Coal gangue stacking problem
Summary
We studied the accumulation of coal gangue. According to the accumulation requirements of coal gangue, the construction of a linear rising orbit of approximately β = 25 ° (the angle is too large, the taking the car cannot be filled), and the tanker is transported to the track in the track. The top of the top is dumped on both sides. After the stack of stones is high, then the orbit is extended by means of the gangue stack, which gradually piles a gangue mountain shown in Figure 1.
First, the geometric analysis of the rules that are finally formed by the stacked coal gangue is the geometry of the mathematics, so the relevant factor in these bodies can be found in accordance with the geometric relationship. In the engineering of the entire transport coal gangue, with the increase in the length of the slope of the coal gangue, the cost of the entire transport coal gangue will also increase accordingly. The entire fee includes land acquisition costs and cost of transporting coal gangue. The land acquisition cost is related to the area coverage of the accumulated coal gangue, and the area of the area is increased to the length of the hillside of the coal specide; at the same time, the electricity charge used in transport coal gangue is also increased as the length of the slope increases. Conversion of cost problems into the length of the slope of coal gangue, and the length of the long slope is a function of time, so the total cost is established with the mining time of coal gangue, and finally we can draw the length of the slope through a given time. Determine the total fees for handling the gangue, when dealing with land acquisition, due to the annual increase of the land price, we use all the land users needed to purchase the life by bank loans to purchase all the lands needed in the date of use, and the results show this operation After that, the annual treatment funds in the design are used to pay the electricity bill. The rest of the remaining repairs will be 1 million yuan / year according to the treatment funds. The application is 10% of the 20% increase, the design fund is Excessive and more than enough. And you can pay off bank loans in the 14th year. After 20 years, you can profit 4,203,37.83 yuan.
Key words
Understanding of the long-term slope length mechanical energy geometry volume treatment total cost land price increase loan interest rate
First, the question is proposed
When the coal mine mining, the beauty of the waste material will be produced in the plain area, and the coal mine has only researched land-stacked coal gangue. When the coal gangue is stacked, it is necessary to set up a track (angle excessive coater can't be filled with the surface angle), and the natural rest horn а is also set, and the gangue car mechanical efficiency of the transport coal gangue also follows the track. Increased and declined, the existing annual increase in land collection fee, under the influence of many factors, we need to solve the following questions:
1. Judging whether the design processing fund is sufficient;
2. Develop a reasonable land acquisition plan according to the relevant data;
3. Predicting minimum costs for treating gangue;
Second, the problem analysis
According to the question, the geometry of the coal gangue is certain (as shown in Figure 1), what we have to solve is the relationship between the volume of coal gangue piles and the geometry of the geometric shape, due to the surrounding price, the land needs to be loan to bank loans However, due to the difference in land prices and bank loan interest rates, the loan plan is also required. According to the volume of the gangue volume after the coal mine, calculate the length of the slope length X determine the floor to deal with the total cost, listed in different 出Minimum cost at the time.
figure 1
Third, model hypothesis
1. The electricity fee is paid in the same year, no arrears;
2. Bank deposit, loan interest rate and land price increase is fixed;
3. The electric energy consumed by the tuning, which is transformed into the gravity potential energy of coal gangue (taking a speed of taking the tanity, and the mesenchym and slope tracks see friction);
4. The land acquisition is once every year from the beginning of the year;
