GATE GUIDE

GATE ME Fluid Mechanics Formula Sheet 2027: Formulas, Examples, Traps

By MD ANISH AHAMADUpdated 4 Oct 202610 min read
GATE ME Fluid Mechanics Formula Sheet 2027: Formulas, Examples, Traps

Which fluid mechanics formulas do you need for GATE ME 2027? A compact set, each safe only when you know the condition it carries. This sheet gives them topic by topic, with a worked example and the trap for each, checked against the last-minute sheet of the GATE ME 2027 book.

In this guide
  1. Key takeaways
  2. The terms and notation this sheet uses
  3. How pressure acts in a fluid at rest
  4. When a floating body is stable
  5. How a fluid moves: acceleration, stream function and velocity potential
  6. Momentum on a control volume and forces on vanes
  7. Bernoulli's equation and flow measurement
  8. Dimensional analysis and model laws
  9. Laminar flow and losses in pipes
  10. Boundary layers on a flat plate
  11. Compressible flow and nozzles
  12. Quick revision

Key takeaways

The terms and notation this sheet uses

A fluid deforms continuously under shear. Density ρ\rho is mass per unit volume, dynamic viscosity μ\mu links shear stress to the velocity gradient, and kinematic viscosity is ν=μ/ρ\nu = \mu/\rho. Units are SI, with g=9.81 m/s2g = 9.81\ \text{m/s}^2.

A flow is steady when nothing changes with time at a point, incompressible when density is constant, inviscid when viscosity is neglected, and irrotational when fluid elements do not spin. A streamline is everywhere tangent to the velocity. Gauge pressure is measured above atmospheric; absolute pressure is gauge plus atmospheric.

The stream function ψ\psi is constant along each streamline; the velocity potential ϕ\phi is a function whose gradient is the velocity. Subscript 1 marks an inlet, 2 an exit, 0 the stagnation state.

How pressure acts in a fluid at rest

Pressure in a still fluid rises linearly with depth. The force on a submerged plane is the centroid pressure times the area, acting lower down at the centre of pressure.

Formula Watch out for
τ=μdudy\displaystyle \tau = \mu \dfrac{du}{dy}; ν=μ/ρ\nu = \mu/\rho Newtonian fluids only
p=patm+ρghp = p_{\text{atm}} + \rho g h 1 atm =101.325= 101.325 kPa, about 10.33 m of water
Drop: Δp=4σ/d\Delta p = 4\sigma/d; soap bubble: 8σ/d8\sigma/d; capillary rise h=4σcos⁡θρgd\displaystyle h = \dfrac{4\sigma\cos\theta}{\rho g d} A bubble has two surfaces
F=ρghˉAF = \rho g \bar{h} A; hcp=hˉ+IGsin⁡2θAhˉ\displaystyle h_{cp} = \bar{h} + \dfrac{I_G \sin^2\theta}{A\bar{h}} hˉ\bar{h} is the centroid depth; θ=90∘\theta = 90^\circ for a vertical wall
Curved surface: FHF_H acts on the vertical projection; FVF_V is the weight of fluid above, up to the free surface Imaginary fluid above counts too
Linear acceleration: tan⁡θ=axg+az\displaystyle \tan\theta = \dfrac{a_x}{g + a_z}; rotation: z=ω2r22g\displaystyle z = \dfrac{\omega^2 r^2}{2g} ω\omega in rad/s

Worked example. A vertical gate 2 m deep and 1 m wide has its top edge at the free surface of water:

F=1000×9.81×1×2=19.62 kN,hcp=1+1×23/122×1=1.333 mF = 1000 \times 9.81 \times 1 \times 2 = 19.62\ \text{kN}, \qquad h_{cp} = 1 + \frac{1 \times 2^3/12}{2 \times 1} = 1.333\ \text{m}

It acts at two-thirds of the depth. For capillary rise of water in a 1 mm tube with σ=0.073\sigma = 0.073 N/m and θ=0\theta = 0, h=4×0.0731000×9.81×0.001=0.0298\displaystyle h = \frac{4 \times 0.073}{1000 \times 9.81 \times 0.001} = 0.0298 m, about 29.8 mm.

