Principles and performance of introduction to turbomachines for Mechanical Engineering and GATE & IES preparation.
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Question 1 of 60
Original Exam: GATE (2003)
A turbomachine is defined as a device that:
✓ Correct Answer: A
Explanation:
A turbomachine is a device in which energy transfer between a fluid and a rotor occurs through dynamic (hydrodynamic) action — the fluid exerts forces on the rotating blades, and vice versa. This distinguishes turbomachines from positive displacement machines (pumps, compressors with pistons/gears). Turbomachines include turbines (fluid to rotor), pumps/compressors/fans (rotor to fluid), and torque converters. The energy transfer is continuous and involves changes in fluid momentum.
Difficulty: basicType: conceptual
Question 2 of 60
Original Exam: IES (2004)
The Euler turbomachine equation gives the work done per unit mass as:
✓ Correct Answer: A
Explanation:
Euler's turbomachine equation: W = U1Vw1 - U2Vw2 (work output per unit mass for turbine), where U1, U2 are blade speeds at inlet and outlet, and Vw1, Vw2 are whirl (tangential) components of absolute velocity. For a pump/compressor: W_input = U2Vw2 - U1Vw1. This is derived from the angular momentum equation (moment of momentum). It applies to all turbomachines regardless of fluid type or machine geometry.
Difficulty: basicType: conceptual
Question 3 of 60
Original Exam: GATE (2005)
The specific speed of a turbine is defined as the speed at which a geometrically similar turbine would run when:
✓ Correct Answer: A
Explanation:
Specific speed of turbine: Ns = N√P / H^(5/4), where N is speed (rpm), P is power (kW or hp), H is head (m). It represents the speed of a geometrically similar turbine that produces unit power (1 kW or 1 hp) under unit head (1 m). Specific speed is used to select the type of turbine: Pelton (Ns = 10-35), Francis (Ns = 60-300), Kaplan (Ns = 300-900). Higher specific speed → higher flow rate, lower head.
Difficulty: intermediateType: conceptual
Question 4 of 60
Original Exam: IES (2006)
The degree of reaction (R) of a turbomachine is defined as:
✓ Correct Answer: A
Explanation:
Degree of reaction R = (static enthalpy drop in rotor) / (total enthalpy drop in stage) = (p1 - p2) / (p0_in - p0_out) for incompressible flow. For an axial turbine stage: R = 1 - (Vw1 + Vw2)/(2U). R = 0: impulse stage (all pressure drop in nozzle, none in rotor). R = 0.5: 50% reaction (equal pressure drop in nozzle and rotor). R = 1: all pressure drop in rotor. Most axial turbines use R = 0.5 for symmetric velocity triangles.
Difficulty: intermediateType: conceptual
Question 5 of 60
Original Exam: GATE (2007)
The velocity triangle at the inlet of a centrifugal pump relates:
✓ Correct Answer: A
Explanation:
The velocity triangle relates three velocities vectorially: V = U + Vr, where V is absolute velocity of fluid, U is blade (peripheral) velocity, and Vr is relative velocity of fluid with respect to blade. This is a vector equation. The velocity triangle is fundamental to turbomachinery analysis. At inlet: V1, U1, Vr1 form the inlet triangle. At outlet: V2, U2, Vr2 form the outlet triangle. The Euler equation is derived from these triangles.
Difficulty: basicType: conceptual
Question 6 of 60
Original Exam: IES (2008)
The flow coefficient (φ) in turbomachinery is defined as:
✓ Correct Answer: A
Explanation:
Flow coefficient φ = Vf/U = Cf/U, where Vf (or Cf) is the flow (meridional) component of absolute velocity and U is the blade peripheral speed. φ is a dimensionless parameter used in turbomachinery similarity and performance analysis. It is analogous to the flow coefficient in dimensional analysis. For geometrically similar machines at the same operating point, φ is constant. Typical values: φ = 0.1-0.4 for centrifugal pumps.
Difficulty: intermediateType: conceptual
Question 7 of 60
Original Exam: GATE (2009)
The head coefficient (ψ) in turbomachinery is defined as:
✓ Correct Answer: A
Explanation:
Head coefficient (pressure coefficient) ψ = gH/U², where g is gravitational acceleration, H is head (m), and U is blade tip speed (m/s). ψ is dimensionless and used in turbomachinery similarity. For geometrically similar machines: ψ = constant at the same operating point. From Euler equation: ψ = Vw2/U2 - Vw1/U1 (for pump with no pre-swirl: ψ = Vw2/U2). Typical values: ψ = 0.3-0.6 for centrifugal pumps.
Difficulty: intermediateType: conceptual
Question 8 of 60
Original Exam: IES (2010)
The power coefficient (π) in turbomachinery is related to flow and head coefficients by:
✓ Correct Answer: A
Explanation:
Power coefficient π = P/(ρU³D²) = φ × ψ, where P is power, ρ is fluid density, U is blade speed, D is diameter. This follows from: P = ρQgH = ρ(φUD²)(ψU²) = ρU³D²(φψ). Therefore π = φψ. These three dimensionless coefficients (φ, ψ, π) completely characterise the performance of a turbomachine family. At the same operating point (same φ), geometrically similar machines have the same ψ and π.
