Thermal Engineering and Gas Dynamics (ME-601) - Unit 3 Short Notes

UNIT 3: Thermal Engineering and Gas Dynamics

I. STEAM POWER PLANT SYSTEMS

A. Steam Generators (Boilers)

Types and Classifications

• Water Tube Boilers: Water flows inside tubes, hot gases outside. Used for high pressure/high capacity.

  • Velox Boiler:

    • Construction: Uses a gas turbine (centrifugal blower) as an air pump. The blower is mounted on the same shaft as a steam turbine, which drives it. The flue gases from the furnace pass through the gas turbine blades, transferring energy to drive the forced draught fan.

    • Significance: Enables very high pressure (up to 70 bar) and high evaporation rate due to high gas velocity (60-90 m/s) and small water content. Rapid steaming.

    • Paths:

      DiagramSEARCH: "Velox boiler flue gas water path diagram"

      1. Water Path: Feedwater → Economiser → Steam Drum (separates steam/water) → Tubes (evaporates) → Steam Drum → Superheater.

      2. Flue Gas Path: Furnace → Gas Turbine (drives blower) → Economiser → Air Preheater → Chimney.

  • Lamont Boiler:

    • Construction: High-pressure, forced circulation water tube boiler. Key feature is a separate steam drum and a centrifugal pump (circulation pump) that forces water from the drum into the evaporator tubes.

    • Salient Features: Handles high pressure (up to 150 bar) safely. Prevents tube overheating by ensuring high water circulation rate independent of steam generation. Evaporator tubes are bent to allow for thermal expansion.

    • DiagramSEARCH: "Lamont boiler diagram with parts labelled"

• General Classifications:

  | Basis | Types | | :--- | :--- | | Water/Gas Path | Fire Tube (flue gases inside tubes), Water Tube (water inside tubes) | | Pressure | Low Pressure (< 15 bar), Medium Pressure (15-32 bar), High Pressure (> 32 bar) | | Furnace Position | Internally Fired, Externally Fired | | Mobility | Stationary, Portable (Locomotive, Marine) |

Boiler Mountings and Accessories

• Essential Mountings (Safety Devices):

  • Safety Valve: Automatically releases steam when pressure exceeds safe limit.

  • Pressure Gauge: Indicates steam pressure.

  • Water Level Indicator: Shows water level in the boiler shell.

  • Fusible Plug: Melts at high temperature to warn of low water level.

  • Steam Stop Valve: Regulates steam flow from boiler to pipe.

  • Blow-off Cock: Removes sediments & empties boiler.

• Important Accessories (Efficiency/Convenience):

  • Feed Pump: Supplies feedwater.

  • Economiser: Recovers heat from flue gases to preheat feedwater.

  • Air Preheater: Recovers heat from flue gases to preheat combustion air.

  • Superheater: Increases steam temperature above saturation (superheats).

  • Feedwater Heater: Uses extracted steam to preheat feedwater (regeneration).

Boiler Performance Analysis

• Boiler Efficiency:

  • Direct Method (Input-Output):

$$\eta_{boiler} = \frac{\text{Heat absorbed by steam}}{\text{Heat supplied by fuel}} = \frac{m_s(h_f - h_{fw})}{m_f \cdot CV}$$

    where $$\displaystyle m_s $$ = steam generated (kg), $$\displaystyle m_f $$ = fuel burnt (kg), $$\displaystyle h_f $$ = final steam enthalpy, $$\displaystyle h_{fw} $$ = feedwater enthalpy, $CV$ = fuel calorific value.

*   **Indirect Method (Heat Loss)**: 

$$\eta_{boiler} = 100 - (q_2 + q_3 + q_4 + q_5 + q_6)\%$$

where $$\displaystyle q_2 $$ to $$\displaystyle q_6 $$ are % losses due to dry flue gas, moisture in fuel/air, unburnt carbon, radiation, etc.

• Heat Balance Sheet: Tabular statement showing heat supplied (from fuel) and heat utilised/absorbed (steam, feedwater heating, etc.) or heat lost. Basis: 1 kg of fuel or 1 kg of steam generated.

• Equivalent Evaporation ($$\displaystyle m_e $$):

  • Definition: The amount of steam (in kg) that would be generated at 100°C from feedwater at 100°C, per kg of fuel burnt, if the steam produced were dry and saturated at 100°C.

  • Purpose: Allows comparison of boilers operating at different pressures/temperatures.

  • Calculation:

$$m_e = \frac{m_s \cdot (h - h_{fw})}{2257 \text{ kJ/kg}}$$

    where $h$ = enthalpy of steam generated (kJ/kg), $$\displaystyle h_{fw} $$ = enthalpy of feedwater (kJ/kg), 2257 kJ/kg = enthalpy of evaporation at 100°C.

    \boxed{m_e = \frac{m_s \cdot (h - h_{fw})}{2257}}

[!TIP] Common Pitfall: In equivalent evaporation, the denominator is always 2257 kJ/kg (latent heat at 100°C), not the actual latent heat at operating pressure.

