accumulation term must equal zero in the absence of mass sources. Therefore Option A is the correct general equation.
వివరణ:
Question 17
ప్రశ్న 17
Which material-derivative form is equivalent to the general continuity equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The conservative continuity equation is ∂ρ/∂t + ∇·(ρV) = 0. Expanding the divergence gives ∂ρ/∂t + V·∇ρ + ρ∇·V = 0. The first two terms combine to form the material derivative Dρ/Dt. Thus the equation becomes Dρ/Dt + ρ∇·V = 0. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
For an incompressible flow, which condition does the velocity field satisfy?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In incompressible flow, the density of a moving fluid element does not change appreciably. Therefore Dρ/Dt = 0 in the continuity relation Dρ/Dt + ρ∇·V = 0. For nonzero density this gives ∇·V = 0. The condition does not require zero vorticity or steady flow. Hence Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
A three-dimensional velocity field is u = 2kx, v = -ky and w = -kz, where k = 1 s^-1. Does this field satisfy the incompressible continuity condition?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For incompressible flow, the required condition is ∂u/∂x + ∂v/∂y + ∂w/∂z = 0. The three derivatives are 2k, -k and -k. Their sum is 2k - k - k = 0. Thus the velocity field is divergence-free. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
A two-dimensional velocity field is u = 3kx and v = -ky, where k = 1 s^-1. Which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a two-dimensional incompressible flow, continuity requires ∂u/∂x + ∂v/∂y = 0. Here ∂u/∂x = 3k and ∂v/∂y = -k. Their sum is 2k, which is nonzero. Therefore the specified velocity field does not satisfy the incompressible continuity condition. Hence Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
Consider a one-dimensional unsteady compressible field with density ρ = ρ0e^(-at) and velocity u = ax, where a is a positive constant with units s^-1. Which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For one-dimensional flow, the general continuity equation is ∂ρ/∂t + ∂(ρu)/∂x = 0. Here ∂ρ/∂t = -aρ because ρ = ρ0e^(-at). Since ρ is independent of x and u = ax, ∂(ρu)/∂x = aρ. The two terms therefore sum to zero exactly. Hence the field satisfies the compressible continuity equation and Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
At a steady junction carrying an incompressible liquid, two inlet branches deliver 0.080 m³/s and 0.050 m³/s. If there is only one outlet and no leakage, what is the outlet discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a steady junction, there is no accumulation of mass within the junction. For the same incompressible liquid, conservation of mass reduces to conservation of volumetric discharge. Therefore Qout = Q1 + Q2. Substituting the values gives Qout = 0.080 + 0.050 = 0.130 m³/s. Hence Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
Two incompressible inlet streams of the same liquid, carrying 0.030 m³/s and 0.020 m³/s, merge into a single circular outlet pipe of diameter 100 mm. Assuming steady flow, what is the mean outlet velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total outlet discharge is 0.030 + 0.020 = 0.050 m³/s. A 100 mm pipe has area A = π(0.10)²/4 = approximately 0.007854 m². The outlet velocity is V = Q/A. Thus V = 0.050/0.007854 = approximately 6.37 m/s. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Two gas streams merge steadily into one outlet. Stream 1 has ρ1 = 1.2 kg/m³, A1 = 0.020 m² and V1 = 10 m/s; Stream 2 has ρ2 = 0.90 kg/m³, A2 = 0.015 m² and V2 = 12 m/s. If the outlet density is 1.0 kg/m³ and its area is 0.030 m², what is the outlet mean velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The first inlet mass-flow rate is 1.2 × 0.020 × 10 = 0.240 kg/s. The second inlet mass-flow rate is 0.90 × 0.015 × 12 = 0.162 kg/s. Steady mass conservation gives an outlet mass-flow rate of 0.240 + 0.162 = 0.402 kg/s. Since ṁout = ρoutAoutVout, Vout = 0.402/(1.0 × 0.030) = 13.4 m/s. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
A tank of constant horizontal plan area 10 m² receives water at 0.12 m³/s while two outlets discharge 0.07 m³/s and 0.03 m³/s. Assuming water is incompressible and the tank level remains horizontal, at what rate is the water depth increasing?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The net volumetric accumulation rate is inflow minus total outflow. Therefore dV/dt = 0.12 - 0.07 - 0.03 = 0.02 m³/s. For a tank of constant plan area Ap, dV/dt = Ap dh/dt. Hence dh/dt = 0.02/10 = 0.002 m/s. Therefore Option B is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: A. Rate of mass accumulation inside the control volume plus net outward mass-flow rate across the control surface equals zero సమాధానం: A.
-
Question 2 ప్రశ్న 2Answer: B. ṁ = ρAVn సమాధానం: B.
-
Question 3 ప్రశ్న 3Answer: B. Q = AV సమాధానం: B.
-
Question 4 ప్రశ్న 4Answer: B. A1V1 = A2V2 సమాధానం: B.
-
Question 5 ప్రశ్న 5Answer: B. 4 సమాధానం: B.
-
Question 6 ప్రశ్న 6Answer: C. 12 m/s సమాధానం: C.
-
Question 7 ప్రశ్న 7Answer: B. 0.0353 m³/s సమాధానం: B.
-
Question 8 ప్రశ్న 8Answer: C. 160 mm సమాధానం: C.
-
Question 9 ప్రశ్న 9Answer: C. 15.08 kg/s సమాధానం: C.
-
Question 10 ప్రశ్న 10Answer: B. ρ1A1V1 = ρ2A2V2 సమాధానం: B.
-
Question 11 ప్రశ్న 11Answer: D. 48 m/s సమాధానం: D.
-
Question 12 ప్రశ్న 12Answer: D. 3.0 సమాధానం: D.
-
Question 13 ప్రశ్న 13Answer: A. Total mass inflow equals total mass outflow సమాధానం: A.
-
Question 14 ప్రశ్న 14Answer: C. +2 kg/s సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: B. Net mass outflow from the control volume సమాధానం: B.
-
Question 16 ప్రశ్న 16Answer: A. ∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0 సమాధానం: A.
-
Question 17 ప్రశ్న 17Answer: A. Dρ/Dt + ρ∇·V = 0 సమాధానం: A.
-
Question 18 ప్రశ్న 18Answer: A. ∇·V = 0 సమాధానం: A.
-
Question 19 ప్రశ్న 19Answer: A. Yes, because its divergence is zero సమాధానం: A.
-
Question 20 ప్రశ్న 20Answer: B. The field does not satisfy incompressible continuity because its divergence is 2k సమాధానం: B.
-
Question 21 ప్రశ్న 21Answer: B. It satisfies ∂ρ/∂t + ∂(ρu)/∂x = 0 సమాధానం: B.
-
Question 22 ప్రశ్న 22Answer: C. 0.130 m³/s సమాధానం: C.
-
Question 23 ప్రశ్న 23Answer: C. 6.37 m/s సమాధానం: C.
-
Question 24 ప్రశ్న 24Answer: C. 13.4 m/s సమాధానం: C.
-
Question 25 ప్రశ్న 25Answer: B. 0.002 m/s సమాధానం: B.