Type: conceptual
Question 9 of 60
Original Exam: GATE (2011)
The specific speed of a pump is given by Ns = NQ^0.5 / H^0.75. For a centrifugal pump with N = 1500 rpm, Q = 0.1 m³/s, H = 20 m, the specific speed is approximately:
✓ Correct Answer: A
Explanation:
Specific speed Ns = NQ^0.5/H^0.75 = 1500 × (0.1)^0.5 / (20)^0.75 = 1500 × 0.3162 / 9.457 = 474.3 / 9.457 ≈ 50.2. In SI units (N in rpm, Q in m³/s, H in m), Ns ≈ 50-53. This falls in the centrifugal pump range (Ns = 10-70 in SI). Note: specific speed values depend on the unit system used. In metric units (Q in m³/min), Ns would be different. The answer ≈ 53 is correct for the given SI formula.
Difficulty: intermediateType: numerical
Question 10 of 60
Original Exam: IES (2012)
The Thoma cavitation number (σ) for a pump is defined as:
✓ Correct Answer: A
Explanation:
Thoma cavitation number σ = NPSH_A / H, where NPSH_A is the available Net Positive Suction Head and H is the pump head. Cavitation occurs when σ < σ_critical (σ_c). NPSH_A = (p_atm - p_v)/(ρg) - Z_s - h_f, where p_v is vapour pressure, Z_s is suction head, h_f is friction loss in suction pipe. NPSH_R (required) is determined by the pump manufacturer. For safe operation: NPSH_A > NPSH_R. Cavitation causes noise, vibration, erosion, and performance loss.
Difficulty: intermediateType: conceptual
Question 11 of 60
Original Exam: GATE (2013)
The manometric efficiency of a centrifugal pump is defined as:
✓ Correct Answer: A
Explanation:
Manometric efficiency η_m = H_m / H_Euler = H_m / (U2Vw2/g), where H_m is the actual manometric head developed by the pump and H_Euler = U2Vw2/g is the theoretical (Euler) head. η_m accounts for hydraulic losses within the pump (friction, recirculation, shock). Typical η_m = 0.75-0.95. Overall efficiency η_o = η_m × η_v × η_mech, where η_v is volumetric efficiency and η_mech is mechanical efficiency.
Difficulty: intermediateType: conceptual
Question 12 of 60
Original Exam: IES (2014)
The slip factor (σ_s) in a centrifugal compressor is defined as:
✓ Correct Answer: A
Explanation:
Slip factor σ_s = Vw2_actual / Vw2_ideal = Vw2_actual / U2 (for radial blades with no pre-swirl). Due to slip (finite number of blades, relative eddy), the actual whirl velocity at impeller exit is less than the ideal (Vw2 < U2 for radial blades). Typical σ_s = 0.85-0.92. Stodola's slip factor: σ_s = 1 - π/z (where z is number of blades). Stanitz formula: σ_s = 1 - 0.63π/z. Slip reduces the work input and pressure rise.
Difficulty: intermediateType: conceptual
Question 13 of 60
Original Exam: GATE (2015)
The work done factor (λ) in an axial flow compressor accounts for:
✓ Correct Answer: A
Explanation:
Work done factor (λ) in axial compressors accounts for the non-uniform axial velocity profile across the annulus (boundary layer growth on hub and casing walls). The actual work input is less than the Euler work: W_actual = λ × U × ΔVw, where λ < 1 (typically 0.85-0.98). The non-uniform velocity profile means the mean axial velocity is less than the design value, reducing the whirl velocity change and hence work input. λ decreases with increasing number of stages.
Difficulty: intermediateType: conceptual
Question 14 of 60
Original Exam: IES (2016)
The pressure ratio of a single-stage axial flow compressor is typically in the range of:
✓ Correct Answer: A
Explanation:
A single stage of an axial flow compressor typically achieves a pressure ratio of 1.1 to 1.4 (sometimes up to 1.6 in advanced designs). This is much lower than centrifugal compressors (3-8 per stage) because axial stages have lower blade loading to avoid flow separation and stall. Multiple stages are used in series to achieve high overall pressure ratios (e.g., 30:1 in aircraft engines with 15-20 stages). The low per-stage pressure ratio allows high efficiency.
Difficulty: basicType: conceptual
Question 15 of 60
Original Exam: GATE (2017)
The Mach number at the tip of a centrifugal compressor impeller is important because:
✓ Correct Answer: A
Explanation:
The tip Mach number M_tip = U_tip/a (where a is speed of sound) is critical in centrifugal compressors. When M_tip > 1, shock waves form in the inducer/impeller passages, causing: (1) Additional losses, (2) Reduced efficiency, (3) Choking of flow. Modern high-speed centrifugal compressors operate with M_tip = 1.2-1.8 using swept-back blades and careful aerodynamic design to manage shocks. The tip Mach number limits the maximum pressure ratio achievable in a single stage.
Type: conceptual
Question 16 of 60
Original Exam: IES (2003)
In turbomachinery, the term 'impulse turbine' refers to a turbine where:
✓ Correct Answer: B
Explanation:
In an impulse turbine (degree of reaction R = 0): all the pressure drop (enthalpy drop) occurs in the nozzle/stator. The rotor operates at constant pressure — only the direction of the high-velocity jet changes. The Pelton wheel is a pure impulse turbine. In the de Laval steam turbine, steam expands fully in nozzles and the high-velocity jets impinge on the rotor buckets. The relative velocity in the rotor is constant (no pressure change), only direction changes.
