Question

Water storage tank Renewable power sources like wind and solar require mechanisms to store energy to use when the wind isn't blowing or the sun isn't shining. One technology is a "water battery," in which water is pumped up to a high elevation when power is available, and then drained through a turbine to generate power later. An open 10 m diameter cylindrical tank will hold 800 m3 of water at maximum fill. The tank will be made of 1015 steel with Sut​=415MPa and Se​=125MPa. What is the minimum thickness of the tank wall to ensure an infinite life with a safety factor on stress of 1.5? Use the Goodman criterion.

Answer 1

To determine the minimum thickness of the water storage tank wall, the Goodman criterion is used. By calculating the alternating stress and plugging in the values, the minimum thickness is determined to be approximately 31.35mm.

Step-by-step explanation:

To determine the minimum thickness of the tank wall, we need to use the Goodman criterion.

The Goodman criterion states that the alternating stress should be less than the endurance limit of the material divided by the safety factor.

In this case, we will use the endurance limit Se​=125MPa and the safety factor of 1.5.

The alternating stress can be calculated using the formula σa=(Sut​-Su)/2, where Sut​ is the ultimate tensile strength and Su​ is the yield strength.

For 1015 steel, Sut​=415MPa and Su​ is typically 0.5 times the ultimate strength. Therefore, Su​=415MPa*0.5=207.5MPa.

Plugging these values into the formula, we get σa=(415MPa-207.5MPa)/2=103.75MPa.

The maximum fill of the cylindrical tank is 800 m3.

The volume of a cylinder can be calculated using the formula V=πr2h, where r is the radius and h is the height.

In this case, the radius is half the diameter, so r=10m/2=5m.

The height can be calculated by rearranging the formula to h=V/(πr2) = 800m3/(π*(5m)2) = 8m.

The weight of the water can be calculated using the formula W=mg, where m is the mass and g is the acceleration due to gravity. The density of water is 1000 kg/m³.

Therefore, the mass is m=V*density = 800m3 * 1000kg/m³ = 800,000kg.

The weight is W=mg=800,000kg * 9.8m/s² = 7,840,000N.

The stress on the tank wall can be calculated using the formula stress=W/(2πrh), where W is the weight and r and h are the radius and height of the tank.

Plugging in the values, we get stress=7,840,000N / (2*π*5m*8m) ≈ 62,549.88Pa.

Now we can apply the Goodman criterion to find the minimum thickness.

The Goodman criterion states that σa/(Sut​-yield stress) + σm/Se = 1, where σa is the alternating stress, Sut​ is the ultimate tensile strength, yield stress is the yield strength, σm is the mean stress, and Se is the endurance limit.

Since we want to calculate the minimum thickness, we will assume a mean stress of zero, which gives us σm=0. Plugging in the values, we get σa/(415MPa-207.5MPa) + 0/125MPa = 1.

Rearranging the equation, we get σa = (415MPa-207.5MPa)*125MPa = 103.75MPa * 125MPa = 12,968.75MPa.

The minimum thickness of the tank wall can be calculated using the formula thickness=σa/(2*σf), where thickness is the minimum thickness, σa is the alternating stress, and σf is the yield strength.

Plugging in the values, we get thickness = 12,968.75MPa/(2*207.5MPa) ≈ 31.35mm.