B. Boiler Draught Systems

Fundamentals

• Definition: The difference in pressure (in mm of water gauge) between the furnace/gas passages and the atmosphere.

• Purpose: To supply controlled air for combustion and to remove flue gases from the furnace to the atmosphere through the chimney.

• Natural Draught: Created by the chimney. Hot flue gases are lighter than cold outside air, setting up a pressure difference.

  • Draught Produced (H):

$$H = 0.353 \times 10^{-3} H_c \left( \frac{1}{T_a} - \frac{1}{T_f} \right) \text{ (m of water)}$$

    where $$\displaystyle H_c $$ = chimney height (m), $$\displaystyle T_a $$ = absolute atmospheric temperature (K), $$\displaystyle T_f $$ = absolute mean flue gas temperature (K).

*   **Chimney Height** depends on draught required, gas temperature, and atmospheric conditions.

Artificial Draught

• Methods:

  | Type | Fan Location | Effect | | :--- | :--- | :--- | | Forced Draught (FD) | Before furnace (pushes air in) | Increases pressure above atmospheric in furnace. | | Induced Draught (ID) | After economiser/air heater (pulls gases out) | Reduces pressure below atmospheric in furnace. | | Balanced Draught | Both FD & ID fans used | Maintains furnace pressure near atmospheric. Most common. |

• Advantages over Natural Draught:

  • Independent of weather/temperature.

  • Higher combustion rates possible.

  • Smaller chimney height possible.

  • Better control.

• Disadvantages: Capital and running cost of fans and power consumption.

Chimney Design Calculations

• Draught Relation:

$$H = H_c \rho_a \left( \frac{1}{T_a} - \frac{1}{T_f} \right)$$

(In consistent units; $$\displaystyle \rho_a $$ = density of air).

• Quantity of Air per kg Fuel:

  From the draught equation, given $H$, $$\displaystyle H_c $$, $$\displaystyle T_a $$, $$\displaystyle T_f $$, we can find the mass of flue gases per kg fuel. Adding the theoretical air and assuming air-fuel ratio, we get actual air used per kg fuel.

  [!TIP] Exam Focus: Problems typically give chimney height, draught (mm w.g.), $$\displaystyle T_{flue} $$, $$\displaystyle T_{atm} $$, $$\displaystyle P_{atm} $$. Use formula to find $$\displaystyle T_f $$ or mass flow, then relate to air-fuel ratio.

C. Steam Power Plant Cycles

Rankine Cycle

• Processes (on T-s and p-v diagrams):

  1. 1-2: Isentropic compression in pump (saturated liquid, $v \approx$ constant).

  2. 2-3: Isobaric heat addition in boiler (water → saturated liquid → saturated vapor → superheated vapor).

  3. 3-4: Isentropic expansion in turbine (work output).

  4. 4-1: Isobaric heat rejection in condenser (wet vapor → saturated liquid).

• Thermal Efficiency:

$$\eta_{thermal} = \frac{W_{net}}{Q_{in}} = \frac{(h_3 - h_4) - (h_2 - h_1)}{h_3 - h_2}$$

For ideal pump work: $$\displaystyle h_2 - h_1 \approx v_f (p_2 - p_1) $$.

• Effect of Parameters:

  • Boiler Pressure ↑: Efficiency ↑ (higher average T of heat addition), but limits due to metallurgy & moisture at turbine exhaust.

  • Condenser Pressure ↓: Efficiency ↑ (lower T of heat rejection), but limits due to vacuum pumps & larger turbine size.

  • Superheat ↑: Efficiency ↑ (higher T of heat addition, reduces moisture at exhaust). No change in pump work.

  DiagramSEARCH: "Rankine cycle T-s diagram effect of pressure"

Cycle Modifications

• Reheat Rankine Cycle:

  • Process: Steam expanded in HP turbine → reheated in boiler → expanded in IP/LP turbines.

  • T-s Diagram: Shows two constant pressure lines in boiler (main & reheater).

  • Advantages: Reduces moisture content at final stages, increases work output, allows higher boiler pressure.

• Regenerative Rankine Cycle:

  • Principle: Use extracted steam from turbine to preheat feedwater (regeneration), reducing $$\displaystyle Q_{in} $$.

  • Open Feedwater Heater (De-aerator): Steam and water mix directly. Serves as de-aerator.

  • Closed Feedwater Heater: Steam heats tube-side feedwater without mixing. Condensate (drain) is throttled to condenser or next heater.

  • T-s Diagram: Shows feedwater line crossing saturation dome via horizontal lines (constant T extraction).

• Binary Vapour Cycle:

  • Uses two different working fluids (e.g., mercury-steam, organic fluid-water).

  • High-temperature fluid (e.g., mercury) expands in a turbine, rejects heat at moderate T to boil the second fluid (water), which then expands in a low-temperature turbine.