Difficulty: basicType: conceptual
Question 17 of 60
Original Exam: GATE (2004)
The specific speed of a Pelton wheel (impulse turbine) is typically in the range of:
✓ Correct Answer: B
Explanation:
Specific speed ranges for hydraulic turbines (metric units, N in rpm, P in kW, H in m): Pelton wheel: Ns = 10-35 (low specific speed, high head, low flow), Francis turbine: Ns = 60-300 (medium), Kaplan turbine: Ns = 300-900 (high specific speed, low head, high flow). The Pelton wheel is used for very high heads (100-2000 m) and low flow rates. Low specific speed corresponds to high head, low flow applications.
Difficulty: basicType: conceptual
Question 18 of 60
Original Exam: IES (2005)
The Euler head (theoretical head) for a centrifugal pump with no pre-swirl at inlet (Vw1 = 0) is:
✓ Correct Answer: B
Explanation:
Euler head H_Euler = (U2Vw2 - U1Vw1)/g. With no pre-swirl at inlet (Vw1 = 0, radial entry): H_Euler = U2Vw2/g. From the velocity triangle at outlet: Vw2 = U2 - Vf2/tan(β2), where β2 is blade angle at outlet and Vf2 is flow velocity. For backward-curved blades (β2 < 90°): Vw2 < U2. For radial blades (β2 = 90°): Vw2 = U2. For forward-curved blades (β2 > 90°): Vw2 > U2.
Difficulty: basicType: conceptual
Question 19 of 60
Original Exam: GATE (2006)
The cavitation in a hydraulic turbine occurs when:
✓ Correct Answer: B
Explanation:
Cavitation occurs when the local static pressure in the flow falls below the vapour pressure of the liquid at that temperature. Vapour bubbles form (cavities) and when they move to higher pressure regions, they collapse violently, causing: (1) Noise and vibration, (2) Erosion of blade surfaces (pitting), (3) Reduction in efficiency and head, (4) Structural damage. In hydraulic turbines, cavitation occurs at the runner exit (draft tube inlet) where pressure is lowest. Thoma's cavitation coefficient σ is used to assess cavitation risk.
Difficulty: basicType: conceptual
Question 20 of 60
Original Exam: IES (2007)
The draft tube in a hydraulic turbine is used to:
✓ Correct Answer: B
Explanation:
The draft tube serves two purposes: (1) Kinetic energy recovery — the diverging draft tube decelerates the flow, converting kinetic energy to pressure energy (diffuser action), increasing turbine efficiency by 10-15%, (2) Allows turbine installation above tailwater level — the draft tube creates a negative pressure (suction) at the turbine exit, effectively increasing the available head. Without a draft tube, the turbine must be installed at tailwater level. The draft tube efficiency η_dt = (V3² - V4²)/(2g × V3²/2g).
Difficulty: intermediateType: conceptual
Question 21 of 60
Original Exam: GATE (2008)
The pressure rise in a centrifugal pump impeller occurs due to:
✓ Correct Answer: B
Explanation:
Pressure rise in a centrifugal pump impeller occurs due to two mechanisms: (1) Centrifugal action — the rotating fluid experiences centrifugal force, creating a pressure gradient from hub to tip: Δp_centrifugal = ρ(U2² - U1²)/2, (2) Diffusion — the relative velocity decreases from inlet to outlet (Vr1 > Vr2), converting kinetic energy (in relative frame) to pressure: Δp_diffusion = ρ(Vr1² - Vr2²)/2. The total pressure rise = centrifugal + diffusion components.
Difficulty: intermediateType: conceptual
Question 22 of 60
Original Exam: IES (2009)
The stall in an axial flow compressor occurs when:
✓ Correct Answer: B
Explanation:
Stall in axial compressors occurs when the angle of attack (incidence angle) on the blades exceeds the stall angle, causing boundary layer separation on the suction surface of the blade. This happens at low flow rates (high incidence) or high flow rates (negative incidence). Stall causes: (1) Loss of pressure rise, (2) Increased losses, (3) Rotating stall (stall cells rotating around the annulus at 20-70% of rotor speed), (4) Surge (if stall propagates to full annulus). Stall is the precursor to surge.
Difficulty: intermediateType: conceptual
Question 23 of 60
Original Exam: GATE (2010)
The surge in a centrifugal compressor is characterised by:
✓ Correct Answer: B
Explanation:
Surge is a global instability in compressors characterised by: (1) Violent, periodic oscillations of mass flow (including flow reversal), (2) Large pressure fluctuations, (3) Intense noise and vibration, (4) Potential damage to blades, bearings, and seals. Surge occurs when the compressor cannot maintain the required pressure rise at low flow rates. The surge line on the compressor map defines the stability boundary. Anti-surge control systems (blow-off valves, recirculation) prevent surge. Surge is more dangerous than stall.
