This tutorial is set by the author's setting without permission.

First check the data of the GM Chemical Energy engine and learn about the burning value of the rocket fuel:

The burning value of the Rocket fuel is not completed at present

Typeburnup Valuetrue Fuel 4500

Purple fuel 61445769.905 Unliked partial two -meta hybrid fuel 9000

Green Fuel 12588

Burning agent-147.39null-147.39

Discussion here: Whether the cricket for hydrogen peroxide should be distilled tower to treat coal coke, or the distillation room.TRUE VALUE is the reduction of the cost of making the production. The production route diagram has not yet been drawn. There is only a process. The combustion agent is made of nitrogen+nitric oxide+hydrogen peroxide lines.

The distilled tower treatment coal coke is 54000EU to produce 10MB of coal coke to produce 50MB 蒽 and 400MB of kerosene and ethylene, stone brain oil, coal taron (although it will not affect the process)

The distillation room treatment coal coke is 27000EU to treat 1B coal coke into 50MB 蒽.

If the distillation tower, consider the problem of kerosene, whether to make RP-1 fuel, and there is also a problem with densely mixed fuel production process. Now discuss the first case, only use the distillery room to ensure that the production is as many as possible.Make hydrogen peroxide.The second case will be discussed in the future.

The main processes of combustion agents are::

Burning agent (this is not considering the use of RP-1 fuel, coal coke distillation room treatment)

(24000EU MV is greatly anticipated)

1B nitric oxide+1B hydrogen peroxide+2 potassium nitrate = 2B combustion agent

122880EU (Agame Anti -HV)

1B nitrogen+3b hydrogen = 1B ammonia gas

2400EU (LV)

1B ammonia qi+2.5B oxygen = 1B nitric oxide+1.5B water

4800EU (LV)

1B nitric oxide+1B oxygen = 1B nitrogen dioxide

(240000EU)

15B air+2B hydrogen+0.05B ​​蒽 = 2B peroxide hydrogen peroxide

14400EU+27000EU

2 coke coal = 0.05B ​​蒽

281,400EU

15B air+2B hydrogen+2 coking coal = 2B hydrogen peroxide

130,080EU

1B nitrogen+3b hydrogen+3.5B oxygen = 1B nitric oxide+1.5B water

294,780EU

7.5B air+4B hydrogen+1 coking coal+1B nitrogen+3.5B oxygen+2 potassium nitrate = 1.5B water+2B combustion agent

If the calculation is obtained, the theoretical consumption per MB combustion agent is 147.39eu, which may actually be slightly higher

Here we discuss the problem of dense rocket fuel power generation

The densely rocket fuel production process is paired with flat

32B hydrogen+6 car carbon powder+6B oxygen+15B air+2 coking coal+2B nitrogen =

10B dense 肼 hybrid fuel

813,960EU

The calculation formula for fuel efficiency is now given (only one kind of fuel, and the 蒽 required for the hydrogen peroxide of the combustion agent is made by the distilled room)

The current fuel flow rate is XMB/T. The flow rate of the combustion agent is YMB/T fuel panel combustion value is Q, the real combustion value is Q1, and 147.39eu/MB required for the burning agent

Then the efficiency of the actual power generation of fuel f (x, y) = [1.5*x*e^(-0.005*x/y)*q1-147.39*y]/(x*Q)

Take densely dense rocket fuel as an example

The dual function and partial guidance written by written

Let f (x, y) conducts 0 on the y to y, and find extreme values.

The relationship between the income is like this

It can be approximated as a straight line y = 5x/13

Bring back the dual function calculation

It is found that this function has nothing to do with X, and the maximum value is 1.42294

Explain that the maximum efficiency value (only integrian) is 142%. At this time, the intensive 肼 hybrid fuel of the general chemical energy engine: combustion agent = 13: 5

In practical applications, the amount of combustion agent can be 10MB/T

At this time, intensive hybrid fuel 26MB/T

Instant output to 113,653eu/T

Obviously, ZPM can't reach it.

In this case, Y is a fixed value, and the function is reduced to a one -dollar function

Next discussion:

Case 1: If the actual power generation efficiency must be above 140%

Consider the intersection of the function and y = 1.4

If you want to ensure that the engine efficiency of the general chemical energy engine is greater than 1.4, then under the condition of 10MB/T combustion agent, the rate of access fuel should be 9MB/T ~ 75MB/T

At this time, the instantaneous output is 322,560EU/T, but it still cannot reach UV

Case 2: Considering the actual power generation efficiency of more than 125%

When the fuel flow is 308MB/T, it can be considered that the efficiency is exactly 1.25

At this moment, the power generation power is 1182720eu/T

Case 3: Considering the actual power generation efficiency of more than 120%or more, the flow of fuel flow is 390MB/T

At this time, the instant output power is 1437696eu/T

Situation 4: Considering the actual power generation efficiency above 100%

Fuel flow is 755MB/T

At this time, the instant output power is 2,319,360EU/T

Special circumstances: common practices in Wiki of GTNH 20: 1 Fuel: Burning agent ratio (y = x/20)

The efficiency is taken to 132%. In actual operation, the 200MB/T densely 肼 rocket fuel+10MB/T combustion agent is used in actual operation. The instant power generation power is 811008eu/T