  • Advantage: Matches heat source/sink temperatures better, reduces irreversibility.

• Modified Rankine (Steam Engine) Cycle:

  • Constant Volume Heat Rejection (release at cut-off): Process 4-1 is constant volume (instead of isobaric). More realistic for steam engines.

  • Efficiency expression differs from simple Rankine.

Performance Analysis

• Overall Plant Efficiency:

$$\eta_{overall} = \eta_{boiler} \times \eta_{turbine} \times \eta_{generator} \times \eta_{cycle}$$

(Note: $$\displaystyle \eta_{cycle} $$ is the thermal efficiency of the ideal cycle, distinct from turbine efficiency).

• Steam Rate (SR): Steam consumed per kWh output.

$$SR = \frac{3600}{W_{net} \text{ (kJ/kg)}} \text{ kg/kWh}$$

• Quality at Turbine Exhaust: Use Mollier chart (h-s). Locate state 3 (inlet), follow isentropic line ($$\displaystyle s_3=s_4s $$) to condenser pressure $$\displaystyle p_4 $$. Read $$\displaystyle x_4s $$. For actual expansion, account for isentropic efficiency: $$\displaystyle h_4 = h_3 - \eta_{turb} (h_3 - h_4s) $$.

[!TIP] Case Study Approach: For given plant data (P_boiler, T_superheat, P_cond, m_condensate/kWh, T_condensate, η_gen):

1. Find $$\displaystyle h_{fw} $$ from $$\displaystyle T_{cond} $$ (approx. saturated liquid at condenser P).

2. $$\displaystyle Q_{in} = h_3 - h_{fw} $$ (from steam tables/Mollier).

3. $$\displaystyle W_{net,actual} = \frac{3600}{m_{condensate}} $$ (kJ/kg).

4. $$\displaystyle \eta_{overall} = \frac{W_{net,actual}}{Q_{in}} \times \frac{1}{\eta_{gen}} $$? No! $$\displaystyle W_{net,actual} $$ is generator output. So:

$$\eta_{overall} = \frac{\text{Generator Output}}{\text{Heat Input}} = \frac{3600}{SR \cdot (h_3 - h_{fw})}$$

(Given $$\displaystyle SR = m_{condensate} $$ kg/kWh).

D. Steam Turbines

Types and Fundamentals

• Impulse Turbine:

  • Working: Pressure drop occurs only in nozzles. No pressure change in moving blades. High-velocity jet impinges on buckets.

  • Velocity Diagram: Shows absolute velocity ($C$), blade velocity ($U$), relative velocity ($V$). Work done per kg = $$\displaystyle U(C_{w1} + C_{w2}) $$.

  • Blade Efficiency (η_b): $$\displaystyle \eta_b = \frac{2 U (C_{w1} + C_{w2})}{C_1^2} $$ (for symmetrical frictionless blades, $$\displaystyle C_{w1}=C_1 \cos\alpha_1 $$, $$\displaystyle C_{w2}=C_1 \cos\alpha_1 $$, $$\displaystyle \eta_b = \cos^2\alpha_1 $$).

• Reaction Turbine:

  • Working: Pressure drop occurs both in fixed nozzles (stators) and moving blades (rotors). Moving blades act as nozzles.

  • Velocity Diagram: Similar but with different pressure conditions. Work done per kg = $$\displaystyle U(V_{w1} + V_{w2}) $$.

• Compounding (Need: To reduce blade speed & size for high pressure drop):

  • Pressure Compounding (Rateau): Multiple stages of impulse turbines with pressure drop divided across nozzle rings.

  • Velocity Compounding (Curtis): Single pressure drop, but velocity is compounded (multiple moving blade rings on same shaft with fixed guide blades in between).

Analysis Using Mollier Chart

1. Locate inlet state (P₁, T₁ or h₁, s₁).

2. For isentropic expansion: move vertically down at constant s to exhaust pressure P₂ → state 2s. Read h₂s, x₂s.

3. Actual expansion: $$\displaystyle h_2 = h_1 - \eta_{turb} (h_1 - h_{2s}) $$. Locate state 2 on constant pressure line P₂ using this h₂. Read x₂.

4. Work output per kg = $$\displaystyle h_1 - h_2 $$.

E. Feedwater and Condensing Systems

Feedwater Heaters (FWH)

• Open-Type (De-aerator): Steam and water mix directly in a tank. Removes dissolved gases (O₂, CO₂). Serves as a storage vessel. Pressure is usually saturation pressure of extraction steam.

• Closed-Type: Shell-and-tube heat exchanger. Steam condenses on tube side (or shell side) heating feedwater in the other side. No mixing. Requires drain cooler to cool condensate before throttling.

• Key Difference: Open-type involves direct contact and mixing; closed-type is indirect heat exchange.

Condensers

• Jet Condensers: Steam jets mix directly with cooling water. Simple, but condensate is contaminated. Low-level counterflow type: Steam enters at top, flows down counterflow to cooling water entering at bottom.