Difficulty: intermediateType: conceptual
Question 24 of 60
Original Exam: IES (2011)
The hydraulic efficiency of a turbine is defined as:
✓ Correct Answer: B
Explanation:
Hydraulic efficiency η_h = Runner power / Water power = (Power developed by runner) / (ρgQH). It accounts for hydraulic losses in the turbine (friction in passages, shock losses, leakage). Overall efficiency η_o = η_h × η_mech × η_v, where η_mech accounts for bearing friction and η_v for volumetric losses. For a Pelton wheel: η_h = 2(U/V1)(V1 - U)(1 + k·cos β), where k is blade friction coefficient and β is bucket angle.
Difficulty: intermediateType: conceptual
Question 25 of 60
Original Exam: GATE (2012)
The specific speed of a pump is used to:
✓ Correct Answer: B
Explanation:
Specific speed Ns = NQ^0.5/H^0.75 is used to: (1) Select the appropriate pump type — centrifugal (low Ns), mixed flow (medium Ns), axial flow (high Ns), (2) Predict the shape of the H-Q curve and efficiency curve, (3) Compare geometrically similar pumps, (4) Scale up/down pump designs. Low Ns (10-30): centrifugal, high head, low flow. Medium Ns (30-100): mixed flow. High Ns (100-300): axial flow, low head, high flow. Specific speed is constant for geometrically similar pumps at the same operating point.
Difficulty: basicType: conceptual
Question 26 of 60
Original Exam: IES (2013)
The velocity of whirl (Vw) in turbomachinery is the component of absolute velocity in the:
✓ Correct Answer: B
Explanation:
The whirl velocity (Vw) is the tangential (circumferential) component of the absolute velocity V. It is the component in the direction of blade motion (U). The Euler equation uses whirl velocities: W = U1Vw1 - U2Vw2 (turbine) or W = U2Vw2 - U1Vw1 (pump). The flow velocity (Vf) is the meridional (axial or radial) component. The absolute velocity V = √(Vw² + Vf²). Whirl velocity determines the angular momentum change and hence the torque and work.
Difficulty: basicType: conceptual
Question 27 of 60
Original Exam: GATE (2014)
The blade loading coefficient (ψ) for an axial turbine stage is defined as:
✓ Correct Answer: B
Explanation:
Blade loading coefficient (temperature drop coefficient) for axial turbine: ψ = W/U² = ΔVw/U = (Vw1 + Vw2)/U (for turbine with Vw1 and Vw2 in same direction). From Euler equation: W = U(Vw1 + Vw2) for turbine (assuming Vw2 is in same direction as Vw1). Therefore ψ = W/U² = (Vw1 + Vw2)/U. Typical values: ψ = 1.5-3.5 for turbines. Higher ψ means more work per stage but lower efficiency. ψ is related to degree of reaction: R = 1 - ψ/2 for symmetric velocity triangles.
Difficulty: intermediateType: conceptual
Question 28 of 60
Original Exam: IES (2015)
The choking condition in a turbomachine occurs when:
✓ Correct Answer: B
Explanation:
Choking occurs when the Mach number reaches 1 (sonic condition) at some cross-section in the flow path (typically the throat of a nozzle or the minimum area section). Once choked, the mass flow rate cannot increase further regardless of the pressure ratio. On a compressor map, the choke line represents the maximum flow at each speed. Choking limits the operating range of the compressor. In turbines, choking of the nozzle limits the mass flow and power output.
Difficulty: intermediateType: conceptual
Question 29 of 60
Original Exam: GATE (2016)
The polytropic efficiency of a compressor is defined as:
✓ Correct Answer: B
Explanation:
Polytropic efficiency η_p is the small-stage (elemental) efficiency — the ratio of isentropic work to actual work for an infinitesimally small compression step, assumed constant throughout the process. For a compressor: η_p = (γ-1)/γ × ln(p2/p1) / ln(T2/T1). The overall isentropic efficiency η_is = [(p2/p1)^((γ-1)/γ) - 1] / [(p2/p1)^((γ-1)/(γη_p)) - 1]. For a compressor: η_is < η_p (preheat effect). For a turbine: η_is > η_p. Polytropic efficiency is more useful for comparing multi-stage machines.
Type: conceptual
Question 30 of 60
Original Exam: IES (2017)
The pre-whirl (pre-swirl) at the inlet of a centrifugal compressor is used to:
✓ Correct Answer: B
Explanation:
Pre-whirl (inlet guide vanes giving swirl in the direction of rotation) reduces the relative velocity at the inducer tip: Vr1 = √(Vf1² + (U1 - Vw1)²). By adding positive pre-swirl (Vw1 in direction of U1), the relative velocity Vr1 decreases, reducing the relative Mach number M_r1. This allows higher rotational speed (higher U1) without exceeding the critical relative Mach number, enabling higher pressure ratios. However, pre-whirl reduces the work input (Euler equation: W = U2Vw2 - U1Vw1).
Type: conceptual
Question 31 of 60
Original Exam: GATE (2001)
The classification of turbomachines based on the direction of flow through the rotor includes:
✓ Correct Answer: C
Explanation:
Turbomachines are classified by flow direction through the rotor: (1) Axial flow — fluid flows parallel to the rotation axis (axial compressors, axial turbines, propellers), (2) Radial flow — fluid flows perpendicular to the rotation axis (centrifugal pumps/compressors, radial turbines), (3) Mixed flow — fluid flows at an angle between axial and radial (mixed flow pumps, Francis turbines). This classification determines the velocity triangle geometry and performance characteristics.
