Loading... please wait.

Simulating a simple sub-critical ORC (v1.0)

Organic Rankine cycle calculation online!


Boundary condition parameters (units in bar and degrees C.)


Optimisation parameters (units in bar and degrees C.)


  Show/hide fluid info
Fire hazard (HMIS)
Health hazard (HMIS)
Physical hazard (HMIS)
Global warming potential (100y)
Ozone depletion potential
Critical pressure (bar)

  •   
  Calculating state points, please wait... If the calculation gets stuck please refresh the web page.
Calculation success!

Results

  Show/hide figures


Turbine power (kW)
Pump power (kW)
Cycle net power (kW)
Cycle efficiency (%)
m T P h s
1
2
3
4
5
6
10
11
12
13
m (kg/s), T (°C), P (bar), h (kJ/kg), s (kJ/kg-K)

Equations

Pressure and heat losses are neglected.
The results are based on the heat source being a flow of 1 kg/s of exhaust gas with a $c_p$=1.1 kJ/kg-K.
$P_4 = P_5 = P_6 = P_1$,
$P_3 = P(T_3, x=0)$,
$P_2 = P_3$,
$h_1 = h(T_1, P_1)$ (must always be a superheated state),
$s_1 = s(T_1, P_1)$,
$h_{2,s} = h(s_1, P_2)$,
$h_2 = h_1 - (h_1-h_{2,s}) \eta_e$,
$h_3 = h(P_3, x=0)$,
$s_3 = s(P_3, x=0)$,
$h_{4,s} = h(s_3, P_4)$,
$h_4 = h_3 + (h_{4,s}-h_3)/\eta_p$,
$h_5 = h(P_5, x=0)$,
$h_6 = h(P_6, x=1)$,
$\dot{m}_{10}=1$ (kg/s),
$h_{10}-h_{13} = c_p(T_{10}-T_{13})$,
$\dot{m}_1 = \dot{m}_{10}(h_{10}-h_{13})/(h_1-h_4)$,
$h_{11} = h_{10} - \dot{m}_1(h_1-h_6)$,
$h_{12} = h_{11} - \dot{m}_1(h_6-h_5)$,
$T_{11} = h_{11}/c_p$,
$T_{12} = h_{12}/c_p$,

Estimation of max. potential

Surprisingly, it turns out that one can estimate the maximum obtainable efficiency of an ORC (optimised with the best fluid and parameters), by using the four equations below here (Link).

For heat sources with an inlet temperature (state 10 in the figure at the top of this page) in the range of 180-360°C, the maximum efficiency for a recuperated ORC is approximately: $\eta_{th,max} = -12.76 + 0.06428 T_{hs,i} + 0.05897 T_{hs,o} + 0.2576 \eta_{p,e} - 0.1727 T_c - 0.1556 \Delta T_{pp}$

$T_{hs,i}$ is the heat source inlet temperature (10),
$T_{hs,o}$ is the heat source outlet temperature (13),
$\eta_{p,e}$ is the polytropic efficiency of the turbine or expander,
$T_c$ is the condensing temperature (3) and
$\Delta T_{pp}$ is the minimum pinch point temperature difference in the boiler and the recuperator. Temperatures are given in degrees Celsius and the efficiencies in percent.

In the same way we can estimate the maximum potential of a non-recuperated ORC: $\eta_{th,max} = -12.33 + 0.05858 T_{hs,i} + 0.03350 T_{hs,o} + 0.2666 \eta _{p,e} - 0.1552 T_c - 0.0810 \Delta T_{pp}$

For heat sources with an inlet temperature in the range of 80-180°C, the maximum efficiency for a recuperated ORC is approximately:
$\eta_{th,max} = -16.32 + 0.08402 T_{hs,i} + 0.08349 T_{hs,o} + 0.1583 \eta _{p,e}$

Note that the equation is only valid when the condensing temperature is 25°C and the minimum pinch point temperature difference is 5°C.

In the same way we can estimate the maximum potential of a non-recuperated ORC in the range of 80-180°C:
$\eta_{th,max} = -14.92 + 0.07339 T_{hs,i} + 0.08363 T_{hs,o} + 0.1464 \eta _{p,e}$

It is the aim of these equations to provide enable the straight-foward estimation of the theoretical maximum potential of an ORC given the design conditions. One can immediately investigate the effect of varying key design parameters, for example the condensing temperature or the turbine efficiency. For more details see here and here.

Calculator for a recuperated ORC, heat source inlet 180-360°C


Max. achievable thermal efficiency: %

Plots are powered by jqplot
Equations formatting powered by mathjax
The site is made by Ulrik Larsen 2015 and is using Coolprop to estimate the fluid properties.
The calculations were checked using the commercial process simulation tool Aspen Plus 8.2.