3.2 First Law Analysis for Open & Closed Thermal Systems

3.2 First Law Analysis for Open & Closed Thermal Systems

The First Law of Thermodynamics is the statement of the principle of conservation of energy. It states that energy can neither be created nor destroyed during a process; it can only change forms. For mechanical engineers analyzing power generation systems, HVAC units, internal combustion engines, and industrial fluid networks, mastering first-law energy balances for closed systems and open control volumes is fundamental.

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1. First Law Formulation for Closed Systems

A closed system (or control mass) consists of a fixed amount of mass, and no mass can cross its boundary. Energy can cross the boundary in two forms: Heat ($Q$) and Work ($W$).

The general First Law balance over a time interval or process from state 1 to state 2 is:

$$Q_{\text{net,in}} - W_{\text{net,out}} = \Delta E_{\text{system}}$$ $$\Delta E_{\text{system}} = \Delta U + \Delta KE + \Delta PE$$

where:

• $\Delta U = m(u_2 - u_1)$ = Change in internal energy
• $\Delta KE = \frac{1}{2} m (\mathbf{V}_2^2 - \mathbf{V}_1^2)$ = Change in kinetic energy
• $\Delta PE = m g (z_2 - z_1)$ = Change in potential energy

For a stationary closed system, changes in kinetic and potential energy are negligible ($\Delta KE = 0, \Delta PE = 0$). Thus, the energy balance simplifies to:

$$Q - W = \Delta U = m (u_2 - u_1)$$

Sign Conventions (Standard Engineering Convention):

• Heat added to the system: $Q > 0$ (+)
• Heat rejected from the system: $Q < 0$ (-)
• Work done by the system: $W > 0$ (+)
• Work done on the system: $W < 0$ (-)

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2. Boundary Work ($P dV$ Work) for Closed System Processes

Boundary work ($W_b$) is the energy transferred when the boundary of a closed system moves during a quasi-equilibrium expansion or compression process:

$$W_b = \int_1^2 P \, dV$$

On a $P-V$ diagram, boundary work equals the area under the process curve.

   P ^
     |   1 (P1, V1)
     |    \   P V^n = C
     |     \  [Work = Area under curve]
     |------\-------> 2 (P2, V2)
     +------------------------> V

Fundamental Process Work Equations for Ideal Gases ($P V = m R T$)

1. Isobaric Process ($P = \text{constant}$)

$$W_b = P \int_1^2 dV = P(V_2 - V_1) = m R (T_2 - T_1)$$ $$Q = \Delta U + W_b = m c_v (T_2 - T_1) + m R (T_2 - T_1) = m c_p (T_2 - T_1) = \Delta H$$

2. Isochoric (Isovolumetric) Process ($V = \text{constant}$)

$$W_b = 0$$ $$Q = \Delta U = m c_v (T_2 - T_1)$$

3. Isothermal Process ($T = \text{constant}$)

For an ideal gas, $P V = C \implies P = C/V$:

$$W_b = \int_1^2 \frac{C}{V} \, dV = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) = m R T \ln\left(\frac{P_1}{P_2}\right)$$

Since $T = \text{constant}$, $\Delta U = 0$; thus $Q = W_b$.

4. Polytropic Process ($P V^n = C$, $n \ne 1$)

$$W_b = \int_1^2 C V^{-n} \, dV = \frac{P_2 V_2 - P_1 V_1}{1 - n} = \frac{P_1 V_1 - P_2 V_2}{n - 1} = \frac{m R (T_1 - T_2)}{n - 1}$$

5. Isentropic (Reversible Adiabatic) Process ($P V^k = C$, $n = k = c_p/c_v$)

$$W_b = \frac{P_1 V_1 - P_2 V_2}{k - 1} = \frac{m R (T_1 - T_2)}{k - 1}$$

Since $Q = 0$, $W_b = -\Delta U = m c_v (T_1 - T_2)$.

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3. Specific Heats, Internal Energy & Enthalpy

Specific heats relate temperature changes to energy changes:

• Constant-volume specific heat: $c_v = \left(\frac{\partial u}{\partial T}\right)_v \implies du = c_v dT$
• Constant-pressure specific heat: $c_p = \left(\frac{\partial h}{\partial T}\right)_P \implies dh = c_p dT$

For ideal gases, $u$ and $h$ depend only on temperature $T$ (Joule's Law). Hence:

$$\Delta u = u_2 - u_1 = c_v (T_2 - T_1), \quad \Delta h = h_2 - h_1 = c_p (T_2 - T_1)$$

Mayer's Relation & Specific Heat Ratio

$$c_p - c_v = R, \quad k = \frac{c_p}{c_v}$$ $$c_v = \frac{R}{k - 1}, \quad c_p = \frac{k R}{k - 1}$$

For Air ($R = 0.2870\text{ kJ/kg}\cdot\text{K}, k = 1.4$):

$$c_v = 0.718\text{ kJ/kg}\cdot\text{K}, \quad c_p = 1.005\text{ kJ/kg}\cdot\text{K}$$

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4. First Law for Open Systems: Steady-Flow Energy Equation (SFEE)

An open system (control volume) involves mass flow across its boundary. Under steady-state steady-flow (SSSF) conditions, mass inside the control volume remains constant ($\,d m_{\text{cv}}/dt = 0$) and total energy inside remains constant ($\,d E_{\text{cv}}/dt = 0$).