Trap: a diameter left in millimetres inside the capillary formula is out by a factor of 1000.

When a floating body is stable

Buoyancy equals the weight of fluid displaced, acting through the centre of buoyancy B. Stability depends on the metacentric height:

GM=BM−BG,BM=IVGM = BM - BG, \qquad BM = \frac{I}{V}

II is the second moment of the waterplane area about the tilt axis; VV is the displaced volume. A floating body is stable if GM>0GM > 0; a submerged one if B lies above G. The rolling period is T=2πk/g GMT = 2\pi k/\sqrt{g\,GM}.

Worked example. A pontoon 10 m long and 4 m wide floats at a 1 m draft, with G 1.2 m above the keel:

BM=10×43/1210×4×1=1.333 m,BG=1.2−0.5=0.7 m,GM=0.633 mBM = \frac{10 \times 4^3/12}{10 \times 4 \times 1} = 1.333\ \text{m}, \quad BG = 1.2 - 0.5 = 0.7\ \text{m}, \quad GM = 0.633\ \text{m}

It is stable. Trap: for rolling, the width is cubed, not the length.

How a fluid moves: acceleration, stream function and velocity potential

A particle accelerates in steady flow by moving to where the velocity differs: the convective part. The local part is change with time at a point.

Formula Watch out for
a⃗=∂V⃗∂t+(V⃗⋅∇)V⃗\displaystyle \vec{a} = \dfrac{\partial \vec{V}}{\partial t} + (\vec{V} \cdot \nabla)\vec{V} Steady flow has no local part; uniform flow has no convective part
∂ρ∂t+∇⋅(ρV⃗)=0\displaystyle \dfrac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{V}) = 0; incompressible: ∇⋅V⃗=0\nabla \cdot \vec{V} = 0; one-dimensional: ρAV=constant\rho A V = \text{constant} ρAV\rho A V stays constant in steady flow even when density changes
u=∂ψ∂y\displaystyle u = \dfrac{\partial \psi}{\partial y}, v=−∂ψ∂x\displaystyle v = -\dfrac{\partial \psi}{\partial x}; flow between streamlines =ψ2−ψ1= \psi_2 - \psi_1 per unit depth Two-dimensional incompressible flow only
ωz=∂v∂x−∂u∂y\displaystyle \omega_z = \dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y} Irrotational when it is zero
V⃗=∇ϕ\vec{V} = \nabla \phi; ∇2ϕ=0\nabla^2 \phi = 0 when also incompressible New in 2027. Some books write −∇ϕ-\nabla \phi; ϕ\phi exists only for irrotational flow

Worked example. Take ψ=3xy\psi = 3xy:

u=∂ψ∂y=3x,v=−∂ψ∂x=−3yωz=0−0=0,so ϕ existsϕ=32(x2−y2),∇2ϕ=3−3=0\begin{aligned} u &= \frac{\partial \psi}{\partial y} = 3x, \qquad v = -\frac{\partial \psi}{\partial x} = -3y \\ \omega_z &= 0 - 0 = 0, \quad \text{so } \phi \text{ exists} \\ \phi &= \tfrac{3}{2}\left(x^2 - y^2\right), \qquad \nabla^2 \phi = 3 - 3 = 0 \end{aligned}

At (1,2)(1, 2): ax=u ∂u/∂x+v ∂u/∂y=9a_x = u\,\partial u/\partial x + v\,\partial u/\partial y = 9 and ay=u ∂v/∂x+v ∂v/∂y=18a_y = u\,\partial v/\partial x + v\,\partial v/\partial y = 18, so ∣a⃗∣=405=20.1 m/s2\vert \vec{a} \vert = \sqrt{405} = 20.1\ \text{m/s}^2. The flow is steady, yet it accelerates.

In one line: The stream function exists for any two-dimensional incompressible flow, the velocity potential only for an irrotational one; where both exist, lines of constant ϕ\phi cross streamlines at right angles.

