Applied Thermodynamics and Heat Transfer
Second Class candidates are expected to apply thermodynamics, not recite it: close an energy balance on a boiler, decide whether a cycle change helps or hurts heat rate, and predict how heat moves through a tube wall or a furnace. Expect steam-table lookups, cycle analysis and heat-transfer calculations where unit consistency earns as many marks as the formula.
Energy balances
- Steady-flow energy: Q = m(h2 - h1) for a boiler, heater or condenser where kinetic and potential energy changes are negligible; W = m(h1 - h2) for a turbine.
- Enthalpy is measured from a datum, so only differences matter. Use hf at the feedwater temperature, not 4.2 x T, unless an estimate is specifically asked for.
- A throttling process (reducing valve, throttling calorimeter) is isenthalpic but irreversible: you cannot extract work from it, and you cannot treat it as a turbine.
Second law and Rankine cycles
- Efficiency rises with higher steam temperature and pressure and lower condenser pressure. Reheat and regenerative feedwater heating raise the mean temperature of heat addition.
- Carnot still sets the ceiling: 1 - Tc/Th, absolute temperatures only.
- Keep turbine exhaust wetness below roughly 12% to limit blade erosion; superheat and reheat are the controls.
Heat transfer
- Conduction Q = kA(dT)/L; series walls add resistances. Convection Q = hA(dT); in a boiler the gas-side film coefficient governs because flue gas is far poorer than water.
- Radiation Q = e(sigma)A(T1^4 - T2^4); the fourth power makes furnace temperature dominant.
- For exchangers use LMTD with a correction factor for multi-pass units, or effectiveness-NTU. Counterflow outperforms parallel flow.
Worked example
A boiler makes 50,000 kg/h of steam at 4 MPa, 400 C from feedwater at 120 C. From tables, h(steam) = 3214 kJ/kg and h(feed) = 504 kJ/kg. Duty = 50,000 x (3214 - 504) = 135.5 x 10^6 kJ/h. Convert: 135.5 GJ/h divided by 3.6 = 37.6 MW. If fuel input is 45 MW, boiler efficiency = 37.6/45 = 83.6%. Note the trap inside the arithmetic: kJ/h divided by 3600 gives kW directly, so 135.5 x 10^6/3600 = 37,640 kW.
Traps
- Mixing gauge and absolute pressure in steam-table lookups.
- Using Celsius in the Carnot equation.
- Assuming the superheater is free duty; it absorbs heat at a lower gas-to-steam temperature difference than the evaporator.
- Applying one LMTD to an exchanger that desuperheats, condenses and subcools, instead of splitting it into zones.