5. The annual processing fund is dial on the year, and in addition to the delivery of electricity bills, the rest is used to land acquire up or repay the loan, strictly achieve special purpose, four, nouns and symbolic agreements
1. Length of X - coal gangue hill;
2. ρ - load weight;
3. а - 石 Nature stack release resting angle;
4. K1 - electricity fee required to use the car;
5. η0 - Treasury car mechanical efficiency initial value;
6. A - Railway extension
10 meters
, The percentage of mechanical efficiency declines on the original basis;
7. The present value of the k2 - land collection fee;
8. b - land price year;
9. C1 - bank deposit interest rate;
10. C2 - Bank's loan interest rate;
11. M - coal mine design of the year production;
12. N - coal mine design life;
13. D - coal mine mining rate;
14. S (x) - When the length of the slope is X, the actual floor area of the gangue;
15. V (X) - When the length of the slope is X, the actual volume of the gangue heap is
16. η (x) - When the length of the slope is X, the mechanical efficiency of the tuning truck;
17. W (x) - When the length of the slope is X, the electric energy to be consumed by the taking the bus;
18. G (N) - Most of the coal mine N years, the total volume of meteorite;
19. h (n) - The total electricity bill used in the coal mine is used;
20. S (X (N)) - The total land area used in the range of use;
Five, the establishment of the model
1) Analysis of geometry
Knowing, the A-SBOD is a pyramid portion, and the A-DCOB is a conical portion, and the SB is tangent to the cone, and the relationships in Figure 1 are as follows:
AO = sin (β) · x Co = sin (β) · COT (а) · x SO = COS (β) · x
∠BOS = Arc Cos (TAN (β) · COT (а))
2) The bottom area of the gangue mountain s (x)
S (x) = s four-sided SDOB S fan-shaped BODC
Si-shaped SDOB = X2 · COS (β) · SiN (β) · COT (а) · SIN [Arc Cos (Tan (β) · COT (а))]
S fan-shaped BODC = X2 · [PI-ARC COS (TAN (β) · COT (а))] · SIN2 (β) · COT2 (а)
S (x) = x2 · {COS (β) · sin (β) · COT (а) · SIN [Arc Cos (Tan (β) · COT (а))] [PI-ARC COS (TAN (β) · COT (а))] · SIN2 (β) · COT2 (а)} = k (а, β) · x2
(1: k (а, β) = COS (β) · SiN (β) · COT (а) · SIN [Arc COS (Tan (β) · COT (а))] [PI-ARC COS (TAN β) · COT (а))] · SIN2 (β) · COT2 (а)
3) Meteori Mountain Volume V (x)
V (x) = 1/3 · s (x) · AO = 1/3 · k (а, β) · sin (β) · x3
4) Mechanical efficiency
From the inscription, the track is extended every
10 meters
The efficiency drops by 2% on the original basis, so η (x) = 30% · (1-2%) X / 10
5) When the mechanical energy can be X, the microme increment DX is present in volume micro-gauge DV (X)
DW (x) = DV (x) · ρ · g · xin (β) / η (x) = k (а, β) · sin2 (β) · ρ · g · x3 · dx / (30% · (1-2%) x / 10) w (x) = ∫0X (k (а, β) · sin2 (β) · ρ · g · x3 / (30% · (1-2%) X / 10) ) DX
6) Mining the total volume of the meteorite during N years;
G (n) = n · m · D / ρ = V (x (n))
X (n) = (3 · N · m · D / ρ · K (, β) · sin (β)) 1/3
7) Total electricity bills used in N years
H (n) = w (x (n)) · 0.5 / (3.6 · 106)
The NAIN is required to pay for H (N) -H (N-1)
8) During use, the total area S (x (n)) = k (а, β) · x2 (N)
Appendix I:
1.
Function [RST] = T1 (NYEAR)
IF Nargin ~ = 1
Error ('Error Input Arguments!');
Else
B = 25/180 * pi; a = 55/180 * pi;
P = 2000; g = 9.8;
sinb = sin (b);
COSB = COS (B);
COTA = COT (a);
Arccos = ACOS (Tan (b) * Cot (a));
Kab1 = COSB * SINB * COTA * SIN (Arccos);
Kab2 = (pi-arccos) * sinb ^ 2 * cota ^ 2;
Kab = kab1 kab2;
XF = (4500000 * Nyear * 0.1 / (kab * sinb)) ^ (1/3);
KW = kab * sinb ^ 2 * p * g / 0.3;
SYM X;
Wintegral = INT ('x ^ 3 / (0.98 ^ (x / 10))', 'x', 0, xf);
W = kW * wintegral;
Monw = 0.5 / (3.6 * 10 ^ 6);
Area = kab * xf ^ 2;
Mona = 1.1 * 80000 / 666.6667;
Allmoney = W * Monw Area * Mona;
% DISP ('The Total Area IS (Unit: MU);'); DISP (Area / 666.6667);
% DISP ('Area Money (Unit: RMB):'); DISP (Area * Mona);
% DISP ('Eletricty Money (UNIT: RMB)'); DISP (W * Monw);
% DISP ('NEED MONEY (Unit: RMB):'); DISP (AllMoney);
RST = W * Monw;
% Rst = allmoney;
END;
2.
Function T2 ()
For iCount = 1: 19
RSTW (ICOUNT, 1) = T1 (iCount 1) -t1 (iCount);
END;
ABC = 7595900 * 1.05- (1000000-73391);
For ncount = 1: 19
DISP ('the year is;'); DISP (Ncount 1);
% DISP ('all Money Have Returned.'); Break;
% ELSE
ABC = ABC * 1.05- (1000000-RSTW (ncount, 1))
% End;
END;
