GATE CE Fluid Mechanics and Hydrology Formula Sheet 2027

This GATE CE fluid mechanics and hydrology formula sheet collects the Water Resources Engineering formulas you need for the 2027 paper, grouped by topic. Each table gives the formula, when it applies and the trap that costs marks. Short worked examples with made-up numbers show every formula in use.

Key takeaways

• Water Resources Engineering averaged 12.6 marks a paper across the sixteen GATE CE papers from 2019 to 2026, with a range of only 3 marks.
• Fluid mechanics and open-channel hydraulics share one set of ideas, so revise them together.
• The hydraulic jump uses momentum for the sequent depth and a closed form for the energy loss.
• Flow past sharp-crested weirs is newly named in the 2027 syllabus, and its whole topic is one formula.
• The well equations use the natural logarithm, and only the ratio of the radii matters.
• Most marks in this section are lost to units: litres against cubic metres, hectares against square metres, kPa against metres of head.

How to read this sheet: notation and units

A few symbols recur throughout. Density $\rho$ is mass per volume; for water take $1000$ kg/m³. Unit weight $\gamma_w = \rho g$ is weight per volume, $9.81$ kN/m³ for water. Dynamic viscosity $\mu$ links shear stress to the velocity gradient. Kinematic viscosity $\nu = \mu/\rho$ is the same property divided by density.

Head means energy per unit weight, measured in metres of the flowing fluid. Hydraulic radius $R = A/P$ is flow area over wetted perimeter. It is not the geometric radius and not the depth. In the sheet, $\log$ means base 10 and $\ln$ means natural, exactly as in the book. Take $g = 9.81$ m/s² unless a question says otherwise.

Fluid properties, hydrostatics and the flow equations

Properties and statics

For example, at $5$ m depth, $p = 1000 \times 9.81 \times 5 = 49.05$ kPa. A block of $0.5$ m³ held fully under water feels $F_B = 9.81 \times 0.5 = 4.905$ kN, whatever it is made of.

Continuity, Bernoulli and momentum

Continuity says the same discharge passes every section of a pipe. Bernoulli says total head stays constant along a streamline when losses are negligible:

Halve a pipe from $300$ mm to $150$ mm at $V_1 = 1.5$ m/s, and $V_2 = 1.5 \times 4 = 6$ m/s. On a horizontal pipe the pressure drop is $\rho(V_2^2 - V_1^2)/2 = 1000 \times 33.75/2 = 16.875$ kPa.

Trap: Velocity scales with area, so it goes as the diameter squared. Scaling with diameter alone, or dropping the half in the kinetic term, are the two classic wrong answers.

For a pipe bend, the momentum equation gives the force in each direction as $pA + \rho Q V$. Add the two perpendicular components as vectors, never arithmetically. The venturimeter combines continuity and Bernoulli:

Here $h$ is head of the flowing fluid. A mercury manometer under water reading $x$ gives $h = 12.6x$.

Pipe flow, drag and model similitude

Reynolds number and Darcy–Weisbach

Water at $1.5$ m/s in a $50$ mm pipe with $\nu = 10^{-6}$ m²/s has $Re = 75{,}000$, so it is turbulent. A $500$ m pipe of $300$ mm diameter, with $f = 0.025$ and $V = 1.5$ m/s, loses $h_f = 28.125/5.886 = 4.78$ m.

Drag on an immersed body

The drag force is $\displaystyle F_D = \tfrac{1}{2} C_D \rho A V^2$, where $A$ is the projected frontal area. A plate of $0.5$ m² normal to water at $2$ m/s, with $C_D = 1.2$, carries $F_D = 0.5 \times 1.2 \times 1000 \times 0.5 \times 4 = 1200$ N. For a sphere, use the area of a disc of the same diameter, not the surface area, which is four times larger.

Froude similitude for models

Free-surface flow is governed by gravity, so a spillway model must match the Froude number $Fr = V/\sqrt{gL}$. Each quantity then scales as a power of the length ratio $L_r$:

At a scale of 1:36, a model discharge of $0.05$ m³/s means $0.05 \times 36^{2.5} = 388.8$ m³/s in the prototype. Reynolds similitude governs enclosed flows instead, and the two generally cannot be satisfied together.

This sheet is a selection. The book's last-minute sheet of formulae and code constants covers all eight technical sections, and every entry traces back to a fully worked solution. It is part of the GATE CE 2027 book.

Open-channel flow: energy, the jump and uniform flow

Specific energy and critical depth

Specific energy is the energy per unit weight measured from the channel bed. For a rectangular channel carrying $q$ per unit width:

Critical depth $y_c$ gives the minimum specific energy and $Fr = V/\sqrt{gy} = 1$. Below it the flow is supercritical; above it, subcritical. For $q = 3$ m²/s, $y_c = (9/9.81)^{1/3} = 0.972$ m and $E_{\min} = 1.458$ m.

Remember: Critical depth depends only on discharge per unit width. Neither Manning's $n$ nor the bed slope appears, so compute $q = Q/b$ first.

