Paddy Rice Storage Steel Silos with Large Capacity

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Product Description:

Paddy Rice Storage Steel Silos with Large Capacity

Specifications for Paddy Rice Storage Silos

1 Galvanized corrugated steel silo 
2 Advanced heat insulation technology 
3 Fast Installation,low cost and less manpower 

Characteristics for Paddy Rice Storage Steel Silos

-New solution for grain storage to replace traditional warehouse

-Low space occupation and high storage capacity

-Patented heat insulation technology to provide grain with safer storage

-Fast loading and unloading system to fill and empty the silo in short time

-Electric control to make the operation easier

Features for Paddy Rice Storage Steel Silos:

1. High strength: The fly ash silo wall is reinforced by vertical stiffener and occluded by screw beam. It has great strength and good resistance to wind, earthquake and snow.

2. Good sealing function: Crimping and seaming of five-layer spiral steel plates ensures airproof, so our fly ash silo can be used to store construction materials such as cement, gypsum, fly ash and slag, as well as liquids.

3. Small area occupancy: The smallest distance between fly ash silo amounts to 600mm.

4. Short construction cycle: High-level automatic construction at site, Need only 5-6 days for a 1,000 tons fly ash silo .

5. Long working life: 25-30 years, achieved by best combination of plates of different thickness for fly ash silo body.

6. Nice appearance: The roof is subulate and is not easy to accumulate dust or water. The bin is shining and observable with silvery lines.

Processing & Assembly for Paddy Rice Storage Steel Silos:

Paddy Rice Storage Steel Silos with Large Capacity


Paddy Rice Storage Steel Silos with Large Capacity

Paddy Rice Storage Steel Silos with Large Capacity

Paddy Rice Storage Steel Silos with Large Capacity

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Q:what did the Monks of Santo Domingo De Silos do with the money they made?
this link provides telephone number and other contact info. do not be intimidated that it is in spanish. there are monks there who speak english. some are from the united states or england. to be prepared, you could ask a fellow student who is studying spanish or a spanish instructor to assist. you might want to have a fax number availble in case they want to fax the info to you. maintaining the vast complex of buildings they have, dating from the medieval times, would be very, very costly and relentless, as well as utilities, and the costs for education, food, clothing, medical, technology and various middle men for their cds are very expensive. there would be other fees and various charities, as you yourself inferred. but i am sure you can get the specifics from them.
Q:Solve this algebra Problem.?
The equation you've stated for the surface area of the silo, including the hemispherical top, is wrong. It should be: S = 2πrh + 2πr² If r = 15 and h = 50, then: S = 2(3.14)(15)(50) + 2(3.14)(15)² = 6,123 ft²
Q:Calculus Help with differentials?
Buy Grain Silo
Q:film about nixon's last days?
That might be Seven Days In May with Burt Lancaster and Kirk Douglas.
Q:Is the story of the sceva house true?
Sceva House
Q:Optimization problem?
The volume of the hemisphere is 2/3*pi*r^3, and the volume of the cylinder is pi*r^2*h. 2/3*pi*r^3+pi*r^2*h=V. pi*r^2*h=V-2/3*pi*r^3. h=V/(pi*r^2)-2/3*r. The lateral surface area of the hemisphere is 2*pi*r^2, and the lateral surface area of the cylinder is 2*pi*r*h. A=2*pi*r^2+2*pi*r*h. A=2*pi*r^2+2*pi*r(V/(pi*r^2)-2/3*r). A=2*pi*r^2+2*V/r-4/3*pi*r^2. dA/dr=4*pi*r-2*V/r^2-8/3*pi*r. dA/dr=4/3*pi*r-2*V/r^2. 4/3*pi*r-2*V/r^2=0. 4/3*pi*r=2*V/r^2. 4/3*pi*r^3=2*V. r^3=3/2*V/pi. r=(3V/(2pi))^(1/3). I'm too lazy to figure out the height in terms of V. It would be really nice if they specified what the fixed volume was!
Q:Meaning of this quote?
Could masculine toys be related to u know sex toys
Q:Can someone solve this very hard math question?
πr²h = π8²*20 = 1280π ≅ 4021 m³
Q:Measuring astronomical objects?
In 240 B.C, Eratosthenes calculated the circumference of the Earth. The same method he used there can be applied here (plus you can google this if you needed additional explanation). We assume the sun is far enough away that the rays of light can be treated as parallel. A little geometry shows that you are separated from the missile silo by 1/10 the circumference of the planet so the planet's circumference if 10,000miles. To get the radius you divide this by 2pi.
Q:Silo calculus question--is this possible to solve?
First, you need to get a function for the volume. Let r= radius of cylinder and hemisphere and h= height of cylinder v= total volume = 12000 The dome has a volume of (2(pi)r^3)/3 The cylinder volume is (pi)r^2*h v= (pi)r^2 * h + (2/3)(pi)r^3 = 12000 Solve to get h in terms of r leaving: h = 12000/((pi)r^2) - (2/3)r Now make a formula for the cost (c): c = 10 (area of hemisphere) + (area of cylinder) = 10 (2(pi)r^2) + 2(pi)rh = 20(pi)r^2 + 2(pi)r(12000/((pi)r^2) - (2/3)r) = (56/3)(pi)r^2 + 24000/r Take the derivative and solve for when dc/dr = 0 to find the the min. dc/dr = (112/3)(pi)r - 24000/r^2 = 0 r = 5.8928 and h = 106.07 I hope you can fill in the missing steps.

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