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Question 1
ప్రశ్న 1
Which statement best describes the operating principle of a Pelton turbine?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A Pelton turbine is an impulse turbine. Most of the available pressure head is converted into kinetic energy before the water reaches the runner. The resulting high-speed free jet strikes the runner buckets. Its direction and tangential momentum change while passing over the buckets. This change of momentum produces force, torque and mechanical power on the runner, so Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
What is the primary function of the nozzle in a Pelton turbine?
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Explanation:
The nozzle receives water under pressure from the penstock. It converts most of the effective head into kinetic energy of a high-speed jet. The jet emerges into the runner region at approximately atmospheric pressure. The runner then extracts energy primarily from the momentum change of this jet. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
If H is the effective head available at a Pelton nozzle and Cv is the nozzle coefficient of velocity, which expression gives the actual jet velocity V1?
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Explanation:
The ideal velocity obtainable from a hydraulic head H is sqrt(2gH). A real nozzle has losses, so its actual velocity is slightly smaller than this ideal value. The coefficient of velocity Cv represents the ratio of actual to theoretical jet velocity. Hence V1 = Cv sqrt(2gH). Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A Pelton nozzle operates under an effective head of 400 m with Cv = 0.98. What is the jet velocity approximately? Take g = 9.81 m/s².
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Explanation:
The actual jet velocity is V1 = Cv sqrt(2gH). Substitution gives V1 = 0.98 sqrt(2 × 9.81 × 400). The ideal velocity is approximately 88.59 m/s. Multiplication by 0.98 gives approximately 86.82 m/s. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
What is the principal function of the central splitter in a Pelton bucket?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A Pelton bucket contains a central ridge or splitter. The incoming jet strikes near this splitter and divides approximately into two streams. The streams then flow around the two curved halves of the bucket. This arrangement enables a large change in jet direction while reducing undesirable side thrust on the runner. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For an ideal tangential Pelton jet striking a bucket moving with peripheral speed U, which inlet velocity relations are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Pelton jet approaches the runner tangentially. Therefore the inlet whirl component of absolute velocity is equal to the jet velocity V1. The bucket moves in the same tangential direction with speed U. Relative inlet velocity is the velocity of the jet relative to the moving bucket and is therefore V1 - U. Hence Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
If bucket friction reduces the magnitude of relative velocity so that Vr2 = k Vr1, what does the factor k represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Water loses some relative kinetic energy as it flows over the bucket surfaces. This reduction is commonly represented by Vr2 = kVr1. The factor k is therefore the ratio of relative outlet speed to relative inlet speed. For a real bucket k is generally less than one, whereas an ideal frictionless bucket would have k equal to one. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
Take the inlet jet direction as positive. If the outlet relative velocity is directed backward at an angle beta from the negative tangential direction, which expression gives the outlet whirl component Vw2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The bucket speed U acts in the positive tangential direction. At outlet, the tangential component of relative velocity acts principally in the opposite direction. Its signed tangential component is therefore -Vr2 cos(beta). Absolute velocity is obtained by adding the bucket velocity to the relative velocity vector. Thus Vw2 = U - Vr2 cos(beta), so Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
For a Pelton runner receiving discharge Q, which expression gives the tangential force on the runner when Vw1 and Vw2 are the inlet and outlet whirl components?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Tangential force follows from the change in tangential momentum of the flowing water. The mass flow rate is rho Q. The change in whirl velocity is Vw1 - Vw2 under the stated sign convention. Therefore the force exerted on the runner is rho Q(Vw1 - Vw2). Hence Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which expression gives the mechanical power transferred from water to a Pelton runner?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Runner power is the product of tangential force and bucket speed. The tangential force is rho Q(Vw1 - Vw2). Multiplication by peripheral speed U gives P = rho Q U(Vw1 - Vw2). The expression is consistent with the Euler turbine work relation for equal inlet and outlet blade speeds. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
Which statement correctly distinguishes Pelton wheel efficiency from hydraulic efficiency when nozzle losses are present?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Wheel or bucket efficiency measures how effectively the kinetic power of the jet striking the runner is converted into runner power. Its denominator is one-half rho Q V1². Hydraulic efficiency based on effective head H at the nozzle entrance compares runner power with rho g Q H. If the nozzle has losses, V1² is less than 2gH, so these two efficiencies are not identical. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For fixed Pelton bucket geometry and a fixed relative-velocity reduction factor, at what theoretical bucket-to-jet speed ratio is wheel efficiency maximum?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fixed outlet geometry and relative-velocity ratio, the idealized wheel-efficiency expression contains the factor U(V1 - U). Differentiating this factor with respect to U gives the condition V1 - 2U = 0. Therefore the theoretical optimum occurs at U = V1/2. The corresponding theoretical speed ratio is 0.50. Hence Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
Why is the practical Pelton bucket-to-jet speed ratio often taken near 0.46 rather than the theoretical wheel-efficiency value of 0.50?