  DiagramSEARCH: "low level counterflow jet condenser diagram"

• Surface Condensers: Steam and cooling water do not mix. Shell-and-tube heat exchanger. Produces pure condensate. Most common in power plants.

• Sources of Air Leakage & Effects:

  • Sources: Joints, glands, seals, vacuum breakdown.

  • Effects:

    1. Increases partial pressure of non-condensable gases → reduces vacuum (increases condenser pressure).

    2. Increases condenser temperature (for given pressure).

    3. Increases pump work (to handle air-water mixture).

    4. Reduces heat transfer coefficient (air film on tubes).

    5. Increases back pressure on turbine → reduces turbine work & efficiency.

• Condenser Performance Parameters:

  • Vacuum: $$\displaystyle P_{atm} - P_{cond} $$ (in mm Hg or bar).

  • Condenser Temperature: Corresponding saturation temperature at $$\displaystyle P_{cond} $$.

  • Condensate Subcooling: $$\displaystyle T_{condensate} - T_{sat} $$ at $$\displaystyle P_{cond} $$. Should be minimal (1-2°C).

II. GAS DYNAMICS AND COMPRESSORS

A. Fundamentals of Compressible Flow

Mach Number (M)

• Definition: $$\displaystyle M = \frac{\text{Local Flow Velocity (C)}}{\text{Local Velocity of Sound (a)}} = \frac{C}{a} $$.

• Significance & Regimes:

  • $$\displaystyle M < 1 $$: Subsonic (pressure disturbances propagate upstream).

  • $$\displaystyle M = 1 $$: Sonic (critical condition, flow choked at throat in C-D nozzle).

  • $$\displaystyle M > 1 $$: Supersonic (pressure disturbances cannot propagate upstream).

  • $$\displaystyle M > 5 $$: Hypersonic (high temperature effects, dissociation).

• Mach Cone: For supersonic flow, disturbances form a conical wavefront. Mach Angle $$\displaystyle \mu = \sin^{-1}(1/M) $$.

• Zone of Action: Region inside the Mach cone where flow is influenced by the body.

• Zone of Silence: Region outside the Mach cone where flow is unaffected by the body.

Velocity of Sound

• For an ideal gas:

$$a = \sqrt{\gamma R T} = \sqrt{\gamma \frac{P}{\rho}}$$

where $$\displaystyle \gamma = c_p/c_v $$, $R$ = gas constant, $T$ = static temperature.

• For steam/real gases, use $$\displaystyle a = \sqrt{\left(\frac{\partial P}{\partial \rho}\right)_s} $$.

Isentropic Flow in Variable Area Ducts

• Governing Equations (1D, steady, adiabatic, no friction, no work):

$$\frac{dA}{A} = \frac{dC}{C} \left( M^2 - 1 \right)$$

**Area-Velocity Relation**:

*   $$\displaystyle M < 1 $$ (subsonic): $dA \uparrow \Rightarrow dC \uparrow$ (divergent duct accelerates).

*   $$\displaystyle M = 1 $$: $$\displaystyle dA = 0 $$ (throat).

*   $$\displaystyle M > 1 $$ (supersonic): $dA \uparrow \Rightarrow dC \downarrow$ (divergent duct decelerates? **No!** For supersonic, divergent duct **accelerates** ($$\displaystyle dC>0 $$), so $$\displaystyle dA>0 $$ requires $$\displaystyle M^2-1>0 $$ → correct. Convergent duct decelerates supersonic flow).

• Stagnation Properties (Total, 0):

  • Stagnation Pressure ($$\displaystyle P_0 $$): Pressure when isentropically brought to rest.

  • Stagnation Temperature ($$\displaystyle T_0 $$): Temperature when isentropically brought to rest. Constant in isentropic flow ($$\displaystyle T_0 = T(1 + \frac{\gamma-1}{2}M^2) $$).

  • Stagnation Density ($$\displaystyle \rho_0 $$).

  • Relations:

$$\frac{P_0}{P} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)}$$

$$\frac{T_0}{T} = 1 + \frac{\gamma-1}{2}M^2$$

$$\frac{\rho_0}{\rho} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{1/(\gamma-1)}$$

B. Flow Through Nozzles and Diffusers

Nozzle Flow

• Nozzle Types:

  • Convergent: For subsonic flow only ($$\displaystyle M_{max}=1 $$ at exit if $$\displaystyle P_e/P_0 $$ ≤ critical).

  • Divergent: For supersonic flow only (decelerates supersonic to subsonic).

  • Convergent-Divergent (C-D): For supersonic flow. Convergent section accelerates to $$\displaystyle M=1 $$ at throat, divergent section accelerates to supersonic.

