GATE ME Thermodynamics Formula Sheet 2027: Cycles to Pumps

Which thermodynamics formulas does a GATE ME candidate actually need? The ones that turn a property table or a cycle diagram into a number: energy balances, entropy, cycle efficiencies, COP, specific humidity and the turbomachinery relations. This sheet gives the most-used of them across Thermodynamics and Applications, each topic with a worked example and its trap.

Key takeaways

• Most thermodynamics questions are an energy balance written for the right system, closed or open.
• Every ratio, power or exponential of temperature needs kelvin; Celsius is safe only in a difference.
• Work and heat are path functions, so you need the process path, not just the end states.
• The 2027 syllabus adds combustion (air-fuel ratio, equivalence ratio) and pumps with their characteristics.
• A turbine's isentropic efficiency is actual over ideal work; a compressor's or a pump's is ideal over actual.
• On a throttle the enthalpy does not change, which fixes the evaporator inlet of every vapour-compression cycle.

The terms and notation this sheet uses

A closed system lets no mass cross its boundary; an open system (a control volume) lets mass flow in and out; an isolated system exchanges neither mass nor energy. Properties such as pressure and temperature are point functions; heat and work are path functions.

Notation: $Q$ is heat added to the system and $W$ is work done by it. Lower case ($q$, $w$, $u$, $h$, $s$, $v$) means per kilogram. A dot means a rate, so $\dot{m}$ is mass flow in kg/s. Subscript 1 is the inlet or initial state, 2 the exit or final state, and 0 the surroundings (the dead state). The quality $x$ is the dryness fraction of a wet mixture. For air, take $R = 0.287$, $c_p = 1.005$ and $c_v = 0.718$ kJ/kg K, with $\gamma = 1.4$, unless the question gives its own values.

Work, heat and the first law

Work is the area under the path on a $p$-$V$ diagram. The first law books the energy: heat in minus work out equals the rise in stored energy.

Worked example. Air at 100 kPa and 0.1 m³ is compressed to 0.05 m³ along $pV^{1.3} = C$. Then $p_2 = 100 \times 2^{1.3} = 246.2$ kPa, and

The sign says work is done on the gas. For a nozzle with a 200 kJ/kg enthalpy drop from rest, $V_2 = 44.72\sqrt{200} = 632$ m/s.

Trap: In a nozzle or a turbine, enthalpy comes from tables in kJ/kg while $V^2/2$ comes out in J/kg. Convert one of them before you add, or the kinetic term is out by a factor of 1000.

Entropy, the second law and exergy

The second law says entropy generated is never negative; exergy measures how much of an energy flow could still become work.

Worked example. Air heated at constant pressure from 300 K to 600 K gains $\Delta s = 1.005 \ln 2 = 0.697$ kJ/kg K. And 100 kJ of heat at 600 K with the surroundings at 300 K carries exergy $100(1 - 300/600) = 50$ kJ.

Remember: If a temperature sits inside a ratio, a power or a logarithm, it must be in kelvin. Celsius survives only in a difference such as $c_p \Delta T$.

Power cycles: Rankine and Brayton

Number the Rankine states as 1 turbine inlet, 2 turbine exit, 3 pump inlet (saturated liquid) and 4 pump exit. Efficiency is net work over heat added in the boiler.

Worked example. Take $h_1 = 3200$, $h_2 = 2200$ and $h_3 = 190$ kJ/kg, with $v_f = 0.001$ m³/kg pumping from 10 kPa to 8 MPa. Pump work is $0.001 \times 7990 = 7.99$ kJ/kg, so $h_4 = 198.0$ kJ/kg.

A Brayton cycle with $r_p = 8$ and $\gamma = 1.4$ gives $\eta = 1 - 1/8^{0.2857} = 1 - 1/1.811 = 0.448$.

Air-standard engine cycles

The compression ratio is $r = (V_c + V_s)/V_c$, clearance plus swept volume over clearance volume. The cut-off ratio $\rho$ is the volume ratio across the constant-pressure heat addition.

Worked example. Otto with $r = 10$: $\eta = 1 - 1/10^{0.4} = 1 - 1/2.512 = 0.602$. Diesel with $r = 16$ and $\rho = 2$: $2^{1.4} = 2.639$, so the bracket is $1.639/1.4 = 1.171$, and $\eta = 1 - 1.171/3.031 = 0.614$. Work such powers on the GATE virtual calculator with its $x^y$ key.

