91Ó°ÊÓ

To understand the energy balance equation in the context of an ideal gas turbine expansion, start with Equation: \( W_{dot} = \dot{m} (h_1 - h_2 + \frac{V_2^2}{2}) \). Here's what each term represents:
- \( W_{dot} \): Power output of the turbine (in Watts).
- \( \dot{m} \): Mass flow rate of the gas (in kg/s).
- \( h_1 - h_2 \): Change in specific enthalpy (in J/kg).
- \( \frac{V_2^2}{2} \): Kinetic energy of the gas at the exit velocity (in J/kg).
This equation simplifies if you assume negligible initial velocity and no heat transfer or potential energy changes. It's a key tool for determining one unknown when others are known. Lower initial velocities simplify the problem by reducing terms, making it easier to calculate mass flow rates and other parameters.

In turbine expansion problems, the mass flow rate (\( \dot{m} \)) tells us how much gas is flowing through the system per second. We calculate this using the energy balance equation.
Substitute known values into \( W_{dot} = \dot{m} (h_1 - h_2 + \frac{V_2^2}{2}) \).
For our case, we substituted:
- \( W_{dot} = 2500000 \) W (or 2500 kW).
- Change in enthalpy \( h_1 - h_2 = 514650 \) J/kg.
- Exit velocity squared term is \( \frac{90^2}{2} = 4050 \) J/kg.
Plugging these in gives:
\[ 2500000 = \dot{m} (514650 + 4050) \dot{m} \ = \frac{2500000}{518700} \approx \ 4.82 \frac{kg}{s}\].
This means roughly 4.82 kg of air flows through the turbine per second.

The exit area (\( A_2 \)) in gas turbine problems denotes the cross-sectional area through which the gas exits. It's found using the continuity equation: \( \dot{m} = \rho_2 A_2 V_2 \).
- \( \dot{m} \) is the mass flow rate.
- \( \rho_2 \) is the gas density at the exit.
- \( A_2 \) is the exit area.
- \( V_2 \) is the exit velocity.
First, find the density and use the ideal gas law, \( \rho_2 = \frac{P_2}{R T_2} \). Given:
- Exit pressure \( P_2 = 100000 \) Pa.
- Gas constant for air \( R = 287 \) J/kg·K.
- Exit temperature \( T_2 = 450 \) K.
Calculate density:
\[ \frac{100000}{287 \cdot 450} = 0.77 \frac{kg}{m^3} \].
Then solve for \( A_2 \):
\[ 4.82 = 0.77 \cdot A_2 \cdot 90 \approx 0.07 \frac{m^2} \].
This shows the exit area is approximately 0.07 square meters.

Specific enthalpy (\( h \)) measures the total energy content per unit mass of a substance. In gas turbine problems, we often need the specific enthalpy change (\( \Delta h = h_1 - h_2 \)). For ideal gases, use:
- \( h_2 - h_1 = c_p (T_2 - T_1) \).
- \( c_p \) is the specific heat at constant pressure.
Given air's specific heat, \( c_p \approx 1.005 \frac{kJ}{kg·K} \).
Substitute initial and final temperatures (\( T_1 \) and \( T_2 \)). For our example:
- \( T_1 = 960 \) K.
- \( T_2 = 450 \) K.
We get:
\[ h_1 - h_2 = 1.005 \cdot (960 - 450) = 514.65 \frac{kJ}{kg} \approx 514650 \frac{J}{kg} \].
This signifies a drop in specific enthalpy by 514650 J/kg from the initial to the exit state. This drop is crucial for energy balance and determining mass flow rate.