Momentum on a control volume and forces on vanes

For steady flow, the net force on the fluid in a control volume equals momentum out minus momentum in, per component:

∑F⃗=∑m˙ V⃗out−∑m˙ V⃗in\sum \vec{F} = \sum \dot{m}\,\vec{V}_{\text{out}} - \sum \dot{m}\,\vec{V}_{\text{in}}
Formula Watch out for
Fixed flat plate normal to a jet: F=ρAV2F = \rho A V^2 Area of the jet, not the plate
Single moving plate: F=ρA(V−u)2F = \rho A (V - u)^2 Mass flow uses the relative velocity
Series of plates: F=ρAV(V−u)F = \rho A V (V - u); best efficiency 12\displaystyle \tfrac{1}{2} at u=V/2u = V/2 Full jet mass flow is used

Worked example. A water jet 50 mm in diameter at 20 m/s has A=1.963×10−3 m2A = 1.963 \times 10^{-3}\ \text{m}^2. On a fixed plate, F=785.4F = 785.4 N. On a single plate moving at 8 m/s, F=282.7F = 282.7 N; on a series of plates at that speed, F=471.2F = 471.2 N.

Trap: the force on the vane is minus the force on the fluid, and pressure forces pApA at each cut section belong in the balance.

Bernoulli's equation and flow measurement

Along a streamline, the pressure, velocity and elevation heads add to a constant:

pρg+V22g+z=constant\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant}

It needs steady, incompressible, inviscid flow with no shaft work. The energy equation adds pump head, turbine head and losses, with kinetic-energy factor α=2\alpha = 2 for laminar pipe flow, about 1.05 for turbulent.

Formula Watch out for
Tank outlet: V=2ghV = \sqrt{2gh} Independent of the fluid when losses are neglected
Pitot tube: V=2Δp/ρV = \sqrt{2\Delta p/\rho} ρ\rho of the flowing fluid
Manometer head: h=x(ρmρ−1)\displaystyle h = x\left(\dfrac{\rho_m}{\rho} - 1\right) Subtract 1 for the fluid above the mercury
Venturi: Q=CdA1A22ghA12−A22\displaystyle Q = C_d \dfrac{A_1 A_2 \sqrt{2gh}}{\sqrt{A_1^2 - A_2^2}} hh is the piezometric head difference

Worked example. A pitot tube in air (ρ=1.2 kg/m3\rho = 1.2\ \text{kg/m}^3) reading 800 Pa gives V=1333.3=36.5V = \sqrt{1333.3} = 36.5 m/s. A 0.1 m mercury deflection under water is 0.1×12.6=1.260.1 \times 12.6 = 1.26 m of water.

Remember: Bernoulli's equation holds along a streamline; it joins points on different streamlines only when the flow is irrotational, and never across a pump, a loss or a shock.

The book's last-minute sheet covers all ten technical sections this way, and the GATE ME 2027 book adds 942 questions with worked solutions and 10 full mock tests, built on all six GATE ME papers 2022-2026 (both 2022 sets).

Dimensional analysis and model laws

Buckingham's π\pi theorem: nn variables in mm fundamental dimensions form n−mn - m dimensionless groups.

Formula Watch out for
Re=ρVLμ\displaystyle \text{Re} = \dfrac{\rho V L}{\mu}, Fr=VgL\displaystyle \text{Fr} = \dfrac{V}{\sqrt{gL}}, Eu=ΔpρV2\displaystyle \text{Eu} = \dfrac{\Delta p}{\rho V^2}, We=ρV2Lσ\displaystyle \text{We} = \dfrac{\rho V^2 L}{\sigma}, Ma=Vc\displaystyle \text{Ma} = \dfrac{V}{c} Viscosity is M L−1T−1\text{M L}^{-1}\text{T}^{-1}; surface tension is M T−2\text{M T}^{-2}
Reynolds similarity: pipes, submerged bodies Match Re, not Fr
Froude similarity, same fluid: Vr=LrV_r = \sqrt{L_r}, Qr=Lr5/2Q_r = L_r^{5/2}, Fr=Lr3F_r = L_r^3, power Lr3.5L_r^{3.5} Free-surface flows

Worked example. A 1:25 spillway model runs at 2 m/s and 0.01 m3/s\text{m}^3/\text{s}. The prototype velocity is 225=102\sqrt{25} = 10 m/s, and the discharge is 0.01×252.5=31.25 m3/s0.01 \times 25^{2.5} = 31.25\ \text{m}^3/\text{s}.

