3 ≈ 2475.03 – 625.42 ≈ 1850 kJ/kg.
qH = h1 – h4 ≈ 2748.79 – 640.16 ≈ 2109 kJ/kg.
wturbine = h1 – h2 ≈ 2748.79 – 2475.03 ≈ 274 kJ/kg.
Carnot: wcompressor = e4 – e3 ≈ 94.78 – 80.05 ≈ 15 kJ/kg, as above.
Rankine: wpump = e4 – e3 ≈ 36.06 – 35.63 ≈ 0.43 kJ/kg, as above.
Carnot: qL = e2 – e3 ≈ 475.19 – 80.05 ≈ 395 kJ/kg.
Rankine: qL = e2 – e3 ≈ 475.19 – 35.63 ≈ 440 kJ/kg.
Carnot: qH = e1 – e4 ≈ 748.95 – 94.78 ≈ 654 kJ/kg.
Rankine: qH = e1 – e4 ≈ 748.95 – 36.06 ≈ 713 kJ/kg.
Carnot and Rankine: wturbine = e1 – e2 ≈ 748.95 – 475.19 ≈ 274 kJ/kg, as above.
The thermal energy efficiency, i.e.
The thermal exergy efficiency, i.e.
is the most efficient process.
between the thermal energy and exergy efficiencies and why are they so poor?
then increases, i.e. more work out.
increase its temperature.
reach efficiencies of about 40%.
H1 – H2 = E1 – E2.
Let us see what this implies for the turbine and the compressor in the Carnot process.
processes. 1-2 gives maximum work, h1 – h2 and 1-2' gives the real work, h1 – h2', see Fig.
which usually is about 0.80-0.85.
which is usually about 0.80-0.85.
Let us now return to the Rankine cycle and see how it usually looks, see Fig. below.
4-1 Liquid water converts to steam at pressure PH, we add heat QH.
1-2 The steam expands adiabatically, i.e. ∆S = 0, to pressure PL, just below saturation.
Work is transferred through the turbine to an electric generator in a power plant.
2-3 The steam is condensed at pressure PL, we remove heat QL.
increases to PH. We feed the pump work Wp.
specific volume v = 0.001 m3/kg.
lower limit of the pressure.
increases, but the exergy efficiency is more or less the same.
plant. Let us just mention some of them in energy units.
  for big plants.
tt ≈ 0.5 for a good design.
m+el ≈ 0.9-0.95 for big plants.
isentropic line until it cross the pressure PL= 6 bar then we have state 2.
In addition to this we also have mechanical and electrical losses, i.e. m+el ≈ 0.90.95, so Pel in reality becomes even less.
heating need is 15 MW, t = 0.82 and el+m = 0.94.
at the real state after the turbine. c) Calculate the exergy output in the two cases.
after the turbine is h2'. If h3 is the state when all the steam is condensed, i.e.
how much steam that must be condensed per second.
Pel  Wel  m (h1  h2 ) tmel  7  (3.37  2.65)      3.9MW.
a) Cogeneration gives the energy power: 3.9 MW electricity and 15 MW heat.
b) The plant gives 7.1 MW electricity and the steam quality is 0.88.
a Carnot cycles going backwards. This is what happens in refrigerators and heat pumps.
will treat the compression process, since it is the most common.
1-2 The refrigerant is adiabatically, i.e. ∆s = 0, compressed to pressure PH.
at temperature TH.
takes up the heat QL at temperature TL.
In the Figs. below we see the process in a Ts and Ph-diagram.
energy is neither added nor removed. However, a lot of exergy is lost in the expansion.
The highest COP we get for a Carnot process.
N.B.! T is the absolute temperature, i.e. in Kelvin.
being used then COPheating = 1 + 0.4 to 0.6×COPcooling, Carnot.
A house is kept at 20˚C by a compressor heat pump a cold winter day (–20˚C).
system works ideally between –5˚C and 45˚C. b) What is then the COPheating,Carnot?
c) Estimate the COP in practice.
State 1, L = –5˚C, saturated gas  PL ≈ 2.61 bar & h1 ≈ 349 kJ/kg.
State 3, H = 45˚C, saturated liquid  PH ≈ 10.84 bar & h3 ≈ 244 kJ/kg.
to PH = 10.8 bar, and find h ≈ 376 kJ/ kg.
after all we did assume no losses.
at 15˚C in the condenser and d) the theoretically highest COP in b) and c).
homogenous body of rectangular shape.
  is the heat conduction [W/m˚C or W/mK] for the body.
Heat is here reagarded both as energy and exergy.
is a material constant. The higher the  the more heat is conducted through the material.
some numerical examples.
through the window, if we only consider heat conduction?
coefficient and indicated by the symbol U.
What is the exergy loss in the example above?
 0.075  7.5%.
hour is ridiculous!
becomes a loss, i.e. no exergy is being transferred.
energy flows and the exergy losses in the two cases?
times if the insulation is wet.
heat conduction is only of interest in gases and liquids if mass transfer is not aloud.
radiator is due to that hot air rises when its density decreases, i.e. the air becomes lighter.
manually “convect” the water, stir it. Thus, we differ between free and forced convection.
A wall of temperature TH is in contact with a gas (or liquid) of temperature TL, TH>TL.
the media has evolved.
and unit area at a temperature difference of 1 K [˚C] when we have convection.
Assume hot water surrounding a pipe with the diameter 55 mm and length 4 m.
Thus more than half of the exergy being transferred is lost in the process.
fluid, e.g. we experience more heat losses when it is windy.
For condensing water we have  ≈ 10000 W/m2K.
The area A becomes: A = dL ≈ 3.14×10×10–2×10 = 3.14 m2.
= 5.8 + 3.95v = 5.8 W/m2K, since v = 0 m/s.
then becomes extra cold.
defined by the color.
and a rough surface reflects the incoming radiation in all directions, as scattered light.
bulb. Usually four wings, like a water-wheel, one side black the other side white.
If you put it in the sunlight it starts to move, in which direction and why?
  is the emission coefficient or absorption coefficient, which is found in tables.
implies that when the temperature doubles the radiation increase by a factor of 16!
temperatures TH, TL and emission coefficients H, L, see below.
which is valid for all colored surfaces (black or white) and liquid surfaces.
have to simplify the problem to be able to solve it.
The person will probably freeze, which we also might guess.
temperature difference TH–TL, which is an important conclusion.
Thus, which form of heat is most ordered?
  .
heat conduction, e.g. in heat exchangers the U-coefficient should be large.
where Uair combine conduction, convection and radiation for each air layer.
A wall with the area 20 m2 has a mineral wool insulation with thickness 170 mm.
5˚C and  = 0.04 W/m˚C.
  , since we only have one layer.