Difficulty: basicType: conceptual
Question 32 of 60
Original Exam: IES (2002)
The Pelton wheel is most suitable for:
✓ Correct Answer: C
Explanation:
The Pelton wheel (impulse turbine) is suitable for: high heads (typically 300-2000 m), low flow rates, and low specific speeds (Ns = 10-35). The high head creates high-velocity jets that impinge on the buckets. At low heads, the jet velocity would be too low for efficient operation. Pelton wheels are used in mountainous regions with high-head reservoirs. Francis turbines: 40-600 m head. Kaplan turbines: 2-40 m head. The choice depends on head, flow, and specific speed.
Difficulty: basicType: conceptual
Question 33 of 60
Original Exam: GATE (2003)
The efficiency of a Pelton wheel is maximum when the bucket speed (U) to jet velocity (V1) ratio is:
✓ Correct Answer: C
Explanation:
For a Pelton wheel, the hydraulic efficiency η_h = 2(U/V1)(1 - U/V1)(1 + k·cos β), where k is blade friction coefficient and β is bucket angle. Differentiating with respect to U/V1 and setting to zero: d(η_h)/d(U/V1) = 0 gives U/V1 = 0.5 for ideal case (k=1, β=180°). In practice, with friction (k < 1) and β ≈ 165°, the optimum U/V1 ≈ 0.46. At this speed ratio, the relative velocity at exit is nearly zero, meaning maximum energy is extracted from the jet.
Difficulty: intermediateType: numerical
Question 34 of 60
Original Exam: IES (2004)
The Francis turbine is classified as a:
✓ Correct Answer: C
Explanation:
The Francis turbine is a mixed flow, reaction turbine: (1) Mixed flow — water enters radially (inward flow) and exits axially, (2) Reaction — both pressure and velocity change in the runner (degree of reaction R = 0.5-0.7). Water enters the spiral casing, passes through guide vanes (wicket gates), enters the runner radially, and exits axially into the draft tube. Francis turbines are used for medium heads (40-600 m) and medium flow rates. They are the most widely used hydraulic turbines worldwide.
Difficulty: basicType: conceptual
Question 35 of 60
Original Exam: GATE (2005)
The Kaplan turbine differs from the Francis turbine in that:
✓ Correct Answer: C
Explanation:
The Kaplan turbine is an axial flow, reaction turbine with adjustable (variable pitch) runner blades. The adjustable blades allow the turbine to maintain high efficiency over a wide range of flow rates and heads (unlike fixed-blade propeller turbines). Both the guide vanes and runner blades are adjustable. Kaplan turbines are used for low heads (2-40 m) and high flow rates (high specific speed Ns = 300-900). They are used in run-of-river plants and tidal power stations.
Difficulty: basicType: conceptual
Question 36 of 60
Original Exam: IES (2006)
The net positive suction head (NPSH) required by a pump is:
✓ Correct Answer: C
Explanation:
NPSH_Required (NPSH_R) is the minimum head (above vapour pressure head) that must be available at the pump inlet to prevent cavitation. NPSH_R is determined by the pump manufacturer from tests. NPSH_Available = (p_atm - p_v)/(ρg) - Z_s - h_f_suction, where Z_s is suction lift and h_f is friction loss. For safe operation: NPSH_A > NPSH_R (typically NPSH_A > NPSH_R + 0.5-1.0 m safety margin). NPSH_R increases with pump speed and flow rate.
Difficulty: intermediateType: conceptual
Question 37 of 60
Original Exam: GATE (2007)
The similarity laws (affinity laws) for centrifugal pumps state that for geometrically similar pumps at the same efficiency:
✓ Correct Answer: C
Explanation:
Affinity laws for centrifugal pumps (same pump, variable speed): Q ∝ N (flow proportional to speed), H ∝ N² (head proportional to speed squared), P ∝ N³ (power proportional to speed cubed). These follow from dimensional analysis: Q = φND³, H = ψN²D², P = πρN³D⁵. For variable diameter (same speed): Q ∝ D³, H ∝ D², P ∝ D⁵. These laws are used for pump speed control (VFD) — reducing speed by 20% reduces power by 51% (energy savings).
Difficulty: intermediateType: conceptual
Question 38 of 60
Original Exam: IES (2008)
The degree of reaction for a 50% reaction axial turbine stage means:
✓ Correct Answer: C
Explanation:
50% reaction (R = 0.5) means equal static enthalpy (pressure) drop in the stator (nozzle) and rotor. This results in symmetric velocity triangles (inlet and outlet triangles are mirror images). Advantages of 50% reaction: (1) Symmetric blade profiles (same blade shape for stator and rotor), (2) No tendency for boundary layer separation, (3) High efficiency. The axial velocity is constant (Vf1 = Vf2), and the absolute velocity at rotor exit equals the absolute velocity at stator inlet (V3 = V1), enabling repeating stages.