• Critical Pressure Ratio ($$\displaystyle r_c $$):

  • Pressure ratio at which $$\displaystyle M=1 $$ at throat (choking occurs). For ideal gas, $$\displaystyle \gamma=1.4 $$: $$\displaystyle r_c = (2/(\gamma+1))^{\gamma/(\gamma-1)} \approx 0.528 $$.

  • Maximum Discharge Condition: When $$\displaystyle P_e/P_0 \leq r_c $$, mass flow rate is maximum and independent of $$\displaystyle P_e $$. Occurs when throat is choked ($$\displaystyle M=1 $$).

  • Derivation: From continuity & isentropic relations, mass flow rate $$\displaystyle \dot{m} = \frac{A_t P_0}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} M \left(1 + \frac{\gamma-1}{2}M^2\right)^{-(\gamma+1)/(2(\gamma-1))} $$. Max w.r.t $M$ at $$\displaystyle M=1 $$ gives $$\displaystyle r_c $$.

• Friction and Nozzle Efficiency:

  • Effect of Friction:

    • Increases entropy → lower exit velocity & enthalpy drop.

    • Increases exit pressure (less than ideal vacuum).

    • Reduces mass flow rate (if not choked, but for choked flow, mass flow may reduce slightly).

  • Nozzle Efficiency ($$\displaystyle \eta_n $$):

$$\eta_n = \frac{\text{Actual kinetic energy at exit}}{\text{Isentropic enthalpy drop}} = \frac{C_a^2/2}{h_0 - h_e} = \frac{h_0 - h_e - \Delta h_f}{h_0 - h_e}$$

    where $$\displaystyle \Delta h_f $$ = friction loss (heat drop loss).

Supersaturated Flow

• Conditions: In steam nozzles, if expansion is very rapid (high velocity), condensation lags behind equilibrium. Steam remains in metastable superheated state even though $P$ and $T$ are in the two-phase region.

• Difference from Isentropic:

  • Isentropic: Condensation occurs at saturation line (Wilson line), follows equilibrium.

  • Supersaturated: No condensation until Wilson line crossed. Entropy is constant (isentropic), but state is superheated.

  • Effect: Higher exit velocity (higher $$\displaystyle h_0 - h_e $$ as no latent heat released), lower exit temperature, higher mass flow rate than isentropic (since $v$ is larger at same $P$).

  • Wilson Line: Empirical line on T-s diagram beyond which condensation begins rapidly.

Diffusers

• Function: Convert high-velocity, low-pressure fluid into low-velocity, high-pressure fluid. Opposite of nozzle.

• Effect:

  • Subsonic Diffuser: Convergent shape. Velocity decreases, pressure increases.

  • Supersonic Diffuser: Convergent-Divergent shape. First convergent section creates a shock (normal/oblique) to decelerate supersonic to subsonic, then divergent section further decelerates to increase pressure. Shock causes total pressure loss.

C. Air Compressors

Classification

Reciprocating Compressors (Single-Stage, Single-Acting)

• Construction & Working: Piston reciprocates in cylinder. Suction & delivery via valves. Single-acting: compression on one side of piston.

• p-v Diagram: Shows suction, compression, delivery, expansion processes.

• Work Input without Clearance:

  For polytropic process $$\displaystyle PV^n = C $$:

$$W = \frac{n}{n-1} P_1 V_1 \left[ \left(\frac{P_2}{P_1}\right)^{(n-1)/n} - 1 \right]$$

For isentropic ($\gamma$ instead of $n$).

• Effect of Clearance Volume ($$\displaystyle V_c $$):

  • Volumetric Efficiency ($$\displaystyle \eta_v $$):

$$\eta_v = 1 + C - C \left(\frac{P_2}{P_1}\right)^{1/n}$$

    where $$\displaystyle C = V_c / V_s $$ (clearance fraction), $$\displaystyle V_s $$ = stroke volume.

*   **p-v Diagram with Clearance**: Shows expansion from $$\displaystyle P_2 $$, $$\displaystyle V_c $$ down to $$\displaystyle P_1 $$ at $$\displaystyle V = V_c + V_s $$ before suction opens. Reduces effective suction volume.

• Polytropic vs Isentropic: Polytropic index $n$ is between 1 (isothermal) and $\gamma$ (isentropic). Actual compression usually polytropic due to heat transfer.

Multistage Compression with Intercooling

• Perfect Intercooling: Intercooler cools air to initial temperature $$\displaystyle T_1 $$ before entering next stage.

• Minimum Work Condition: For perfect intercooling and same pressure ratio per stage, total work is minimum when pressure ratios are equal in all stages.

  • For $n$ stages: $$\displaystyle r_p = (P_{out}/P_{in})^{1/n} $$ per stage.

  • Derivation: Total work $$\displaystyle W_{total} = \sum \frac{n}{n-1} P_1 V_1 \left( r^{n-1/n} - 1 \right) $$. For fixed $$\displaystyle P_{in}, P_{out} $$, minimize w.r.t $$\displaystyle r_i $$ using calculus or AM-GM inequality → $$\displaystyle r_1 = r_2 = ... = r_n $$.

