8.3 Thermal-Fluid Engineering — Thermodynamics & Fluid Mechanics

8.3 Thermal-Fluid Engineering — Thermodynamics & Fluid Mechanics

Thermal-Fluid Engineering integrates thermodynamics, heat transfer, and fluid mechanics. For electrical engineers, these principles are indispensable when analyzing thermal power generation systems (steam, gas turbine, geothermal, diesel), transformer cooling, electrical enclosure heat dissipation, hydraulic power generation, and pump/fan fluid dynamics.

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1. First & Second Laws of Thermodynamics & Ideal Gas Processes

The First Law of Thermodynamics (Conservation of Energy)

• Closed System (Non-Flow Process): $Q - W = \Delta U$ where $Q$ is heat added to system, $W$ is boundary work done by system, and $\Delta U = m c_v (T_2 - T_1)$ is change in internal energy.
• Open System (Steady-Flow Energy Equation - SFEE): $q - w = (h_2 - h_1) + \frac{v_2^2 - v_1^2}{2} + g (z_2 - z_1)$ where $h = u + P v$ is specific enthalpy ($\Delta h = c_p \Delta T$).

Ideal Gas Equation & Specific Heat Relations

$P V = m R T \quad \text{or} \quad P v = R T$

where $R = \frac{\bar{R}}{M}$ is the specific gas constant ($R_{\text{air}} = 287\ \text{J/(kg}\cdot\text{K)}$).

• Specific Heat Capacity Identity: $c_p - c_v = R, \quad \gamma = \frac{c_p}{c_v}$ $c_v = \frac{R}{\gamma - 1}, \quad c_p = \frac{\gamma R}{\gamma - 1}$ (For air: $c_p = 1.005\ \text{kJ/(kg}\cdot\text{K)}$, $c_v = 0.718\ \text{kJ/(kg}\cdot\text{K)}$, $\gamma = 1.4$).

Summary of Ideal Gas Non-Flow Processes

Second Law of Thermodynamics & Entropy

• Kelvin-Planck Statement: No heat engine operating in a cycle can convert all absorbed heat into useful work (100% thermal efficiency is impossible).
• Clausius Statement: Heat cannot spontaneously flow from a colder body to a hotter body without external work input.
• Thermal Efficiency of Heat Engine: $\eta_{\text{th}} = \frac{W_{\text{net}}}{Q_H} = 1 - \frac{Q_L}{Q_H}$
• Carnot Cycle Efficiency Limit: Operating between absolute temperatures $T_H$ and $T_L$ (in Kelvin): $\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$
• Entropy Change (Isentropic / Reversible): $dS = \frac{\delta Q_{\text{rev}}}{T}$.

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2. Power Cycles & Heat Transfer Fundamentals

Standard Thermodynamic Power Cycles

1. Rankine Cycle: Standard ideal cycle for steam power plants (boilers, steam turbines, condensers, feed pumps).
2. Otto Cycle: Ideal air-standard cycle for spark-ignition engines: $\eta_{\text{Otto}} = 1 - \frac{1}{r_v^{\gamma - 1}}$ where $r_v = V_1 / V_2$ is the volumetric compression ratio.
3. Diesel Cycle: Ideal cycle for compression-ignition engines: $\eta_{\text{Diesel}} = 1 - \frac{1}{r_v^{\gamma - 1}} \left[ \frac{r_c^\gamma - 1}{\gamma (r_c - 1)} \right]$ where $r_c = V_3 / V_2$ is the fuel cut-off ratio.

Modes of Heat Transfer

• Conduction (Fourier's Law): Heat diffusion through solid media: $q = -k A \frac{dT}{dx} \implies Q = \frac{k A (T_1 - T_2)}{L}$ where $k$ is thermal conductivity (W/m·K).
• Convection (Newton's Law of Cooling): Heat transfer between solid surface and moving fluid: $Q = h A (T_s - T_\infty)$ where $h$ is convection heat transfer coefficient (W/m²·K).
• Radiation (Stefan-Boltzmann Law): Electromagnetic wave thermal emission: $Q = \epsilon \sigma A (T_s^4 - T_{\text{surr}}^4)$ where $\sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\cdot\text{K}^4)$ and $\epsilon$ is surface emissivity.

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3. Fluid Statics, Hydrostatic Pressure & Buoyancy

Fluid Properties

• Density ($\rho$): Mass per unit volume (Water $\rho_w = 1000\ \text{kg/m}^3$).
• Specific Weight ($\gamma$): Weight per unit volume: $\gamma = \rho g \quad (\text{Water } \gamma_w = 9810\ \text{N/m}^3 = 9.81\ \text{kN/m}^3)$
• Specific Gravity ($SG$): Ratio of fluid density to standard water density: $SG = \frac{\rho_{\text{fluid}}}{\rho_w} = \frac{\gamma_{\text{fluid}}}{\gamma_w} \quad (\text{Mercury } SG_{\text{Hg}} = 13.6)$\n

Hydrostatic Pressure Equation

$P = P_0 + \rho g h = P_0 + \gamma h$

• Absolute vs. Gage Pressure: $P_{\text{abs}} = P_{\text{gage}} + P_{\text{atm}}$ (Standard Atmospheric Pressure $P_{\text{atm}} = 101.325\ \text{kPa} = 1.01325\ \text{bar} = 14.7\ \text{psi} = 760\ \text{mmHg}$).

Archimedes' Buoyancy Principle

Any body submerged partially or fully in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced:

$F_B = \rho_{\text{fluid}} g V_{\text{displaced}} = \gamma_{\text{fluid}} V_{\text{displaced}}$

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4. Fluid Dynamics, Bernoulli's Equation & Pipe Flow Head Loss

Mass Continuity Equation (Incompressible Flow)

$\dot{m} = \rho A_1 v_1 = \rho A_2 v_2 \implies Q = A_1 v_1 = A_2 v_2$

where $Q$ is volumetric flow rate (m³/s) and $v$ is mean flow velocity (m/s).

