Draw an ERD for each of the following situations. (If you believe that you need to make additional assumptions, clearly state th
1 answer:
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Answer:
yield strength before cold work = 370 MPa
yield strength after cold work = 437.87 MPa
ultimate strength before cold work = 440 MPa
ultimate strength after cold work = 550 MPa
Explanation:
given data
AISI 1018 steel
cold work factor W = 20% = 0.20
to find out
yield strength and ultimate strength before and after the cold-work operation
solution
we know the properties of AISI 1018 steel is
yield strength σy = 370 MPa
ultimate tensile strength σu = 440 MPa
strength coefficient K = 600 MPa
strain hardness n = 0.21
so true strain is here ∈ = = 0.223
so
yield strength after cold is
yield strength =
yield strength =
yield strength after cold work = 437.87 MPa
and
ultimate strength after cold work is
ultimate strength =
ultimate strength =
ultimate strength after cold work = 550 MPa
Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate =
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.
Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is
As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%
Answer:
import numpy as np
import time
def matrixMul(m1,m2):
if m1.shape[1] == m2.shape[0]:
t1 = time.time()
r1 = np.zeros((m1.shape[0],m2.shape[1]))
for i in range(m1.shape[0]):
for j in range(m2.shape[1]):
r1[i,j] = (m1[i]*m2.transpose()[j]).sum()
t2 = time.time()
print("Native implementation: ",r1)
print("Time: ",t2-t1)
t1 = time.time()
r2 = m1.dot(m2)
t2 = time.time()
print("\nEfficient implementation: ",r2)
print("Time: ",t2-t1)
else:
print("Wrong dimensions!")
Explanation:
We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.