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Explanation:
The theoretical value 0.50 maximizes an idealized runner-efficiency expression. A real turbine also has bearing friction, windage and other mechanical losses. Several of these losses increase as runner speed increases. The maximum overall efficiency therefore commonly occurs at a speed ratio somewhat below the theoretical wheel-efficiency value. A practical value near 0.46 is widely used, so Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a Pelton wheel with speed ratio phi = U/V1, relative-velocity ratio k = Vr2/Vr1, and bucket outlet angle beta measured from the backward tangential direction, which wheel-efficiency expression follows from the velocity triangles?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The relative inlet velocity is V1 - U. The outlet whirl component is U - k(V1 - U)cos(beta). Therefore Vw1 - Vw2 equals (V1 - U)[1 + k cos(beta)]. Substitution into runner power and division by the jet kinetic power gives eta_w = 2phi(1 - phi)[1 + k cos(beta)]. Hence Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
For fixed k and beta in the standard Pelton velocity-triangle analysis, what is the maximum theoretical wheel efficiency?
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Explanation:
Wheel efficiency is eta_w = 2phi(1 - phi)[1 + k cos(beta)]. The factor phi(1 - phi) has its maximum value of one-quarter at phi = one-half. Substitution gives eta_w,max = 2 × one-quarter × [1 + k cos(beta)]. This simplifies to [1 + k cos(beta)]/2. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A Pelton wheel has speed ratio phi = 0.46, relative-velocity ratio k = 0.90, bucket outlet angle beta = 15 degrees from the backward tangent, and nozzle coefficient Cv = 0.98. What is the hydraulic efficiency based on the effective head H at the nozzle entrance approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
First compute wheel efficiency as eta_w = 2phi(1 - phi)[1 + k cos(beta)]. This gives approximately 0.9287, or 92.87 percent. Since V1 = Cv sqrt(2gH), jet kinetic power is Cv² times rho g Q H. Therefore hydraulic efficiency based on nozzle-entry head is eta_h = Cv² eta_w = 0.98² × 0.9287. This gives approximately 0.8919, or 89.2 percent, so Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
A Pelton wheel receives a jet at V1 = 50 m/s and its buckets move at U = 23 m/s. If k = 0.90 and beta = 15 degrees from the backward tangent, what is the outlet whirl velocity Vw2 approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The inlet relative velocity is Vr1 = 50 - 23 = 27 m/s. The outlet relative speed is Vr2 = 0.90 × 27 = 24.3 m/s. The outlet whirl component is Vw2 = 23 - 24.3 cos(15 degrees). This evaluates to approximately -0.472 m/s. The negative sign indicates a small whirl component opposite to the direction of bucket motion, so Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
For a Pelton wheel with V1 = 50 m/s, U = 23 m/s, Q = 0.20 m³/s, k = 0.90 and beta = 15 degrees, what tangential force acts on the runner? Take rho = 1000 kg/m³.