Bernoulli's Energy Equation (Inviscid, Incompressible Flow)

Along a streamline in steady friction-free flow, total energy head remains constant:

$\frac{P_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{v_2^2}{2g} + z_2$

where $\frac{P}{\gamma}$ is pressure head, $\frac{v^2}{2g}$ is velocity head, and $z$ is elevation head.

Extended Energy Equation with Pipe Friction Head Loss ($h_f$) & Pump Head ($H_p$)

$\frac{P_1}{\gamma} + \frac{v_1^2}{2g} + z_1 + H_p = \frac{P_2}{\gamma} + \frac{v_2^2}{2g} + z_2 + h_f$

Darcy-Weisbach Equation for Pipe Head Loss

$h_f = f \left(\frac{L}{D}\right) \left(\frac{v^2}{2g}\right)$

where $f$ is the Darcy friction factor, $L$ is pipe length, $D$ is pipe inside diameter, and $v$ is fluid velocity.

• Reynolds Number ($Re$): Characterizes flow regime: $Re = \frac{\rho v D}{\mu} = \frac{v D}{\nu}$
  • Laminar Flow ($Re < 2300$): Friction factor $f = \frac{64}{Re}$.
  • Turbulent Flow ($Re > 4000$): $f$ depends on relative roughness $\epsilon/D$ and $Re$ (Moody Chart or Colebrook equation).

Hydraulic & Electrical Pump Power

• Hydraulic Power Output ($P_{\text{hyd}}$): $P_{\text{hyd}} = \gamma Q H_p = \rho g Q H_p$
• Electrical Motor Input Power ($P_{\text{elec}}$): $P_{\text{elec}} = \frac{P_{\text{hyd}}}{\eta_{\text{pump}} \cdot \eta_{\text{motor}}} = \frac{\gamma Q H_p}{\eta_{\text{overall}}}$

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Solved Board Exam Examples

Example 1: Polytropic Gas Compression Work Calculation

Problem: Air ($R = 287\ \text{J/(kg}\cdot\text{K)}$, $\gamma = 1.4$) with mass $m = 2.0\ \text{kg}$ is compressed polytropically with exponent $n = 1.30$ from initial conditions $P_1 = 100\ \text{kPa}$ and $T_1 = 300\ \text{K}$ to a final pressure $P_2 = 800\ \text{kPa}$. Calculate the final temperature $T_2$, final volume $V_2$, and boundary work $W_{1-2}$.

Solution:

1. Calculate initial volume $V_1$ using Ideal Gas Law: $V_1 = \frac{m R T_1}{P_1} = \frac{2.0 \times 287 \times 300}{100,000} = \frac{172,200}{100,000} = 1.722\ \text{m}^3$
2. Determine final temperature $T_2$ using polytropic relation: $T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} = 300 \left(\frac{800}{100}\right)^{\frac{1.3 - 1}{1.3}} = 300 \left(8\right)^{0.230769}$ $T_2 = 300 \times 1.6146 = 484.39\ \text{K} \quad (211.24^\circ\text{C})$
3. Calculate final volume $V_2$: $V_2 = \frac{m R T_2}{P_2} = \frac{2.0 \times 287 \times 484.39}{800,000} = \frac{278,039.86}{800,000} = 0.34755\ \text{m}^3$
4. Calculate polytropic boundary work $W_{1-2}$: $W_{1-2} = \frac{m R (T_1 - T_2)}{n - 1} = \frac{2.0 \times 287 \times (300 - 484.39)}{1.3 - 1} = \frac{574 \times (-184.39)}{0.3}$ $W_{1-2} = \frac{-105,839.86}{0.3} = -352,799.5\ \text{J} = -352.80\ \text{kJ}$ (Negative sign indicates work is done ON the air during compression).

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Example 2: Bernoulli Pipeline Flow, Friction Head Loss & Pump Power

Problem: Water ($\gamma = 9810\ \text{N/m}^3$) is pumped from an open intake reservoir (surface elevation $z_1 = 10\ \text{m}$) to an elevated storage tank (surface elevation $z_2 = 45\ \text{m}$) at a rate of $Q = 0.05\ \text{m}^3/\text{s}$ through a pipeline of diameter $D = 0.15\ \text{m}$ and length $L = 200\ \text{m}$. Total pipe friction head loss is calculated to be $h_f = 8.5\ \text{m}$. If the combined pump-motor overall efficiency is $\eta_{\text{overall}} = 75%$, calculate the total pump dynamic head $H_p$ required and the electrical power input $P_{\text{elec}}$.

Solution:

1. Apply extended Bernoulli equation between free water surfaces (where $P_1 = P_2 = P_{\text{atm}}$ and $v_1 = v_2 \approx 0$): $z_1 + H_p = z_2 + h_f \implies H_p = (z_2 - z_1) + h_f$ $H_p = (45 - 10) + 8.5 = 35 + 8.5 = 43.5\ \text{m}$
2. Calculate hydraulic power delivered to water ($P_{\text{hyd}}$): $P_{\text{hyd}} = \gamma Q H_p = 9810 \times 0.05 \times 43.5 = 21,336.75\ \text{W} = 21.34\ \text{kW}$
3. Calculate electrical power input required ($P_{\text{elec}}$): $P_{\text{elec}} = \frac{P_{\text{hyd}}}{\eta_{\text{overall}}} = \frac{21.34\ \text{kW}}{0.75} = 28.45\ \text{kW}$