The hydraulic jump

A hydraulic jump turns supercritical flow into subcritical flow. Check $Fr_1 > 1$ first, because no jump forms otherwise.

With $y_1 = 0.4$ m and $Fr_1 = 4$, $y_2 = 0.2(\sqrt{129} - 1) = 2.072$ m. The loss is $\Delta E = 1.672^3/(4 \times 0.4 \times 2.072) = 1.409$ m.

Manning, efficient sections and profiles

Manning's equation gives the velocity of uniform flow, where bed slope, water-surface slope and energy slope are equal:

With $n = 0.02$, $R = 0.8$ m and $S = 0.0004$, $V = 50 \times 0.862 \times 0.02 = 0.862$ m/s.

The most efficient rectangular section has $b = 2y$, so $R = y/2$. A trapezoid is most efficient when its sloping side equals half the top width, again giving $R = y/2$. An area of $18$ m² therefore needs $y = 3$ m and $b = 6$ m.

Water-surface profiles take a letter from the slope and a number from the zone. Mild means $y_n > y_c$; steep means $y_n < y_c$. Zone 1 lies above both depths, zone 2 between them, zone 3 below both.

Sharp-crested weirs

The discharge over a sharp-crested rectangular weir comes from integrating $\sqrt{2gh}$ over the head:

For $L = 2$ m, $H = 0.4$ m and $C_d = 0.6$, $Q = 0.896$ m³/s. End contractions shorten the effective length to $L - 0.1nH$. Flow past sharp-crested weirs had no question in any of the sixteen counted papers; the 2027 syllabus changes explain why it still deserves your time.

Trap: An orifice uses $Q = C_d A \sqrt{2gH}$ with a half power. A weir uses the three-halves power. Mixing them up is the standard error.

Hydrology and groundwater

Risk, rainfall and runoff

A structure designed for a $100$-year flood over a $25$-year life carries $R = 1 - 0.99^{25} = 0.222$, about 22 per cent. A $3$-hour unit hydrograph peaking at $40$ m³/s, with $2.5$ cm of effective rain in $3$ hours and $10$ m³/s base flow, peaks at $2.5 \times 40 + 10 = 110$ m³/s.

The φ-index is the constant loss rate that makes rainfall above it equal the observed runoff volume. Subtract it before applying the unit hydrograph. Muskingum routing uses $S = K[xI + (1 - x)O]$, and $x = 0$ reduces it to a reservoir. Streamflow measurement is newly named, yet it was examined in 2023 Set 2, Q.63.

Darcy's law and well hydraulics

With $k = 20$ m/day, $b = 15$ m, a head difference of $1.5$ m and radii of $5$ and $50$ m, a confined well gives $Q = 2827/2.303 = 1228$ m³/day. An unconfined well with $k = 15$ m/day, heads of $20$ and $18$ m and the same radius ratio gives $1555$ m³/day.

In one line: A given thickness means confined and a linear head term; saturated depths mean unconfined and squared heads.

Irrigation: duty, delta and the exit gradient

Delta is the total depth of water a crop needs over its base period. Duty is the area one cumec can irrigate. With base period $B$ in days and duty $D$ in hectares per cumec, $\Delta = 8.64\,B/D$ metres. For $B = 100$ days and $D = 1200$ ha/cumec, $\Delta = 0.72$ m, or $72$ cm.

To find what must be released at the canal head, divide the field requirement by the efficiency. Multiplying is the wrong direction.

For a weir on a permeable foundation, Khosla's exit gradient is:

With $H = 3$ m, a $4$ m downstream cut-off and a $24$ m floor, $\alpha = 6$, $\lambda = 3.541$ and $G_E = 0.127$, about 1 in 7.9. As $d$ shrinks towards zero, $G_E$ grows without limit, so the downstream cut-off is essential. Lacey's regime equations were examined in 2025 Set 1, Q.43, but they are not on this sheet.

To see how these topics weigh against the rest of the paper, read the GATE CE subject-wise weightage. Practise the arithmetic on the GATE virtual calculator, and fit the revision into the last two months strategy for GATE CE.

More formula sheets: all of GATE Civil · Environmental and Transportation · Geotechnical · Structural Engineering

Quick revision

1. Continuity first, then Bernoulli; velocity scales with the diameter squared.
2. Use the Darcy friction factor in $h_f = fLV^2/(2gD)$, and $f = 64/Re$ only for laminar flow.
3. Drag uses the projected frontal area, and Froude models scale discharge as $L_r^{5/2}$.
4. Critical depth is $(q^2/g)^{1/3}$, and $E_{\min} = 1.5\,y_c$ for a rectangular channel.
5. Jump: check $Fr_1 > 1$, use Bélanger for $y_2$, then the cube formula for the loss.
6. Manning uses $R^{2/3}S^{1/2}$; the efficient rectangle has $b = 2y$ and $R = y/2$.
7. Weir discharge goes as $H^{3/2}$; risk is $1 - (1 - 1/T)^n$; well equations use $\ln$.
8. Delta in metres is $8.64\,B/D$, and the exit gradient needs a downstream cut-off.