Thermal efficiency formula

The thermal efficiency formula itself is related to the order index "entropy change" (represented by simplified S), that is,

ηs=A/Q= 1 -(T2/T 1) editing is not standard.

= 1 -(T2/Q 1)S ⑷

If the microscopic particles in the heat engine move in order and develop (do work) according to the macro order, that is, the entropy S→0, then (T2/Q 1)S→0.

ηs→ 1

If the movement of microscopic particles is disordered, 0 ≤η

If q in equation (4) is expressed by the total energy that the system can do work, that is,

Q=3PV or Q=U=3PV.

Thermal efficiency of traditional heat engine

η0=A/Q=PV/3PV

= 1/3

It is a limit of the efficiency of the traditional heat engine, which is the fundamental reason why the efficiency of the traditional heat engine is not easy to improve.

When the micro-motion is orderly, A=3PV can be known from υ and υ, so the efficiency of the new ordered power machine is improved.

ηs=A/Q=3PV/3PV

= 1

Obviously, the efficiency of "thermal" engine can reach or tend to the ideal value of 100%.

Extended data:

Ways to improve efficiency

The efficiency η of an energy substance or engine can be expressed as the ratio of work W or A to energy E or heat Q, that is

η= W/E = A/E

E=Q+W=PE+( 1-P)E, W=A=( 1-P)E, then

η= 1-p = 1-wi/ω= q⒁

or

η= 1-lnW/lnω=-lnP/lnω⒂

= 1s/klnω⒃

Through statistical entropy s = k'-'b'! lnW，p = w/ω。

W=EXP(S/k`-`B `！ `)

P=EXP(S/k`-`B `！ `)/Ω

Efficiency can also be expressed by entropy.

η= 1-EXP(S/k`-`B `！ `)/Ω ⒄

Substitute P=2/3 into the formula [13], and you will get the sum of η = 1-q `-` 2'! `/Q`-` 1 `！ ` = 1/3 Same result

η= 1-P = 1-2/3 = 1/3

That is, the efficiency limit of single-stage disordered heat engine is 1/3. For multi-stage heat engines, the total energy Ei+ 1 of the latter stage heat engine is the heat Qi discharged by the former stage heat engine, ei+1= qi; Its efficiency is 1/3 of that of the previous engine, ηi+ 1=ηi( 1/3), so the compound efficiency of the N-class engine.

ηn =∑ηI

For an N-class heat engine with ηi= 1/3, the limit of its compound efficiency.

limηn = lim∑( 1/3)n = 1/2

n→∞ n→∞

Only when P=0 and η= 1-P= 1, the efficiency of the engine is 100%, which is the efficiency of a single-stage engine.

If multi-stage engines are used, in order to achieve the engine efficiency of 1, only the efficiency of each single-stage engine, that is, the order is P= 1/2,

limηn=lim∑( 1/2)n= 1

Find a solution

If the efficiency of 100% can be achieved by using only a limited number of engines, then the efficiency or order p required by a single-stage engine can be derived by using the compound efficiency formula and the summation formula of its proportional series S = A [(1-qn)/(1-q)]. Generally, there should be a=q=η, S= 1. If only two-stage engines are used, that is, n=2, the efficiency of the unit will tend to 100%, then the formula of s = a [(1-Q2)/(1-q)] is as follows.

η2+ η - 1 = 0

` .` Solving

η 1=-( 1+5 1/2)/2

η2=(5 1/2- 1)/2

Because η≯ 1, ┪ 0, ┪1=-(1+51/2)/2 was discarded, η = (51/2-/kloc-0. That is, only the single-stage efficiency η=(5 1/2- 1)/2 or p = 1-η=(3-5 1/2)/2 can make the comprehensive efficiency of a two-stage orderly engine reach 100%.

The incomplete order of this combination is much smaller than the complete order P = (3-5 1)/2, so it is easier to realize than P= 1. Other stages of the engine can also imitate this treatment, and its single-stage efficiency is usually (3-5 1/2)/2.