• Imperfect Intercooling: Intercooler does not bring temperature back to $$\displaystyle T_1 $$. Work is higher than perfect case.

Rotary Compressors (Centrifugal)

• Construction & Working: Impeller with radial blades rotates at high speed. Air enters axially at centre, gains kinetic energy, passes through diffuser (volute) where kinetic energy converts to pressure.

• Velocity Triangles:

  • Inlet (Radius $$\displaystyle r_1 $$): $$\displaystyle C_1 $$ (axial), $$\displaystyle \alpha_1 \approx 90^\circ $$; $$\displaystyle U_1 = 2\pi r_1 N $$; $$\displaystyle V_1 $$ (relative).

  • Outlet (Radius $$\displaystyle r_2 $$): $$\displaystyle C_2 $$ (radial/backward curved), $$\displaystyle \alpha_2 $$; $$\displaystyle U_2 $$; $$\displaystyle V_2 $$.

  • Work Input (Euler's Equation): $$\displaystyle \Delta h = U_2 V_{w2} - U_1 V_{w1} $$. For axial inlet ($$\displaystyle V_{w1}=0 $$): $$\displaystyle \Delta h = U_2 V_{w2} $$.

  • Blade Shape: Backward curved (most common, efficient, stable), radial, forward curved.

D. Compressor Performance and Efficiencies

Key Parameters

• Mean Effective Pressure (MEP):

  • Definition: Hypothetical constant pressure that, if acted on the piston during the power stroke, would produce the same net work as the actual cycle.

  • For reciprocating compressor (work input per cycle): $$\displaystyle W_{cycle} = MEP \times V_s $$.

$$MEP = \frac{W_{cycle}}{V_s}$$

• Volumetric Efficiency ($$\displaystyle \eta_v $$):

$$\eta_v = \frac{\text{Actual volume of air drawn at inlet conditions}}{\text{Swept volume } V_s}$$

With clearance: $$\displaystyle \eta_v = 1 + C - C \left(\frac{P_2}{P_1}\right)^{1/n} $$.

• Power Required:

$$P = \frac{W_{cycle} \times N}{60} \text{ (Watts)}$$

where $N$ = rpm (for single-acting). For double-acting, double the work per revolution.

Or: $$\displaystyle P = \frac{m \cdot w}{\eta_{mech}} $$ where $m$ = mass flow rate, $w$ = specific work.

• Temperature of Delivered Air (Polytropic):

$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{(n-1)/n}$$

(Assuming no intercooling for single-stage).

Efficiencies

• Isentropic Efficiency ($$\displaystyle \eta_s $$ or $$\displaystyle \eta_{isen} $$):

$$\eta_{isen} = \frac{\text{Isentropic work input}}{\text{Actual work input}} = \frac{w_s}{w_a} = \frac{(h_{2s} - h_1)}{(h_2 - h_1)}$$

For ideal gas: $$\displaystyle \eta_{isen} = \frac{\gamma-1}{\gamma} \frac{r_p^{(\gamma-1)/\gamma} - 1}{r_p^{(n-1)/n} - 1} $$.

• Isothermal Efficiency ($$\displaystyle \eta_{iso} $$):

$$\eta_{iso} = \frac{\text{Isothermal work input}}{\text{Actual work input}} = \frac{w_{iso}}{w_a}$$

Isothermal work (min possible): $$\displaystyle w_{iso} = RT_1 \ln(r_p) $$.

• Mechanical Efficiency ($$\displaystyle \eta_{mech} $$):

$$\eta_{mech} = \frac{\text{Indicated work (inside cylinder)}}{\text{Brake work (input power)}} = \frac{W_i}{W_b}$$

Accounts for friction, windage.

• Volumetric Efficiency ($$\displaystyle \eta_v $$): As defined above. Affected by clearance, pressure drop in valves/pipes, heating during suction.

E. Gas Turbine Cycles (Brayton Cycle and Modifications)

Ideal Brayton (Joule) Cycle

• Processes:

  1. 1-2: Isentropic compression in compressor.

  2. 2-3: Isobaric heat addition in combustion chamber.

  3. 3-4: Isentropic expansion in turbine.

  4. 4-1: Isobaric heat rejection to atmosphere.

• T-s & p-v Diagrams: Two vertical lines (isentropic), two horizontal lines (isobaric).

• Thermal Efficiency:

$$\eta_{Brayton} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}$$

where $$\displaystyle r_p = P_2/P_1 = P_3/P_4 $$ (pressure ratio).

*Depends only on $$\displaystyle r_p $$ and $\gamma$.*

Cycle Improvements

• Reheat:

  • Working: Turbine expansion in stages with reheating between stages (like steam Rankine).

  • Advantage: Increases work output, reduces moisture (for steam), allows higher pressure ratio.

  • T-s Diagram: Shows two constant pressure lines in combustion chamber (main & reheater).