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Explanation:
The outlet whirl velocity for the stated condition is approximately -0.472 m/s. The inlet whirl component is Vw1 = 50 m/s. Therefore F = rho Q(Vw1 - Vw2) = 1000 × 0.20 × [50 - (-0.472)]. This gives approximately 10094 N. Thus the tangential force is about 10.09 kN and Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
For the Pelton wheel with V1 = 50 m/s, U = 23 m/s, Q = 0.20 m³/s, k = 0.90 and beta = 15 degrees, what power is transferred to the runner approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The tangential force for the stated condition is approximately 10094.4 N. Runner power is force multiplied by bucket peripheral speed. Hence P = 10094.4 × 23 W. This gives approximately 232171 W. Therefore the runner receives about 232 kW and Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
A Pelton runner develops approximately 232.17 kW. If the mechanical efficiency from runner to shaft is 96%, what shaft power is obtained?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Mechanical efficiency is defined as shaft power divided by runner power. Therefore shaft power equals mechanical efficiency multiplied by runner power. Substitution gives 0.96 × 232.17 kW. The result is approximately 222.88 kW. Thus Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
A Pelton turbine operates under H = 400 m with Cv = 0.98. If its practical speed ratio is U/V1 = 0.46 and the mean runner diameter is 1.75 m, what rotational speed is required approximately? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The jet velocity is V1 = 0.98 sqrt(2 × 9.81 × 400), which is approximately 86.82 m/s. The bucket speed is U = 0.46 × 86.82 = approximately 39.94 m/s. Peripheral speed is related to rotational speed by U = pi DN/60. Therefore N = 60U/(pi D) = approximately 435.8 rpm. Rounding gives 436 rpm, so Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
A single Pelton nozzle supplies Q = 1.74 m³/s at a jet velocity of 86.82 m/s. Assuming a circular jet, what nozzle diameter is required approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a circular jet, discharge is Q = (pi d²/4)V1. Rearranging gives d = sqrt[4Q/(pi V1)]. Substituting Q = 1.74 m³/s and V1 = 86.82 m/s gives approximately 0.1597 m. This rounds to about 0.160 m. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
The jet ratio of a Pelton wheel is defined as D/d, where D is the mean runner diameter and d is the jet diameter. If D = 2.40 m and d = 0.15 m, what is the jet ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Jet ratio compares mean runner diameter with jet diameter. It is calculated directly as D/d. Here D/d = 2.40/0.15. This gives a value of 16. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement gives the best engineering interpretation of Pelton jet ratio D/d?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The jet ratio relates the pitch diameter of the wheel to the jet diameter. If the ratio is too small, the jet is large relative to the runner and efficient bucket arrangement becomes difficult. If it is too large, the runner becomes unnecessarily large and its rotational speed is reduced for a given peripheral velocity. Practical high-efficiency Pelton designs therefore use a moderate jet ratio, commonly around the mid-teens. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Why may a Pelton turbine use multiple jets instead of a single jet?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each Pelton jet supplies additional mass flow and momentum to the runner. Multiple jets therefore permit greater total discharge and power for a given head. This can avoid using one jet that is excessively large relative to the runner diameter. The jets must be positioned so that their interactions with successive buckets remain satisfactory. The turbine nevertheless remains an impulse turbine, so Option A is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. A high-speed free jet strikes the buckets and transfers energy mainly through change of momentum సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. To convert the available hydraulic head mainly into jet kinetic energy సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: B. V1 = Cv sqrt(2gH) సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. 86.8 m/s సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. To divide the incoming jet approximately into two streams and guide them around the bucket సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. Vw1 = V1 and Vr1 = V1 - U సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. The ratio of outlet to inlet relative velocity magnitude through the bucket సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. Vw2 = U - Vr2 cos(beta) సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. F = rho Q(Vw1 - Vw2) సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. P = rho Q U(Vw1 - Vw2) సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. Wheel efficiency is runner power divided by jet kinetic power, whereas hydraulic efficiency based on nozzle-entry head is runner power divided by rho g Q H సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. U/V1 = 0.50 సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: A. Bearing, windage and other practical losses alter the speed corresponding to maximum overall efficiency సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: A. eta_w = 2phi(1 - phi)[1 + k cos(beta)] సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. eta_w,max = [1 + k cos(beta)]/2 సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: B. 89.2% సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. -0.472 m/s సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. 10.09 kN సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. 232 kW సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: C. 222.9 kW సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 436 rpm సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 0.160 m సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 16 సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. It relates runner size to jet size; an excessively small value causes bucket-spacing and interference difficulties, while an excessively large value leads to a comparatively large and slower runner సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: A. To increase total discharge and power for a given head without requiring one excessively large jet సమాధానం: A.