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Question 1
ప్రశ్న 1
For steady incompressible frictionless flow along a streamline, which expression represents Bernoulli's equation in head form?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Bernoulli's equation expresses conservation of mechanical energy per unit weight under its ideal-flow assumptions. The pressure-energy term per unit weight is p/γ. The kinetic-energy term per unit weight is V²/(2g), while the gravitational potential-energy term per unit weight is z. Each of these three quantities therefore has dimensions of length and is called a head. Hence Option A gives the correct head form.
వివరణ:
Question 2
ప్రశ్న 2
Which interpretation of the three terms p/γ, V²/(2g) and z in Bernoulli's equation is correct?
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Explanation:
The quantity p/γ is pressure divided by specific weight and therefore represents pressure head. The quantity V²/(2g) is the kinetic energy per unit weight and is called velocity head. The elevation z measured above a selected datum is the gravitational potential head or elevation head. Their sum is commonly called total head when the usual Bernoulli assumptions apply. Therefore Option A correctly identifies all three terms.
వివరణ:
Question 3
ప్రశ్న 3
Which set of conditions is appropriate for applying the classical Bernoulli equation directly between two points on the same streamline without additional machine or loss terms?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The familiar Bernoulli equation is a mechanical-energy relation for an idealized flow. Its basic introductory form assumes steady flow, constant density and negligible viscous dissipation along the streamline considered. If a pump adds energy or a turbine extracts energy between the two sections, additional head terms are required. Similarly, significant friction requires a head-loss term rather than being silently neglected. Therefore Option A states the appropriate conditions.
వివరణ:
Question 4
ప్రశ్న 4
Which statement correctly describes the streamline restriction on the classical Bernoulli equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a steady inviscid flow, Bernoulli's constant is conserved along an individual streamline. In a rotational flow, different streamlines can generally have different Bernoulli constants. If the flow is irrotational, the Bernoulli constant can be common throughout a connected region under the other appropriate assumptions. This distinction prevents an unjustified comparison of points located on unrelated streamlines. Hence Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A liquid is at rest under gravity. What does Bernoulli's relation reduce to between two points in the same constant-density liquid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A fluid at rest has zero velocity, so its velocity-head term vanishes. Bernoulli's mechanical-energy relation then reduces to p/γ + z = constant within the connected static liquid. This is equivalent to the familiar hydrostatic pressure-elevation relationship. A decrease in elevation must therefore be accompanied by an increase in pressure head. Hence Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
Water has a gauge pressure of 196.2 kPa at a point. Taking ρ = 1000 kg/m³ and g = 9.81 m/s², what is the corresponding pressure head?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure head is calculated from p/γ = p/(ρg). The specific weight of water is 1000 × 9.81 = 9810 N/m³. Dividing 196200 Pa by 9810 N/m³ gives 20 m. The use of gauge pressure simply expresses the head relative to atmospheric pressure. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
What is the velocity head corresponding to a flow velocity of 6.0 m/s when g = 9.81 m/s²?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Velocity head is V²/(2g). Substituting V = 6.0 m/s gives 36/(2 × 9.81). This evaluates to approximately 1.835 m. Velocity head represents kinetic energy per unit weight expressed as an equivalent fluid-column height. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
At a point in a water pipeline, the gauge pressure is 150 kPa, velocity is 4.0 m/s and elevation above the selected datum is 10 m. Taking ρ = 1000 kg/m³ and g = 9.81 m/s², what is the total head using the usual Bernoulli form?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pressure head is 150000/(1000 × 9.81) = approximately 15.29 m. The velocity head is 4²/(2 × 9.81) = approximately 0.815 m. The elevation head is given directly as 10 m. Adding the three terms gives 15.29 + 0.815 + 10 = approximately 26.11 m. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
Water flows steadily and without head loss through a horizontal pipe that contracts from 200 mm diameter at section 1 to 100 mm diameter at section 2. If V1 = 2.0 m/s and p1 = 250 kPa gauge, what is p2? Take ρ = 1000 kg/m³.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Continuity gives V2/V1 = A1/A2 = (D1/D2)² = 4, so V2 = 8 m/s. Because the pipe is horizontal and losses are neglected, Bernoulli gives p1/ρ + V1²/2 = p2/ρ + V2²/2. Hence p2 - p1 = 0.5ρ(V1² - V2²). This equals 0.5 × 1000 × (4 - 64) = -30000 Pa. Therefore p2 = 250 - 30 = 220 kPa gauge, making Option B correct.