• Regeneration:

  • Working: Use turbine exhaust heat (before entering combustion chamber) to preheat compressed air from compressor outlet using a regenerator (heat exchanger).

  • Advantage: Reduces fuel consumption ($$\displaystyle Q_{in} $$) → higher efficiency.

  • Regenerative Brayton Efficiency:

$$\eta_{regen} = 1 - \frac{T_1}{T_3} \frac{r_p^{(\gamma-1)/\gamma} - 1}{\epsilon - 1}$$

    where $\epsilon$ = regenerator effectiveness = $$\displaystyle (T_5 - T_2)/(T_3 - T_2) $$.

• Derivation of Conditions for Maximum Output with Reheat & Regenerator:

  • For maximum net work output (not necessarily max efficiency), the intermediate pressure $$\displaystyle P_x $$ (between HP & LP turbines) should be such that the temperature after reheating equals the maximum cycle temperature $$\displaystyle T_3 $$.

  • For maximum thermal efficiency with given $$\displaystyle T_{max} $$ and $$\displaystyle r_p $$, the pressure ratio for maximum efficiency is:

$$r_{p,opt} = \left(\frac{T_3}{T_1}\right)^{\gamma/(2(\gamma-1))}$$

    (This is the pressure ratio that makes the temperatures at compressor exit and turbine inlet equal in a Brayton cycle with regeneration, minimizing exergy destruction).

III. SUPPORTING ANALYSES AND APPLICATIONS

A. Steam Property Determination

• Wet Steam:

  • Quality (x): Mass fraction of vapor. $$\displaystyle x = \frac{m_v}{m_v + m_f} $$.

  • Specific Volume: $$\displaystyle v = v_f + x v_{fg} $$.

  • Enthalpy: $$\displaystyle h = h_f + x h_{fg} $$.

  • Entropy: $$\displaystyle s = s_f + x s_{fg} $$.

  • Internal Energy: $$\displaystyle u = u_f + x u_{fg} $$ or $$\displaystyle u = h - Pv $$.

  • From Steam Tables: Find $P$ or $T$, locate saturation values, interpolate if needed.

  • From Mollier Chart (h-s): Locate pressure line, move horizontally at constant $s$ to find $h$, $x$.

• Superheated Steam:

  • Read directly from superheated tables at given $P$ and $T$.

  • On Mollier chart: Locate $P$ and $T$ (or $h$), read other properties.

B. Problem-Solving Methodologies

1. Boiler Draught & Chimney

• Given: $$\displaystyle H_c $$, $H$ (mm w.g.), $$\displaystyle T_{flue} $$, $$\displaystyle T_{atm} $$, $$\displaystyle P_{atm} $$.

• Find: Mass of air/kg fuel.

• Steps:

  1. Convert $H$ to meters of flue gas: $$\displaystyle H_{fg} = H \times \frac{\rho_{water}}{\rho_{fg}} $$.

  2. Use: $$\displaystyle H_{fg} = H_c \rho_a \left( \frac{1}{T_a} - \frac{1}{T_f} \right) $$ (in consistent units). Solve for $$\displaystyle T_f $$ or $$\displaystyle \rho_{fg} $$.

  3. Mass of flue gas/kg fuel: $$\displaystyle m_{fg} = \frac{\text{Air used} + \text{Fuel mass} + \text{Moisture in fuel/air}}{} $$.

  4. Usually, assume air-fuel ratio from fuel analysis or given. Then: Air used = $$\displaystyle m_{fg} - \text{fuel mass} - \text{moisture} $$. Often simplified: Air used ≈ Mass of flue gas (if fuel mass & moisture small).

2. Compressor Cycles

• Single-stage:

  • Given: $$\displaystyle P_1, T_1, P_2, V_s, N, C, n $$.

  • Find MEP, Power, $$\displaystyle \eta_v $$, $$\displaystyle T_2 $$.

  • Steps:

    1. $$\displaystyle T_2 = T_1 (P_2/P_1)^{(n-1)/n} $$.

    2. Work/cycle: $$\displaystyle W = \frac{n}{n-1} P_1 V_1 \left( r_p^{(n-1)/n} - 1 \right) $$, where $$\displaystyle V_1 = V_s - V_c $$? Better: Use swept volume $$\displaystyle V_s $$. Indicated work per cycle = area on p-v = $$\displaystyle \frac{n}{n-1} P_1 (V_1 - V_2) $$ but $$\displaystyle V_1 $$ is volume at start of compression. For clearance, $$\displaystyle V_1 = V_c + \eta_v V_s $$. Standard formula: $$\displaystyle W = \frac{n}{n-1} P_1 V_s \left[ 1 + C - C r_p^{1/n} \right] \left( r_p^{(n-1)/n} - 1 \right) $$.

    3. MEP = $$\displaystyle W / V_s $$.

    4. Power = $(W \times N) / 60$ (single-acting).

    5. $$\displaystyle \eta_v = 1 + C - C r_p^{1/n} $$.