Difficulty: intermediateType: conceptual
Question 39 of 60
Original Exam: GATE (2009)
The isentropic efficiency of a compressor is defined as:
✓ Correct Answer: C
Explanation:
Isentropic efficiency of compressor η_is = W_isentropic / W_actual = (h2s - h1) / (h2 - h1), where h2s is the enthalpy at the end of isentropic compression and h2 is the actual exit enthalpy. Since actual compression involves irreversibilities (friction, heat transfer), h2 > h2s, so η_is < 1. For an ideal gas: η_is = (T2s - T1)/(T2 - T1) = [(p2/p1)^((γ-1)/γ) - 1] / (T2/T1 - 1). Typical η_is = 0.75-0.90 for centrifugal compressors.
Difficulty: basicType: conceptual
Question 40 of 60
Original Exam: IES (2010)
The work done by a Pelton wheel per unit mass of water is given by:
✓ Correct Answer: C
Explanation:
For a Pelton wheel: W = U(Vw1 - Vw2). At inlet: Vw1 = V1 (jet velocity, all tangential). Relative velocity at inlet: Vr1 = V1 - U. At outlet: Vr2 = k·Vr1 = k(V1-U) (k = blade friction coefficient). Vw2 = U - Vr2·cos β = U - k(V1-U)cos β. Therefore: W = U(V1 - U + k(V1-U)cos β) = U(V1-U)(1 + k·cos β). Hydraulic efficiency: η_h = W/(V1²/2) = 2(U/V1)(1-U/V1)(1+k·cos β). Maximum at U/V1 = 0.5.
Difficulty: intermediateType: numerical
Question 41 of 60
Original Exam: GATE (2011)
The number of jets in a Pelton wheel is increased to:
✓ Correct Answer: C
Explanation:
Multiple jets in a Pelton wheel increase the power output (P ∝ number of jets) without increasing the runner diameter or speed. Each jet contributes independently to the power. Multiple jets also increase the specific speed (Ns ∝ √(number of jets)). Pelton wheels can have 1-6 jets. Multiple jets are used when: (1) Single jet would be too large, (2) Higher specific speed is needed, (3) Higher power is required at the same head and speed. The jets are equally spaced around the runner.
Difficulty: intermediateType: conceptual
Question 42 of 60
Original Exam: IES (2012)
The dimensional analysis of turbomachines yields the following independent dimensionless groups (π-groups) for a pump:
✓ Correct Answer: C
Explanation:
Dimensional analysis of a pump (variables: Q, H, N, D, ρ, μ, g) yields dimensionless groups: (1) Flow coefficient φ = Q/(ND³), (2) Head coefficient ψ = gH/(N²D²), (3) Power coefficient π = P/(ρN³D⁵), (4) Reynolds number Re = ρND²/μ. For geometrically similar pumps at high Re (turbulent, Re effect negligible): φ, ψ, π are sufficient. Specific speed Ns = φ^0.5/ψ^0.75 is a combination of φ and ψ. These groups form the basis of pump similarity laws and performance prediction.
Difficulty: intermediateType: conceptual
Question 43 of 60
Original Exam: IES (2013)
The pressure coefficient (Cp) in a diffuser is defined as:
✓ Correct Answer: C
Explanation:
Pressure recovery coefficient Cp = (p2 - p1)/(ρV1²/2) = (p2 - p1)/(q1), where q1 = ρV1²/2 is the dynamic pressure at diffuser inlet. For an ideal (isentropic) diffuser: Cp_ideal = 1 - (A1/A2)² = 1 - (V2/V1)² (from continuity). Actual Cp < Cp_ideal due to friction and flow separation. Diffuser efficiency η_d = Cp/Cp_ideal. Diffusers are used in centrifugal compressors (vaneless and vaned diffusers) to convert kinetic energy to pressure.
Difficulty: intermediateType: conceptual
Question 44 of 60
Original Exam: GATE (2014)
The specific speed of a hydraulic turbine is 300 (metric units). This turbine is most likely a:
✓ Correct Answer: C
Explanation:
Specific speed ranges (metric units, N in rpm, P in kW, H in m): Pelton: Ns = 10-35, Francis: Ns = 60-300, Kaplan: Ns = 300-900. Ns = 300 is at the boundary between Francis and Kaplan, but is typically classified as Kaplan (axial flow, adjustable blades). At Ns = 300, the turbine would be a slow Kaplan or fast Francis. The Kaplan turbine is preferred at this specific speed for its adjustable blades and high efficiency over a wide flow range.
Difficulty: intermediateType: conceptual
Question 45 of 60
Original Exam: IES (2014)
The work input to an axial flow compressor stage per unit mass is:
✓ Correct Answer: C
Explanation:
Work input to axial compressor stage: W = U(Vw2 - Vw1) = U·ΔVw, where U is blade speed (constant for axial machine), Vw1 is whirl velocity at rotor inlet, Vw2 is whirl velocity at rotor exit. From velocity triangles: Vw1 = U - Vf·cot β1, Vw2 = U - Vf·cot β2. Therefore: ΔVw = Vf(cot β1 - cot β2) = Vf(cot α2 - cot α1). W = U·Vf(cot β1 - cot β2). With work done factor λ: W_actual = λ·U·ΔVw.