Laminar flow and losses in pipes

In laminar pipe flow, viscosity alone sets the pressure drop, so ff depends only on Re.

Formula Watch out for
Δp=32μVLD2=128μQLπD4\displaystyle \Delta p = \dfrac{32 \mu V L}{D^2} = \dfrac{128 \mu Q L}{\pi D^4}; Vmax⁡=2VV_{\max} = 2V; τw=8μV/D\tau_w = 8\mu V/D Laminar below Re≈2000\text{Re} \approx 2000 (2300 in some texts)
Darcy: f=64/Ref = 64/\text{Re} Independent of roughness
Parallel plates: Δp=12μVLh2\displaystyle \Delta p = \dfrac{12 \mu V L}{h^2}; Vmax⁡=1.5VV_{\max} = 1.5V hh is the gap
hf=fLV22gD\displaystyle h_f = \dfrac{f L V^2}{2 g D}; Blasius f=0.316/Re0.25f = 0.316/\text{Re}^{0.25} up to Re=105\text{Re} = 10^5 At fixed ff, hf∝Q2/D5h_f \propto Q^2/D^5
Minor loss KV2/(2g)K V^2/(2g); sudden expansion (V1−V2)2/(2g)(V_1 - V_2)^2/(2g); exit K=1K = 1; sharp entrance K=0.5K = 0.5 One KK per fitting
Series: losses add; parallel: flows add, same hfh_f; greatest power when hf=H/3h_f = H/3 Efficiency is then 2/32/3

Worked example. Oil with ρ=900 kg/m3\rho = 900\ \text{kg/m}^3 and μ=0.1\mu = 0.1 Pa s flows at 1 m/s in a 50 mm pipe:

Re=900×1×0.050.1=450,f=64450=0.142,Δp10 m=32×0.1×1×100.052=12.8 kPa\text{Re} = \frac{900 \times 1 \times 0.05}{0.1} = 450, \quad f = \frac{64}{450} = 0.142, \quad \Delta p_{10\,\text{m}} = \frac{32 \times 0.1 \times 1 \times 10}{0.05^2} = 12.8\ \text{kPa}

Check: Darcy-Weisbach gives hf=1.450h_f = 1.450 m and ρghf=12.8\rho g h_f = 12.8 kPa. A sudden expansion from 4 m/s to 1 m/s loses 9/19.62=0.4599/19.62 = 0.459 m.

Trap: halving the diameter at fixed flow and ff raises the head loss 25=322^5 = 32 times.

Boundary layers on a flat plate

A boundary layer is the thin region near a wall where viscosity slows the flow. On a flat plate it turns turbulent near Rex=5×105\text{Re}_x = 5 \times 10^5.

Formula Watch out for
Laminar: δx≈5Rex\displaystyle \dfrac{\delta}{x} \approx \dfrac{5}{\sqrt{\text{Re}_x}}; Cf=0.664Rex\displaystyle C_f = \dfrac{0.664}{\sqrt{\text{Re}_x}}; CD=1.328ReL\displaystyle C_D = \dfrac{1.328}{\sqrt{\text{Re}_L}} Local CfC_f against mean CDC_D
Turbulent: δx=0.37Rex0.2\displaystyle \dfrac{\delta}{x} = \dfrac{0.37}{\text{Re}_x^{0.2}}; CD=0.074ReL0.2\displaystyle C_D = \dfrac{0.074}{\text{Re}_L^{0.2}} Check Rex\text{Re}_x first
δ∗=∫0δ(1−uU)dy\displaystyle \delta^* = \int_0^\delta \left(1 - \dfrac{u}{U}\right) dy; θ=∫0δuU(1−uU)dy\displaystyle \theta = \int_0^\delta \dfrac{u}{U}\left(1 - \dfrac{u}{U}\right) dy; H=δ∗/θ=2.59H = \delta^*/\theta = 2.59 (Blasius) Linear profile: δ∗=δ/2\delta^* = \delta/2, θ=δ/6\theta = \delta/6
Separation where ∂u/∂y=0\partial u/\partial y = 0 at the wall Needs an adverse pressure gradient

Worked example. Air with ν=1.5×10−5 m2/s\nu = 1.5 \times 10^{-5}\ \text{m}^2/\text{s} at 3 m/s, 0.5 m from the leading edge: Rex=105\text{Re}_x = 10^5, laminar, so δ=5×0.5/316.2=7.9\delta = 5 \times 0.5/316.2 = 7.9 mm. Trap: laminar δ\delta grows as x\sqrt{x}; wall shear falls as 1/x1/\sqrt{x}.

