Question

Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at \(1 \mathrm{MPa}\) and \(1500 \mathrm{K}\) at a rate of \(0.1 \mathrm{kg} / \mathrm{s}\), and exit at \(0.2 \mathrm{MPa}\) and \(900 \mathrm{K} .\) If heat is lost from the turbine to the surroundings at a rate of \(15 \mathrm{kJ} / \mathrm{s}\), the power output of the gas turbine is \((a) 15 \mathrm{kW}\) (b) \(30 \mathrm{kW}\) \((c) 45 \mathrm{kW}\) \((d) 60 \mathrm{kW}\) \((e) 75 \mathrm{kW}\)

Short answer

Answer: The power output of the gas turbine is approximately 75 kW.

Step-by-step solution

Write down the given information

We are given the following information: - Inlet state: \(P_1 = 1 \,\text{MPa}\) and \(T_1 = 1500 \,\text{K}\) - Outlet state: \(P_2 = 0.2 \,\text{MPa}\) and \(T_2 = 900 \,\text{K}\) - Mass flow rate: \(m = 0.1 \,\text{kg/s}\) - Heat loss rate: \(Q = 15 \,\text{kJ/s}\)

Apply the first law of thermodynamics for an open system

We will use the following form of the first law for an open system in steady-state condition with no work inputs: \(W = m \cdot (h_1 - h_2) - Q\) Where \(W\) is the power output, \(m\) is the mass flow rate, \(h_1\) and \(h_2\) are the specific enthalpy at the inlet and outlet states, and \(Q\) is the heat loss rate. However, we do not have the specific enthalpy values. We will use the specific heat capacity at constant pressure \(c_p\) instead, as it relates to specific enthalpy with: \(h_2 - h_1 = c_p \cdot (T_2 - T_1)\)

Calculate the specific heat capacity of air at room temperature

We assume the hot combustion gases have the properties of air at room temperature. The specific heat capacity of air at constant pressure \(c_p\) is approximately \(1.005 \, \text{kJ/(kg·K)}\).

Calculate the power output

Now, we will plug in the given values and calculate the power output \(W\): \(W = m \cdot (h_1 - h_2) - Q = m \cdot c_p \cdot (T_2 - T_1) - Q\) \(W = 0.1 \,\text{kg/s} \cdot 1.005 \,\text{kJ/(kg·K)} \cdot (900 \,\text{K} - 1500 \,\text{K}) - 15 \,\text{kJ/s}\) \(W= 0.1 \cdot 1.005\cdot(-600)-15\) \(W= -60.3 - 15\) \(W = -75.3 \,\text{kW}\) Since this value is negative, the power output of the gas turbine is actually \(75.3 \,\text{kW}\) (positive value) as the work is being done by the turbine (output).

Choose the correct answer

Among the given options, the closest value to our calculated power output of \(75.3 \,\text{kW}\) is: \((e) 75 \mathrm{kW}\) Hence, the correct answer is option \((e)\).

Key concepts

First Law of Thermodynamics

Understanding the first law of thermodynamics is crucial when analyzing energy systems, such as gas turbines. This principle, often described as the conservation of energy, states that energy cannot be created or destroyed in an isolated system. It can only change from one form to another.

In the context of a gas turbine, the first law is applied to a control volume that includes the turbine. Energy enters the system in the form of heat from the combustion of fuel and exits both as mechanical work output and heat lost to the surroundings. Mathematically, we express this using the formula \(W = m \times (h_1 - h_2) - Q\), where \(W\) is the power output, \(m\) the mass flow rate, \(h_1\) and \(h_2\) are the specific enthalpies at the inlet and outlet respectively, and \(Q\) is the rate of heat loss to the surroundings.

The application of the first law allows us to calculate the energy output of gas turbines and other thermodynamic systems, ensuring that the energy balance is maintained. It's important to note that the specific enthalpies, \(h_1\) and \(h_2\), are linked to the state of the gas—its pressure and temperature—at those two points in the system's flow.

Specific Enthalpy

Specific enthalpy, denoted as \(h\), is a property of a substance that measures the total heat content per unit mass. It is a concept often used in thermodynamics to bridge the gap between heat transfer and work done in a system. For a gas turbine, the specific enthalpy change between the inlet and outlet—\(h_2 - h_1\)—tells us how much energy is transferred out of the system as work and heat.

To calculate specific enthalpy without direct measurements, we rely on the specific heat capacity of a substance, which measures how much energy is required to raise the temperature of a unit mass by one degree. For air or combustion gases in a gas turbine, the formula \(h_2 - h_1 = c_p \times (T_2 - T_1)\) uses the specific heat capacity at constant pressure \(c_p\) to determine the change in specific enthalpy based on inlet and outlet temperatures, \(T_1\) and \(T_2\) respectively.

Specific enthalpy is a key parameter in the energy balance equations for thermodynamic cycles. It enables engineers to predict the performance and efficiency of systems such as gas turbines by taking into account both the sensible and latent heat changes, especially in processes like heating, cooling, or phase changes.

Specific Heat Capacity

Specific heat capacity, symbolized as \(c_p\) when measured at constant pressure, involves the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It's a property intrinsic to the material and determines how temperature changes affect the energy stored in a substance.

In the situation of a gas turbine, knowing the specific heat capacity of air (or combustion gases) at room temperature, which is approximately \(1.005 \, \text{kJ/(kg\cdot K)}\), allows us to link temperature changes to heat energy changes. This relationship is vital for step 3 of the solution process where we calculate the power output of the turbine using the specific heat capacity:\[W = m \cdot c_p \cdot (T_1 - T_2) - Q\]

The correct understanding of specific heat capacity dictates the energy efficiency of a gas turbine and also has a direct influence on the specific enthalpy. This parameter is central to the design and operational control of thermal systems, playing a significant role in predicting how systems respond to thermal inputs and losses.