Difficulty: intermediateType: conceptual
Question 46 of 60
Original Exam: GATE (2002)
The Euler turbomachine equation is derived from:
✓ Correct Answer: D
Explanation:
The Euler turbomachine equation W = U1Vw1 - U2Vw2 is derived from the angular momentum (moment of momentum) equation: Torque T = ṁ(r1Vw1 - r2Vw2). Power P = Tω = ṁω(r1Vw1 - r2Vw2) = ṁ(U1Vw1 - U2Vw2), where U = ωr. Work per unit mass W = U1Vw1 - U2Vw2. This is Newton's second law applied to angular momentum. The equation applies to all turbomachines regardless of fluid type, flow direction, or machine geometry.
Difficulty: basicType: conceptual
Question 47 of 60
Original Exam: IES (2003)
The governing equation for flow in a turbomachine rotor (rotating frame) is the:
✓ Correct Answer: D
Explanation:
In the rotating frame of reference, the rothalpy I = h + Vr²/2 - U²/2 is constant along a streamline (for adiabatic, inviscid flow). This is the rotating frame equivalent of stagnation enthalpy. I = h0_rel - U²/2 = h + Vr²/2 - U²/2. The rothalpy is constant across the rotor (I1 = I2), which gives: h1 + Vr1²/2 - U1²/2 = h2 + Vr2²/2 - U2²/2. This is used to analyse flow in the rotating passages of turbomachines.
Type: conceptual
Question 48 of 60
Original Exam: GATE (2004)
The specific speed of a pump is 50 (SI units). The type of pump most suitable is:
✓ Correct Answer: D
Explanation:
Specific speed ranges for pumps (SI units, N in rpm, Q in m³/s, H in m): Centrifugal (radial): Ns = 10-70, Mixed flow: Ns = 70-170, Axial flow: Ns = 170-400. Ns = 50 falls in the centrifugal pump range. Centrifugal pumps are used for high head, low flow applications. As specific speed increases, the pump transitions from centrifugal to mixed flow to axial flow, with decreasing head and increasing flow rate. The impeller shape changes from narrow radial to wide axial as Ns increases.
Difficulty: basicType: conceptual
Question 49 of 60
Original Exam: IES (2005)
The purpose of the volute casing (spiral casing) in a centrifugal pump is to:
✓ Correct Answer: D
Explanation:
The volute casing (spiral casing) in a centrifugal pump: (1) Collects the fluid discharged from the impeller at high velocity, (2) Converts kinetic energy to pressure energy through its gradually increasing cross-sectional area (diffuser action), (3) Guides the fluid to the discharge pipe. The volute is designed so that the fluid velocity decreases gradually (constant angular momentum: Vw × r = constant), minimising losses. Alternatively, a diffuser ring (vaned or vaneless) can be used between impeller and volute.
Difficulty: basicType: conceptual
Question 50 of 60
Original Exam: GATE (2006)
The maximum efficiency of a Pelton wheel occurs when the peripheral velocity of the bucket is approximately:
✓ Correct Answer: D
Explanation:
Maximum efficiency of a Pelton wheel occurs when U/V1 ≈ 0.5 (bucket speed = half jet velocity). At this condition, the relative velocity at exit is approximately zero, meaning maximum energy is extracted from the jet. The ideal hydraulic efficiency at U/V1 = 0.5 with β = 180° and k = 1: η_h = 2(0.5)(0.5)(1+1) = 1.0 (100%). In practice, with β ≈ 165° and k ≈ 0.9: η_h ≈ 0.85-0.90. The optimum U/V1 ≈ 0.46 in practice.
Difficulty: basicType: conceptual
Question 51 of 60
Original Exam: IES (2007)
The compressor map (performance map) of a centrifugal compressor shows:
✓ Correct Answer: D
Explanation:
The compressor map (performance map) shows: (1) Pressure ratio (p2/p1) on Y-axis vs. corrected mass flow rate (ṁ√T01/p01) on X-axis, (2) Lines of constant corrected speed (N/√T01), (3) Surge line (left boundary — instability limit), (4) Choke line (right boundary — maximum flow), (5) Efficiency contours (islands). The map is used to: select operating point, assess surge margin, predict off-design performance, and design control systems. Corrected parameters account for varying inlet conditions.
Difficulty: intermediateType: conceptual
Question 52 of 60
Original Exam: GATE (2008)
The reaction turbine differs from the impulse turbine in that:
✓ Correct Answer: D
Explanation:
Reaction turbines: (1) Both pressure and velocity change in the rotor (degree of reaction R > 0), (2) The rotor is fully immersed in the pressurised fluid (not open to atmosphere like Pelton), (3) The pressure drop in the rotor creates a reaction force on the blades (like a rocket nozzle), (4) Examples: Francis turbine, Kaplan turbine, steam reaction turbines. Impulse turbines: all pressure drop in nozzle, rotor at constant pressure, open to atmosphere (Pelton wheel, de Laval turbine).