Compressible flow and nozzles

Every row here is New in 2027: the syllabus now reads "one dimensional high speed compressible fluid flow; flow through converging and converging-diverging nozzles".

Formula Watch out for
c=γRTc = \sqrt{\gamma R T}; M=V/cM = V/c Incompressible is fair below about M=0.3M = 0.3
T0T=1+γ−12M2\displaystyle \dfrac{T_0}{T} = 1 + \dfrac{\gamma - 1}{2} M^2; p0p=(T0T)γ/(γ−1)\displaystyle \dfrac{p_0}{p} = \left(\dfrac{T_0}{T}\right)^{\gamma/(\gamma - 1)} T0T_0 constant in adiabatic flow; p0p_0 also in isentropic
γ=1.4\gamma = 1.4: T∗/T0=0.8333T^*/T_0 = 0.8333, p∗/p0=0.5283p^*/p_0 = 0.5283, ρ∗/ρ0=0.6339\rho^*/\rho_0 = 0.6339 Absolute values only
dAA=(M2−1)dVV\displaystyle \dfrac{dA}{A} = (M^2 - 1)\dfrac{dV}{V} Supersonic flow speeds up in a diverging duct
AA∗=1M[2γ+1(1+γ−12M2)]γ+12(γ−1)\displaystyle \dfrac{A}{A^*} = \dfrac{1}{M}\left[\dfrac{2}{\gamma + 1}\left(1 + \dfrac{\gamma - 1}{2}M^2\right)\right]^{\frac{\gamma + 1}{2(\gamma - 1)}} Two roots, one subsonic and one supersonic
m˙max⁡=A∗p0γRT0(2γ+1)γ+12(γ−1)≈0.0404 A∗p0T0\displaystyle \dot{m}_{\max} = A^* p_0 \sqrt{\dfrac{\gamma}{R T_0}} \left(\dfrac{2}{\gamma + 1}\right)^{\frac{\gamma + 1}{2(\gamma - 1)}} \approx 0.0404\,\dfrac{A^* p_0}{\sqrt{T_0}} for air Choked when back pressure ≤0.528 p0\le 0.528\,p_0
Normal shock: M22=M12+2/(γ−1)2γM12/(γ−1)−1\displaystyle M_2^2 = \dfrac{M_1^2 + 2/(\gamma - 1)}{2\gamma M_1^2/(\gamma - 1) - 1}; p2p1=1+2γγ+1(M12−1)\displaystyle \dfrac{p_2}{p_1} = 1 + \dfrac{2\gamma}{\gamma + 1}\left(M_1^2 - 1\right) T0T_0 unchanged, p0p_0 falls, entropy rises

Worked example 1. Air at 300 K moves at M=2M = 2. Then c=1.4×287×300=347.2c = \sqrt{1.4 \times 287 \times 300} = 347.2 m/s and V=694.4V = 694.4 m/s. Also T0/T=1+0.2×4=1.8T_0/T = 1 + 0.2 \times 4 = 1.8, so T0=540T_0 = 540 K and p0/p=1.83.5=7.82p_0/p = 1.8^{3.5} = 7.82.

Worked example 2. A nozzle with a 10 cm2\text{cm}^2 throat draws air from p0=500p_0 = 500 kPa absolute and T0=400T_0 = 400 K:

m˙max⁡=0.0404×10−3×500 000400=1.01 kg/s\dot{m}_{\max} = 0.0404 \times \frac{10^{-3} \times 500\,000}{\sqrt{400}} = 1.01\ \text{kg/s}

It chokes at back pressures up to 0.528×500=2640.528 \times 500 = 264 kPa. For a normal shock at M1=2M_1 = 2: M2=0.577M_2 = 0.577, p2/p1=4.5p_2/p_1 = 4.5, T2/T1=1.6875T_2/T_1 = 1.6875 and p02/p01=0.721p_{02}/p_{01} = 0.721. Practise powers such as 1.83.51.8^{3.5} on the GATE virtual calculator.