The last-minute formula sheet in the GATE ME 2027 book covers all ten technical sections with each formula's conditions, alongside 942 questions with worked solutions and 10 full mock tests.

Combustion: air-fuel ratio and equivalence ratio (New in 2027)

Air is 21 per cent oxygen by volume (3.76 mol nitrogen per mol oxygen) and about 23.2 per cent oxygen by mass.

Worked example. Methane has $x = 1$, $y = 4$, so it needs 2 mol of oxygen, or 64 kg per 16 kg of fuel. The air is $64/0.232 = 275.9$ kg, and $A/F = 275.9/16 = 17.2$. Burned at $A/F = 21.5$, it has $\phi = 17.2/21.5 = 0.8$, a lean mixture with $(1/0.8 - 1) \times 100 = 25$ per cent excess air.

In one line: The equivalence ratio compares fuel to fuel, actual over stoichiometric, so a mixture with more air than it needs has $\phi$ below 1.

Refrigeration and psychrometry

Number the vapour-compression states as 1 evaporator exit, 2 compressor exit, 3 condenser exit and 4 after the throttle.

Worked example. With $h_1 = 400$, $h_2 = 430$ and $h_3 = h_4 = 250$ kJ/kg, the refrigerating effect is 150 kJ/kg and the work 30 kJ/kg, so $\text{COP} = 5$ and $\text{COP}_{HP} = 6$. Moist air at 30 $^\circ$C with $p_v = 3$ kPa at 101.325 kPa has $\omega = 0.622 \times 3/98.325 = 0.0190$ kg/kg and $h = 30.15 + 0.0190 \times 2557.4 = 78.7$ kJ/kg dry air.

Trap: The throttle keeps enthalpy constant, not entropy. Put $h_4 = h_3$, and never draw the expansion as a vertical line on the $T$-$s$ diagram.

Turbomachinery, compressors and pumps

Every rotor obeys Euler's equation: the work per kilogram is the change in blade speed times whirl velocity.

Worked example. A pump speeded up from 1440 to 1800 rpm has $N_2/N_1 = 1.25$, so flow rises by 1.25, head by $1.25^2 = 1.5625$ and power by $1.25^3 = 1.953$. If a pump curve $H = 30 - 300Q^2$ meets a system curve $H = 10 + 200Q^2$, then $500Q^2 = 20$, $Q = 0.2$ m³/s and $H = 18$ m. A two-stage compressor from 1 bar to 16 bar intercools best at $\sqrt{1 \times 16} = 4$ bar.

Remember: Pumps in series add heads at one flow; pumps in parallel add flows at one head.

The mistakes that cost marks under time pressure

Check these four before you type an answer.

1. Units: kJ/kg times kg/s is kW; rpm becomes rad/s through $\omega = 2\pi N/60$.
2. Temperature: kelvin in every ratio, power or exponential.
3. Pressure: gauge or absolute; the ideal-gas law and isentropic ratios need absolute.
4. Efficiency: turbine actual over ideal; compressor and pump ideal over actual.

Numerical-answer questions carry no negative marking, while a wrong MCQ does; the marking scheme, which is common to every GATE paper, sets out both. You cannot carry this sheet into the hall, so read the admit card and exam-day rules and keep the formulas in memory.

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

Quick revision

1. Closed system $Q - W = \Delta U$; steady flow adds enthalpy, kinetic and potential terms per kilogram.
2. Polytropic work is $(p_1V_1 - p_2V_2)/(n - 1)$; isothermal work uses a logarithm of the volume ratio.
3. Carnot, COP and isentropic relations all take kelvin and absolute pressure.
4. Rankine heat added runs from $h_4$, after the pump; specific steam consumption is $3600/w_{\text{net}}$.
5. At equal compression ratio, Otto beats dual beats Diesel; at equal peak pressure the order reverses.
6. The equivalence ratio is $(A/F)_{\text{stoich}}/(A/F)_{\text{actual}}$; below 1 means lean.
7. In refrigeration $h_4 = h_3$ across the throttle, and $\text{COP}_{HP} = \text{COP}_R + 1$.
8. Pump affinity laws scale flow with $N$, head with $N^2$ and power with $N^3$ at a fixed diameter.