వివరణ:
Question 10
ప్రశ్న 10
A large reservoir discharges water through a small nozzle whose outlet is 12 m below the reservoir free surface. Both the free surface and jet are exposed to atmosphere, the reservoir-surface velocity is negligible, and losses are neglected. What is the ideal jet velocity? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Apply Bernoulli between the reservoir free surface and the free jet. Both points are at atmospheric pressure, so their pressure heads cancel when gauge pressure is used. The velocity at the large reservoir surface is negligible compared with the jet velocity. Therefore the 12 m elevation difference becomes velocity head, giving V = sqrt(2g × 12). This is approximately 15.34 m/s, so Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
A Pitot arrangement brings an ideal water stream of velocity 10 m/s to rest at a stagnation point at essentially the same elevation and with negligible loss. What is the stagnation-pressure rise above the local static pressure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At the stagnation point the local velocity is reduced ideally to zero. Bernoulli between the moving stream and the stagnation point converts the original velocity head into pressure head. The pressure rise is therefore the dynamic pressure, Δp = ρV²/2. For water at 1000 kg/m³ and V = 10 m/s, Δp = 0.5 × 1000 × 100 = 50000 Pa. Hence the pressure rise is 50 kPa and Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A horizontal Venturi carries water with negligible loss. Its inlet diameter is 200 mm and throat diameter is 100 mm. If the inlet pressure exceeds the throat pressure by 45 kPa, what is the discharge approximately? Take ρ = 1000 kg/m³.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The area ratio is A1/A2 = (0.20/0.10)² = 4, so continuity gives V2 = 4V1. For the horizontal lossless flow, the pressure difference equals ρ(V2² - V1²)/2. Thus 45000 = 0.5 × 1000 × 15V1², giving V1 = approximately 2.449 m/s. The inlet area is π(0.20)²/4 = approximately 0.031416 m², so Q = A1V1 = approximately 0.07695 m³/s. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
Water flows steadily through a pipe with the same mean velocity at sections 1 and 2, and head loss is negligible. Section 2 is 8 m higher than section 1. If p1 = 200 kPa gauge, what is p2? Take ρ = 1000 kg/m³ and g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because the mean velocities are equal, the two velocity-head terms cancel in Bernoulli's equation. With no head loss, an 8 m rise in elevation must be balanced by an 8 m reduction in pressure head. The corresponding pressure decrease is ρgΔz = 1000 × 9.81 × 8 = 78480 Pa. Therefore p2 = 200 - 78.48 = 121.52 kPa gauge. Hence Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
Using the usual one-dimensional head notation, what does the hydraulic grade line represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The hydraulic grade line represents the sum of elevation head and pressure head. Its elevation relative to the chosen datum is therefore z + p/γ. In a piezometer connected to a liquid pipe, the liquid level ideally indicates this piezometric head. The velocity-head contribution is not part of the hydraulic grade line. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
Under the usual Bernoulli convention with kinetic-energy correction factor taken as unity, what is the vertical separation between the energy grade line and hydraulic grade line at a flowing section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The energy grade line represents total head, z + p/γ + V²/(2g). The hydraulic grade line represents z + p/γ. Subtracting the hydraulic grade elevation from the energy grade elevation leaves V²/(2g). Thus the separation between the two lines is the local velocity head under the stated convention. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Water flows steadily through a horizontal constant-diameter pipe with no pump or turbine, but with frictional head loss. Which statement about the EGL and HGL is correct if the mean velocity remains constant?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In a constant-diameter incompressible pipe, steady continuity makes the mean velocity constant. Therefore the velocity-head separation between the EGL and HGL remains constant. Friction dissipates mechanical energy, so total head decreases in the direction of flow. With no change in velocity head, the hydraulic grade also falls by the corresponding amount. Hence Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
Water flows steadily through a horizontal pipe of constant diameter. The gauge pressure falls from 300 kPa to 260 kPa between two sections, while the velocity is unchanged and there is no pump or turbine. What is the head loss? Take ρ = 1000 kg/m³ and g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The elevation heads are equal because the pipe is horizontal. The velocity heads also cancel because the diameter and mean velocity are unchanged. The loss in total head is therefore simply the decrease in pressure head. Thus hL = (300000 - 260000)/(1000 × 9.81) = approximately 4.077 m. Rounded appropriately, the head loss is 4.08 m and Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which form of the extended energy equation has the correct sign convention when hp is head added by a pump, ht is head extracted by a turbine, and hL is irreversible head loss between sections 1 and 2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Let H denote the mechanical head z + p/γ + V²/(2g) at a section. A pump supplies mechanical energy to the fluid, so its head is added on the upstream side of the balance. A turbine removes mechanical energy from the fluid, while irreversible losses also consume mechanical head. Therefore the balance can be written H1 + hp = H2 + ht + hL. Hence Option A uses the correct engineering sign convention.
