A grain bucket conveyor (grain leg) raises soybeans from the ground level up to the peak of its height, h= 150 ft. A bushel of soybean weight, W=60 lb/bu. The bucket elevator can deliver, Q= 120 bu/min, to the top spout. The motor shaft power to drive the elevator is 40 hp. Determine the power (hp) required to elevate the grain.
Answer: 32.7 hp
show work
Bushels?? Really.
P = mgh/t
m/t is your converted Bu/min.
Then maybe convert the whole thing into hogsheads and find a different class.
To determine the power required to elevate the grain, we need to consider the work done to lift the soybeans to the peak height. The power can be calculated using the formula:
Power (hp) = (Work done (ft-lb/min)) / (Time (min))
First, we need to calculate the work done. The work done to lift an object is given by the formula:
Work done (ft-lb) = Force (lb) x Distance (ft)
The force required to lift the soybeans is equal to the weight of the soybeans. The weight of the soybeans can be calculated using the formula:
Weight (lb) = Weight per bushel (lb/bu) x Number of bushels (bu)
In this case, the weight per bushel is given as 60 lb/bu. The number of bushels can be calculated as:
Number of bushels = Delivery rate (bu/min) x Time (min)
Now, let's plug in the values:
Number of bushels = 120 bu/min x 1 min = 120 bu
Weight (lb) = 60 lb/bu x 120 bu = 7200 lb
Next, we need to calculate the distance lifted. The distance lifted is the height of the grain leg, which is given as 150 ft.
Work done (ft-lb) = Force (lb) x Distance (ft) = 7200 lb x 150 ft = 1,080,000 ft-lb
Now that we have the work done, we can calculate the power:
Power (hp) = (Work done (ft-lb/min)) / (Time (min)) = 1,080,000 ft-lb / 1 min = 1,080,000 ft-lb/min
We know that 1 horsepower is equal to 550 ft-lb/min, so we can convert the power to horsepower:
Power (hp) = (1,080,000 ft-lb/min) / (550 ft-lb/min/hp) = 1963.636 hp
However, the given motor shaft power is 40 hp. This means that there are additional losses in the system, such as friction and inefficiencies. Therefore, the actual power required to elevate the grain is less than the motor shaft power.
To determine the actual power required, we can use the efficiency of the system. Efficiency is given by the formula:
Efficiency = (Output power) / (Input power)
In this case, the output power is the power required to elevate the grain, and the input power is the motor shaft power. Rearranging the formula, we have:
(Output power) = Efficiency x (Input power)
Since the output power is the power required to elevate the grain, and the input power is given as 40 hp, we can determine the power required:
(Output power) = Efficiency x 40 hp = (1963.636 hp) x Efficiency
Since the efficiency is not given, we cannot determine the exact power required. However, if we assume an efficiency of 95%, the power required would be:
(Output power) = (1963.636 hp) x 0.95 = 1865.4542 hp
Rounding to the nearest tenth, the power required to elevate the grain is approximately 1865.5 hp.