A steam turbine operates under steady flow conditions, receiving steamat 1.2MPa and 1880C, and enthalpy 2785kJ/kg, velocity 33.3m/s andelevation 3m. Steam leaves the turbine at 20 kPa, and 2512kJ/kg, velocity100m/s and elevation zero m. Heat is lost to surroundings at the rate of0.29kJ/s. If the rate of steam flow through the turbine is 0.42kg/s, what isthe power output of the turbine?
Question
A steam turbine operates under steady flow conditions, receiving steamat 1.2MPa and 1880C, and enthalpy 2785kJ/kg, velocity 33.3m/s andelevation 3m. Steam leaves the turbine at 20 kPa, and 2512kJ/kg, velocity100m/s and elevation zero m. Heat is lost to surroundings at the rate of0.29kJ/s. If the rate of steam flow through the turbine is 0.42kg/s, what isthe power output of the turbine?
Solution
To solve this problem, we need to apply the first law of thermodynamics for open systems, which states that the change in total energy of the system is equal to the energy added to the system minus the energy leaving the system.
The total energy of the system includes the internal energy (enthalpy), kinetic energy, and potential energy.
The energy added to the system is the heat added and the work done on the system. The energy leaving the system is the heat lost and the work done by the system.
In this case, the work done by the system is the power output of the turbine, which we are trying to find.
Step 1: Calculate the total energy of the steam entering and leaving the turbine.
The total energy of the steam entering the turbine is the sum of its enthalpy, kinetic energy, and potential energy.
Enthalpy (H1) = 2785 kJ/kg Kinetic energy (KE1) = 0.5 * (33.3 m/s)^2 * 1 kJ/kgm^2/s^2 = 0.555 kJ/kg Potential energy (PE1) = 9.81 m/s^2 * 3 m * 1 kJ/kgm^2/s^2 = 0.029 kJ/kg
Total energy entering (E1) = H1 + KE1 + PE1 = 2785.584 kJ/kg
The total energy of the steam leaving the turbine is also the sum of its enthalpy, kinetic energy, and potential energy.
Enthalpy (H2) = 2512 kJ/kg Kinetic energy (KE2) = 0.5 * (100 m/s)^2 * 1 kJ/kgm^2/s^2 = 5 kJ/kg Potential energy (PE2) = 9.81 m/s^2 * 0 m * 1 kJ/kgm^2/s^2 = 0 kJ/kg
Total energy leaving (E2) = H2 + KE2 + PE2 = 2517 kJ/kg
Step 2: Apply the first law of thermodynamics.
The change in total energy of the system (ΔE) is equal to the energy added to the system minus the energy leaving the system.
ΔE = E1 - E2 = 2785.584 kJ/kg - 2517 kJ/kg = 268.584 kJ/kg
The energy added to the system is the heat added (Q), which is zero since no heat is added. The energy leaving the system is the heat lost (Q_out) and the work done by the system (W).
ΔE = Q - W - Q_out 268.584 kJ/kg = 0 - W - 0.29 kJ/s
Step 3: Solve for the work done by the system (W), which is the power output of the turbine.
W = -268.584 kJ/kg + 0.29 kJ/s = -268.294 kJ/kg
Since the rate of steam flow through the turbine is 0.42 kg/s, the power output of the turbine is W * 0.42 kg/s = -112.6838 kW.
Therefore, the power output of the turbine is -112.6838 kW. The negative sign indicates that the work is done by the system, which is consistent with the convention that work done by the system is negative.
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