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Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
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.