Conservation of Mass

$$\dot{m}_1 = \dot{m}_2 = \dot{m} = \frac{A_1 \mathbf{V}_1}{v_1} = \frac{A_2 \mathbf{V}_2}{v_2}$$

Steady-Flow Energy Equation (SFEE) per unit mass (kJ/kg)

$$q - w = (h_2 - h_1) + \frac{\mathbf{V}_2^2 - \mathbf{V}_1^2}{2000} + \frac{g(z_2 - z_1)}{1000}$$

Rate Form of SFEE (kW or MW)

$$\dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{\mathbf{V}_2^2 - \mathbf{V}_1^2}{2000} + \frac{g(z_2 - z_1)}{1000} \right]$$

where $\mathbf{V}$ is fluid velocity (m/s), $z$ is elevation (m), $g = 9.81\text{ m/s}^2$, $\dot{Q}$ is heat transfer rate (kW), and $\dot{W}$ is power output/input (kW).

SFEE Applications to Engineering Control Volume Devices

1. Nozzles & Diffusers:

   • Devices that accelerate or decelerate a fluid. $\dot{W} = 0$, $\dot{Q} \approx 0$, $\Delta PE = 0$.
   $$h_1 + \frac{\mathbf{V}_1^2}{2000} = h_2 + \frac{\mathbf{V}_2^2}{2000}$$

2. Turbines (Steam / Gas / Hydraulic):

   • Devices that produce shaft power. $\dot{Q} \approx 0$, $\Delta KE \approx 0$, $\Delta PE \approx 0$.
   $$w_{\text{out}} = h_1 - h_2 \implies \dot{W}_{\text{out}} = \dot{m} (h_1 - h_2)$$

3. Compressors, Pumps & Fans:

   • Devices that increase pressure using shaft work. $\dot{Q} \approx 0$, $\Delta KE \approx 0$, $\Delta PE \approx 0$.
   $$w_{\text{in}} = h_2 - h_1 \implies \dot{W}_{\text{in}} = \dot{m} (h_2 - h_1)$$

4. Throttling Valves & Expansion Tubes:

   • Flow-restricting devices causing a large pressure drop. $\dot{W} = 0$, $\dot{Q} \approx 0$, $\Delta KE \approx 0$, $\Delta PE = 0$.
   $$h_1 = h_2 \quad (\text{Isenthalpic Process})$$

5. Heat Exchangers & Condensers:

   • Devices transferring thermal energy between fluid streams without work. $\dot{W} = 0$, $\Delta KE \approx 0$, $\Delta PE \approx 0$.
   $$\dot{m}_A (h_{A,1} - h_{A,2}) = \dot{m}_B (h_{B,2} - h_{B,1})$$

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5. Step-by-Step Worked Numerical Calculation

Problem Statement

Steam enters an adiabatic steam turbine operating at steady state at $P_1 = 3.0\text{ MPa}$ and $T_1 = 400.0^\circ\text{C}$ with an inlet enthalpy of $h_1 = 3230.9\text{ kJ/kg}$ and an inlet velocity of $\mathbf{V}_1 = 50.0\text{ m/s}$. The steam exits the turbine at $P_2 = 30.0\text{ kPa}$ with a quality of $x_2 = 0.920$ and an exit velocity of $\mathbf{V}_2 = 180.0\text{ m/s}$. The mass flow rate of steam is $\dot{m} = 12.0\text{ kg/s}$. Potential energy change is negligible.

Saturated steam data at $P_2 = 30.0\text{ kPa}$:

• $h_f = 289.23\text{ kJ/kg}$
• $h_{fg} = 2336.1\text{ kJ/kg}$

Calculate:

1. The exit enthalpy $h_2$ of steam in kJ/kg.
2. The change in kinetic energy $\Delta ke$ per unit mass in kJ/kg.
3. The work produced per unit mass $w_{\text{out}}$ in kJ/kg.
4. The total shaft power output $\dot{W}_{\text{out}}$ of the turbine in Megawatts (MW).

Step-by-Step Solution

Step 1: Calculate exit enthalpy ($h_2$)

$$h_2 = h_f + x_2 h_{fg} = 289.23 + (0.920 \times 2336.1) = 289.23 + 2149.21 = 2438.44\text{ kJ/kg}$$

Step 2: Calculate change in kinetic energy ($\Delta ke$)

$$\Delta ke = \frac{\mathbf{V}_2^2 - \mathbf{V}_1^2}{2000} = \frac{(180.0)^2 - (50.0)^2}{2000} = \frac{32400 - 2500}{2000} = \frac{29900}{2000} = 14.95\text{ kJ/kg}$$

Step 3: Apply SFEE to calculate work output per unit mass ($w_{\text{out}}$) For an adiabatic turbine ($\dot{Q} = 0$) with negligible potential energy change ($\Delta pe = 0$):

$$q - w_{\text{out}} = (h_2 - h_1) + \Delta ke$$ $$0 - w_{\text{out}} = (2438.44 - 3230.9) + 14.95$$ $$-w_{\text{out}} = -792.46 + 14.95 = -777.51\text{ kJ/kg}$$ $$w_{\text{out}} = 777.51\text{ kJ/kg}$$

Step 4: Calculate total turbine power output ($\dot{W}_{\text{out}}$)

$$\dot{W}_{\text{out}} = \dot{m} w_{\text{out}} = 12.0\text{ kg/s} \times 777.51\text{ kJ/kg} = 9330.12\text{ kW}$$

Convert to Megawatts (MW):

$$\dot{W}_{\text{out}} = \frac{9330.12\text{ kW}}{1000\text{ kW/MW}} = 9.330\text{ MW}$$

Final Summary of Results:

• Exit enthalpy $h_2 = 2438.44\text{ kJ/kg}$
• Kinetic energy change $\Delta ke = +14.95\text{ kJ/kg}$
• Work per unit mass $w_{\text{out}} = 777.51\text{ kJ/kg}$
• Turbine Power Output $\dot{W}_{\text{out}} = 9.330\text{ MW}$