• Multistage with Perfect Intercooling:

  • Minimum work: $$\displaystyle W_{min} = \frac{n}{n-1} P_1 V_1 \left[ \left(\frac{P_2}{P_1}\right)^{(n-1)/n} - 1 \right] \times \text{number of stages} $$? No. For equal pressure ratios $r$ per stage, $$\displaystyle r^n = P_{out}/P_{in} $$.

  • Work per kg: $$\displaystyle w = \frac{n}{n-1} R T_1 \left( r^{(\gamma-1)/\gamma} - 1 \right) \times n $$? Better: Total work = $$\displaystyle n \times \frac{n}{n-1} R T_1 \left( r^{(n-1)/n} - 1 \right) $$ with $$\displaystyle r = (P_{out}/P_{in})^{1/n} $$.

  • Heat rejected to intercooler: $$\displaystyle Q_{intercooler} = m c_p (T_2 - T_1) $$ per stage (since cooled to $$\displaystyle T_1 $$).

3. Nozzle Flow

• Given: Inlet $$\displaystyle P_0, T_0 $$ (or $$\displaystyle h_0 $$), exit $$\displaystyle P_e $$, nozzle type, friction (η_n or % loss).

• Find: Exit velocity, mass flow rate, critical pressure.

• Steps:

  1. Check if choked: If $$\displaystyle P_e/P_0 \leq r_c $$ (for given fluid), then $$\displaystyle M=1 $$ at throat, $$\displaystyle P^* = P_0 \cdot r_c $$, $$\displaystyle T^* = T_0 \cdot \frac{2}{\gamma+1} $$.

  2. Isentropic exit velocity: $$\displaystyle C_{e,s} = \sqrt{2 (h_0 - h_e)} $$ where $$\displaystyle h_e $$ from isentropic expansion to $$\displaystyle P_e $$.

  3. Actual velocity: $$\displaystyle C_e = \eta_n \cdot C_{e,s} $$ (if nozzle efficiency given) or account for heat loss.

  4. Mass flow rate:

     • If not choked: $$\displaystyle \dot{m} = \frac{A_e P_e}{R T_e} C_e $$ (use ideal gas law at exit).

     • If choked: $$\displaystyle \dot{m} = \frac{A_t P_0}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \left(\frac{2}{\gamma+1}\right)^{(\gamma+1)/(2(\gamma-1))} $$.

• Supersaturated Flow: Use Wilson line (approx. $$\displaystyle T_{Wilson} \approx 0.92 T_{sat} $$ for steam) to find effective $$\displaystyle h_e $$ or treat as isentropic superheated expansion.

4. Steam Power Plant Performance

• Given: Plant data (boiler P,T; condenser P; m_condensate/kWh; T_condensate; η_gen).

• Find: Overall efficiency, steam rate, quality.

• Steps:

  1. Heat Input per kg steam: $$\displaystyle q_{in} = h_3 - h_{fw} $$. $$\displaystyle h_3 $$ from boiler conditions (use steam tables/Mollier). $$\displaystyle h_{fw} $$ ≈ $$\displaystyle h_f $$ at $$\displaystyle T_{cond} $$ (saturated liquid at condenser P).

  2. Net Work Output per kg steam: $$\displaystyle w_{net} = \frac{3600}{SR} $$ kJ/kg, where $$\displaystyle SR = m_{condensate} $$ kg/kWh.

  3. Cycle Efficiency: $$\displaystyle \eta_{cycle} = w_{net} / q_{in} $$.

  4. Overall Efficiency: $$\displaystyle \eta_{overall} = \eta_{cycle} \times \eta_{generator} $$? No: $$\displaystyle w_{net} $$ is turbine work output. Generator output = $$\displaystyle w_{net} \times \eta_{gen} $$. So:

$$\eta_{overall} = \frac{\text{Generator Output}}{\text{Heat Input}} = \frac{w_{net} \times \eta_{gen}}{q_{in}} = \eta_{cycle} \times \eta_{gen}$$

5. **Quality at Turbine Exhaust**: Use Mollier chart. From $$\displaystyle P_4 $$ (condenser P) and $$\displaystyle s_4 = s_3 $$ (isentropic), find $$\displaystyle h_{4s} $$, $$\displaystyle x_{4s} $$. Then $$\displaystyle h_4 = h_3 - \eta_{turb} (h_3 - h_{4s}) $$. On $$\displaystyle P_4 $$ line, find state with $$\displaystyle h_4 $$, read $$\displaystyle x_4 $$.

[!TIP] Common Pitfall: In overall efficiency calculation, ensure consistent basis. Steam rate is per kWh output. Work from steam tables is per kg steam. $$\displaystyle w_{net} = 3600 / SR $$ gives net work in kJ per kg steam if SR is in kg/kWh. Then $$\displaystyle \eta_{overall} = (w_{net} \cdot \eta_{gen}) / q_{in} $$.