A choked converging-diverging nozzle delivers shock-free supersonic flow at its design back pressure. At higher back pressures that still choke it, a normal shock stands in the diverging part, moving towards the throat as back pressure rises.

Trap: A converging nozzle never gives an exit Mach number above 1, and every ratio here needs absolute pressure and kelvin; a gauge or Celsius value gives a plausible wrong answer.

More formula sheets: all of GATE ME · Heat Transfer · Strength of Materials · Thermodynamics

Quick revision

  1. Force on a plane: ρghˉA\rho g \bar{h} A at depth hˉ+IGsin⁡2θ/(Ahˉ)\bar{h} + I_G \sin^2\theta/(A\bar{h}).
  2. Floating stability: GM=I/V−BG>0GM = I/V - BG > 0, with II about the tilt axis.
  3. u=∂ψ/∂yu = \partial \psi/\partial y, v=−∂ψ/∂xv = -\partial \psi/\partial x; ϕ\phi exists only if ωz=0\omega_z = 0, and then ∇2ϕ=0\nabla^2 \phi = 0 for incompressible flow.
  4. Bernoulli: along a streamline; across streamlines only if irrotational; never through a pump, loss or shock.
  5. Laminar pipe: f=64/Ref = 64/\text{Re} and Vmax⁡=2VV_{\max} = 2V; at fixed ff, hf∝Q2/D5h_f \propto Q^2/D^5.
  6. Laminar plate: δ/x≈5/Rex\delta/x \approx 5/\sqrt{\text{Re}_x}, transition near 5×1055 \times 10^5.
  7. Nozzles: choke at 0.528 p00.528\,p_0; at M1=2M_1 = 2 a normal shock gives M2=0.577M_2 = 0.577 and p2/p1=4.5p_2/p_1 = 4.5.
  8. Recheck units, kelvin and absolute pressure before typing a NAT answer; NATs carry no negative marking under the marking scheme common to every GATE paper. See the exam-day rules and the GATE 2027 exam dates.

Frequently asked questions

Is velocity potential new in the GATE ME 2027 syllabus?

Yes. The 2027 Fluid Mechanics syllabus adds the phrase "concept of velocity potential". The potential ϕ\phi is defined so that V⃗=∇ϕ\vec{V} = \nabla \phi, and it exists only when the flow is irrotational. When the flow is also incompressible, ϕ\phi satisfies Laplace's equation, and lines of constant ϕ\phi cross the streamlines at right angles.

When can Bernoulli's equation be used between two different streamlines?

Bernoulli's equation holds along a streamline for steady, incompressible, inviscid flow with no shaft work. It may join points on different streamlines only when the flow is also irrotational, so check the vorticity first. Never apply it across a pump, a turbine, a friction loss or a shock; use the energy equation with head terms there instead.

What is the critical pressure ratio for a converging nozzle?

For air with γ=1.4\gamma = 1.4, the critical ratio is p∗/p0=0.528p^*/p_0 = 0.528. A converging nozzle chokes once the back pressure falls to 0.528 p00.528\,p_0. Lowering the back pressure further does not raise the mass flow, and the exit Mach number stays at 1. Use absolute pressures in this ratio, never gauge readings.

What is the friction factor for laminar flow in a pipe?

For fully developed laminar flow the Darcy friction factor is f=64/Ref = 64/\text{Re}, independent of wall roughness. It follows from the Hagen-Poiseuille result for pressure drop. Put into the Darcy-Weisbach equation, it gives the same head loss. The laminar limit is about Re=2000\text{Re} = 2000, or 2300 in some texts, so use the value the question gives.

Which model law applies to a spillway or ship model?

Free-surface flows follow Froude similarity. With the same fluid in model and prototype, velocity scales as Lr\sqrt{L_r}, discharge as Lr5/2L_r^{5/2} and force as Lr3L_r^3, where LrL_r is the length scale ratio. Flow in pipes and around fully submerged bodies follows Reynolds similarity instead, so match the Reynolds number there.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

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