Thermodynamics Question (Power Output)

[WA2000]

This is not really homework, it's just something I don't understand and I will be assessed on it soon, so I need some help on it. First off our teacher is the worst teacher you will ever encounter, he gives us the solution but doesn't tell us how (tells us to figure it out ourselves). I tried looking at books but all of them are for advanced thermodynamics (Our thermodynamics is introductary, it's all in pre-calculas level), I really don't understand it and I really need help on this. I'm in semester 3 in the Mechanical Engineering Diploma program at the Australian College in Kuwait.

1.
A steam turbine takes in steam with specific enthalpy 3095 kJ/kgm at a rate of 80 kg/min. The steam leaves the turbine with specific enthalpy 2660 kJ/kg. If heat losses from the turbine are 120 kW, Determine the power output. (Assume that kinetic-energy and potential energy changes are negligible).
All he has given us so far are these equations:
Q-W =H2-H1
Q-W= U2-U1
H= U+pV
H2-H1 =mc(T2-T1)

3. I know for a fact that the answer is 460 kW, But how do you get the answer, can you explain it please?

Any help would be greatly appreciated.

 

[Valkarie]

The equation you need to solve this problem is the energy balance equation: Q - W = H₂ - H₁where Q is the heat input, W is the work output, H₂ is the specific enthalpy of the steam leaving the turbine, and H₁ is the specific enthalpy of the steam entering the turbine.We can also use the ideal gas law to convert the specific enthalpy into temperature. The equation for this is: H = U + pVwhere U is the internal energy, p is the pressure, and V is the specific volume.Using these equations, we can solve for the work output of the turbine: W = Q - (H₂ - H₁) = Q - [mc(T₂ - T₁)] = Q - m[(U₂ + pV₂) - (U₁ + pV₁)] = Q - m[U₂ + pV₂ - U₁ - pV₁]Where m is the mass flow rate of the steam and c is the specific heat capacity of the steam.We know that the heat input to the turbine is 120 kW, and the mass flow rate of the steam is 80 kg/min. We also know from the problem statement that the specific enthalpy of the steam entering and leaving the turbine are 3095 kJ/kgm and 2660 kJ/kgm respectively.Plugging all the known values into the equation, we get: W = 120 kW - 80 kg/min * [(3095 kJ/kgm - 2660 kJ/kgm) - (U₂ + pV₂ - U₁ - pV₁)] = 120 kW - 80 kg/min * (435 kJ/kgm - (U₂ + pV₂ - U₁ - pV₁))We can assume that U₂, U₁, pV₂, and pV₁ are negligible compared to the 435 kJ/kgm, so we can simplify the equation to