Power Plant Analysis (Vapour Cycles)

This is a topic from Thermodynamics:

Introduction

A common exam question which is a culmination of the material covered in the course is an analysis of the efficiency of the steam power plant, which is a form of vapour cycle. In vapour cycles the ideal gas laws do not apply because the fluid tends to change phase during the various processes which comprise the cycle such that it is not always a gas. There are many different vapour cycles which you may be asked to analyse, but two important cycles are covered in this section to exemplify the method of analysis.

The Basic Rankine Cycle

The Rankine cycle is as follows: There is work in during the pumping phase, heat in during the boiling phase ('vaporiser'), useful work out during the turbine phase ('expander') and heat out during the condensor phase. Recalling that thermal efficiency equals net work out divided by heat in, we need to consider the boiler heat and both the work in and work out to calculate efficiency.

The Rankine Regeneration Cycle

This cycle is a type of Rankine cycle where the mass flow rate into the turbine is split into two components, one which passes through the high pressure turbine and into the open feedwater heater (this is path 5→6) and one which passes through the low pressure turbine and is compressed and pumped back into the open feedwater heater (this is path 5→7). A proportion of ṁ equal to y follows path 6→7→3 and the remaining proportion equal to 1-y follows path 5→7→1→2→3. Analysis for the Ideal Case

• The thermal efficiency is defined as ηth = Wnet / Qin
• The heat and work flows are as follows:
1. The work out of the system (Ẇout) occurs during the expansion through the turbine.
2. The work into the system (Ẇin) occurs during the two pumping phases (Ẇ12 and Ẇ34).
3. The heat out of the system (Q̇out) occurs during the condensing phase.
4. The heat into the system (Q̇in) occurs during the boiling phase.

Step 1: Find the value of y

• Considering the first law of thermodynamics for the open feedwater heater:
Q̇ - Ẇ = ṁout(h + V2/2 + gz)out - ṁin(h + V2/2 + gz)in
0 = ṁ3h3 - ṁ2h2 - ṁ6h6...........(There is no work or heat flow in the open feedwater heater)
• Considering the proportions of the mass flow that follow each path:
0 = ṁh3 - ṁ(1-y)h2 - ṁ(y)h6
0 = h3 - (1-y)h2 - (y)h6
• Rearranging gives:
y = (h2 - h3) / (h2 - h6)

Step 2: Find the value of Ẇnet

• Considering work flow rates in and out:
net = Ẇout - Ẇin
= Ẇturbine - Ẇ12 - Ẇ34
• Applying the first law for Ẇturbine gives us:
net = ṁ(h5 - h6) - ṁ(1-y)(h6 - h7) - ṁ(1-y)w12 - ṁw34

Step 3: Find Q̇in

• Applying the first law for the boiler gives:
in = ṁ(h5 - h4)

Step 4: Substitute and solve for ηth

Analysis for the Case with Irreversibilities

If you are told that there are irreversibilities in one of the components you will need to apply the efficiency of these components and recalculate values of enthalpy. For example if the low pressure turbine segment has irreversibilities such that the turbine efficiency ηturbine = 0.9 you will have to recalculate the actual enthalpy after the expansion using the definition of turbine efficiency and resubstitute to find the new, slightly reduced ηth.

The Rankine Reheat Cycle

This cycle is a type of Rankine cycle where the fluid passes through the high pressure turbine, only to be reheated and pass through the low pressure turbine. There is one mass flow, but there are two processes of heat addition. Analysis for the Ideal Case

• The thermal efficiency is defined as ηth = Wnet / Qin
• The heat and work flows are as follows:
1. The work out of the system (Ẇout) occurs during the expansion through the turbine (Ẇ34 and Ẇ56). .
2. The work into the system (Ẇin) occurs during the one pumping phase (Ẇ12).
3. The heat out of the system (Q̇out) occurs during the condensing phase.
4. The heat into the system (Q̇in) occurs during the two boiling phases (Q̇23 and Q̇45).

Step 1: Find the value of Ẇnet

• Considering work flow rates in and out:
net = Ẇout - Ẇin
= Ẇturbine - Ẇpump
= Ẇ34 and Ẇ56 - Ẇ12
• Applying the first law for Ẇturbine gives us:
net = ṁ(h3 - h4) + ṁ(h5 - h6) - ṁw12

Step 2: Find Q̇in

• Applying the first law for the boiler gives:
in = Q̇23 + Q̇45
= ṁ(h3 - h2) and ṁ(h5 - h4)

Step 4: Substitute and solve for ηth

The case with irreversibilities is approached in the same way as it is with the Rankine regeneration cycle.