వివరణ:
Question 19
ప్రశ్న 19
Water is pumped steadily from a large lower reservoir whose free surface is at elevation 10 m to a large upper reservoir whose free surface is at elevation 35 m. Both reservoirs are open to atmosphere, free-surface velocities are negligible, and total head loss is 4 m. What pump head is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At each large reservoir free surface, pressure is atmospheric and the velocity is negligible. The mechanical-head increase required between the free surfaces is therefore the elevation difference, 35 - 10 = 25 m. The pump must also supply the 4 m of head that will be dissipated as loss. Hence hp = 25 + 4 = 29 m. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
The total mechanical head immediately upstream of a turbine is 80 m and the total head at a downstream section is 25 m. If 5 m of head is lost between these sections in addition to the turbine extraction, what head is extracted by the turbine?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The extended energy equation without a pump is H1 = H2 + ht + hL. Substituting the values gives 80 = 25 + ht + 5. Therefore ht = 80 - 25 - 5 = 50 m. The remaining 5 m of head is dissipated rather than converted into turbine output. Hence Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
How is the energy grade line expected to behave across idealized hydraulic machines when the direction of flow is followed?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A pump transfers mechanical energy from a shaft to the fluid. The fluid total head therefore increases across the pump, appearing as an upward jump in the EGL. A turbine performs the reverse energy transfer by extracting mechanical energy from the fluid. The total head consequently falls across the turbine by the turbine head, apart from any additional losses. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
In a horizontal, frictionless, incompressible flow along a streamline with no machine between two sections, what generally happens to static pressure when the velocity increases?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a horizontal ideal flow, elevation head is unchanged. Bernoulli's equation therefore requires the sum of pressure head and velocity head to remain constant along the streamline. If velocity increases, V²/(2g) increases. The pressure head must consequently decrease by a corresponding amount. Hence Option B correctly describes the pressure-velocity trade-off.
వివరణ:
Question 23
ప్రశ్న 23
Two engineers analyse the same Bernoulli problem but select different horizontal reference datums for elevation z. If both remain internally consistent, what happens to the predicted pressure and velocity differences?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Elevation head is measured relative to an arbitrary reference datum. Moving that datum shifts the numerical elevation of every point by the same constant amount. Bernoulli calculations depend on differences in total head between physical locations, so this common shift cancels. Pressure and velocity predictions therefore do not depend on where the zero-elevation datum is placed. Hence Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
When Bernoulli's equation is applied between two points using the same atmospheric pressure as the reference, which statement about gauge and absolute pressures is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Absolute pressure equals gauge pressure plus the applicable atmospheric reference pressure. If the same atmospheric pressure applies to both sections, converting both pressures to gauge values subtracts the same constant from each. That constant cancels when the two Bernoulli expressions are equated. Either absolute or consistently referenced gauge pressures therefore give the same pressure difference and velocity result. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
A large reservoir discharges water through a nozzle 10 m below its free surface. Both the reservoir surface and nozzle jet are exposed to atmosphere, the reservoir velocity is negligible, and the total head loss is specified as hL = 0.20[V²/(2g)], where V is the nozzle velocity. What is V? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The extended energy equation between the reservoir surface and the nozzle exit gives 10 = V²/(2g) + hL. With hL = 0.20[V²/(2g)], this becomes 10 = 1.20[V²/(2g)]. Thus V²/(2g) = 10/1.20 = 8.333 m. The resulting velocity is V = sqrt(2 × 9.81 × 8.333) = approximately 12.79 m/s. Therefore Option C is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. p/γ + V²/(2g) + z = constant సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. Pressure head, velocity head and elevation head respectively సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. Steady, incompressible flow with negligible viscous losses, with no pump or turbine between the points సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. For steady inviscid flow it is ordinarily applied along a streamline; if the flow is also irrotational, the Bernoulli constant is the same throughout a connected flow region సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. p/γ + z = constant సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 20 m సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. 1.83 m సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. 26.11 m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 220 kPa సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 15.34 m/s సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 50 kPa సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. 0.0770 m³/s సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. 121.52 kPa సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. z + p/γ సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: C. Velocity head V²/(2g) సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. Both EGL and HGL fall in the flow direction by the accumulated head loss and remain separated by the constant velocity head సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. 4.08 m సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. H1 + hp = H2 + ht + hL సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. 29 m సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. 50 m సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. It rises across a pump and falls across a turbine సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. Static pressure decreases because part of the pressure head is converted into velocity head సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: B. They remain unchanged because shifting the datum adds the same constant elevation to all relevant sections సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. Gauge pressures may be used consistently because subtracting the same atmospheric pressure from both absolute pressures does not change their pressure-head difference సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: C. 12.79 m/s సమాధానం: C.