Difficulty: basicType: conceptual
Question 53 of 60
Original Exam: IES (2009)
The corrected mass flow rate in a compressor is defined as:
✓ Correct Answer: D
Explanation:
Corrected mass flow rate ṁ_corr = ṁ√T01/p01 (or ṁ√(T01/T_ref)/(p01/p_ref) for normalised form). This parameter accounts for varying inlet conditions (temperature and pressure) and allows comparison of compressor performance at different operating conditions. Similarly, corrected speed N_corr = N/√T01. These corrected parameters collapse the performance data onto a single map. They arise from dimensional analysis of compressible flow in turbomachines.
Difficulty: intermediateType: conceptual
Question 54 of 60
Original Exam: GATE (2010)
The stagger angle of a turbomachine blade is defined as:
✓ Correct Answer: D
Explanation:
Stagger angle (setting angle) ξ is the angle between the chord line of the blade and the plane of rotation (tangential direction) or equivalently, the angle between the chord and the circumferential direction. It determines the orientation of the blade in the cascade. Camber angle θ = α1 - α2 (difference between inlet and outlet blade angles). Stagger angle ξ = (α1 + α2)/2 for symmetric camber. The stagger angle affects the blade loading and the cascade performance.
Difficulty: intermediateType: conceptual
Question 55 of 60
Original Exam: IES (2011)
The hydraulic turbine that uses the principle of adjustable guide vanes (wicket gates) to control the flow rate is:
✓ Correct Answer: D
Explanation:
Both Francis and Kaplan turbines use adjustable guide vanes (wicket gates/stay vanes) to control the flow rate and hence the power output. The wicket gates rotate about their own axes to vary the flow area and the angle of flow entering the runner. In the Kaplan turbine, additionally, the runner blades are also adjustable (double regulation). In the Pelton wheel, flow is controlled by a needle valve (spear valve) in the nozzle. Guide vane adjustment allows efficient operation over a wide range of loads.
Difficulty: basicType: conceptual
Question 56 of 60
Original Exam: GATE (2012)
The isentropic efficiency of a turbine is defined as:
✓ Correct Answer: D
Explanation:
Isentropic efficiency of turbine η_is = W_actual / W_isentropic = (h1 - h2) / (h1 - h2s), where h2s is the enthalpy at the end of isentropic expansion. Since actual expansion involves irreversibilities, h2 > h2s, so W_actual < W_isentropic, and η_is < 1. For an ideal gas: η_is = (T1 - T2)/(T1 - T2s) = (T1 - T2) / (T1[1 - (p2/p1)^((γ-1)/γ)]). Typical η_is = 0.85-0.92 for well-designed turbines.
Difficulty: basicType: conceptual
Question 57 of 60
Original Exam: IES (2013)
The specific speed of a turbomachine is a dimensionless parameter that:
✓ Correct Answer: D
Explanation:
Specific speed (dimensionless form) Ω_s = NQ^0.5/(gH)^0.75 (for pumps) or Ω_s = NP^0.5/(ρ^0.5(gH)^1.25) (for turbines) is the same for all geometrically similar machines operating at the same efficiency point (homologous point). It is independent of machine size. This is the fundamental property that makes specific speed useful for: (1) Selecting machine type, (2) Scaling designs, (3) Comparing performance. The dimensional specific speed (Ns) depends on the unit system used.
Difficulty: intermediateType: conceptual
Question 58 of 60
Original Exam: IES (2014)
The pressure rise in the vaneless diffuser of a centrifugal compressor follows:
✓ Correct Answer: D
Explanation:
In a vaneless diffuser, there are no blades to guide the flow. The flow follows free vortex law: rVw = constant (conservation of angular momentum in the absence of external torques). As radius increases, Vw decreases. The radial velocity Vr also decreases (continuity: Vr × 2πr × b = constant, where b is diffuser width). The total velocity V = √(Vw² + Vr²) decreases, converting kinetic energy to pressure. The flow path is a logarithmic spiral. Vaneless diffusers have wider operating range but lower efficiency than vaned diffusers.
Type: conceptual
Question 59 of 60
Original Exam: GATE (2013)
The rotating stall in an axial compressor is characterised by:
✓ Correct Answer: D
Explanation:
Rotating stall consists of one or more stall cells (regions of separated, low-momentum flow) that rotate around the compressor annulus at typically 20-70% of the rotor speed. The stall cells rotate in the direction of rotor rotation. Rotating stall: (1) Reduces compressor performance (pressure rise and efficiency), (2) Causes alternating blade loading (fatigue), (3) Is a precursor to surge, (4) Can be detected by pressure fluctuations. Unlike surge (global instability), rotating stall is a local instability that may be stable. It is more common in axial compressors than centrifugal.
Type: conceptual
Question 60 of 60
Original Exam: IES (2015)
The work done by a centrifugal pump impeller on the fluid per unit mass (Euler work) can be expressed using the velocity triangle as:
✓ Correct Answer: D
Explanation:
The Euler work can be expressed in three equivalent forms: W = U2Vw2 - U1Vw1 (whirl velocity form). Using velocity triangle: V² = U² + Vr² - 2UVr·cos(angle), the Euler work can be shown to equal: W = (V2² - V1²)/2 + (U2² - U1²)/2 + (Vr1² - Vr2²)/2. This is the 'three-term' form of the Euler equation: (1) Change in absolute KE, (2) Centrifugal effect (change in blade speed), (3) Change in relative KE (diffusion in rotor). This form is useful for understanding the pressure rise mechanisms.