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The steady-state energy equation is essential in evaluating the performance of systems like turbines. It ensures that energy inputs and outputs remain balanced over time. For turbines, this equation becomes critical because it helps in determining how much work is produced and how much energy is lost. The general steady-state energy equation for an open system is: \[ h_{\text{in}} + \frac{V_{\text{in}}^2}{2} + q = h_{\text{out}} + \frac{V_{\text{out}}^2}{2} + w_s \] In this equation:

• \(h_{\text{in}}\): Enthalpy at the inlet
• \(h_{\text{out}}\): Enthalpy at the outlet
• \(V_{\text{in}}\): Velocity at the inlet
• \(V_{\text{out}}\): Velocity at the outlet
• \(q\): Heat transfer per unit mass
• \(w_s\): Specific work developed

By inputting the known values of enthalpies, velocities, and heat transfer into this equation, we can solve for the specific work developed by the turbine. This method assures that all inflows and outflows of energy are accurately accounted for.

Enthalpy is a measure of the total heat content in a thermodynamic system and plays a crucial role in energy analysis. It combines internal energy with the product of pressure and volume, allowing us to account for energy in terms of pressure and temperature. For air, enthalpy values can be derived from thermodynamic tables or ideal gas assumptions. In our turbine exercise:

• Inlet enthalpy: \(h_{\text{in}} = 158 \text{ BTU/lb}\) at \(75 \text{ lbf/in}^2\) and \(800^{\text{o}}R\)
• Outlet enthalpy: \(h_{\text{out}} = 124 \text{ BTU/lb}\) at \(15 \text{ lbf/in}^2\) and \(600^{\text{o}}R\)

Accurate data from these tables ensure that our energy balance calculations are precise, hence correctly determining the turbine's performance.

Exergy destruction quantifies the loss of potential work in a system due to irreversibilities such as friction, heat loss, and unrestrained expansion. Exergy destruction is critical because it measures efficiency loss. The exergy destruction can be calculated using the exergy balance equation: \[ \text{Exergy destruction} ( \text{B}) = T_0 \frac{q}{T_{\text{surface}}} - w_s \] Here:

• The terms are divided into positive contributions from surroundings heat (first term) and negative work output (second term).
• In our problem: \[ q = 2 \text{ Btu/lb} \] \[ T_{\text{surface}} = 620^{\text{o}}R \] \[ T_0 = 500^{\text{o}}R \]

By plugging the known values into the equation, we find the exergy destruction, helping us understand the inefficiencies within the turbine system.

Specific work is the work done by a turbine per unit mass of the working fluid (in this case, air). It is derived from the steady-state energy equation and represents the amount of mechanical energy generated by the turbine. The specific work is calculated from our energy balance equation: In the given exercise, specific work is calculated as follows: \[ h_{\text{in}} + \frac{V_{\text{in}}^2}{2} + q = h_{\text{out}} + \frac{V_{\text{out}}^2}{2} + w_s \] Plugging in the known values: \[ 158 + \frac{400^2}{2 \times 32.174} + 2 = 124 + \frac{100^2}{2 \times 32.174 } + w_s \] After simplifying, the specific work \(w_s\) was found to be \(19.38 \text{ Btu/lb} \). This value signifies how effectively the turbine converts heat and kinetic energy into mechanical work.