A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
1 answer:
Answer:
ηa=0.349
ηb=0.345
Explanation:
The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:
The quality at state 4 is determined from the condition and the entropies of the components at the condenser pressure taken from table:
The enthalpy at state 4 then is:
Part A
In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:
The enthalpy at state 2 is determined from an energy balance on the pump:
=346.67 kJ/kg
The thermal efficiency is then determined from the heat input and output in the cycle:
Part B
In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:
The enthalpy at state 2 is then determined from an energy balance on the pump:
=299.79 kJ/kg
The thermal efficiency in this case then is:
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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.
Answer:
1) A=282.6 mm
2)
3)T=0.42 sec
4)f= 2.24 Hz
Explanation:
Given that
V=3.5 m/s at x=150 mm ------------1
V=2.5 m/s at x=225 mm ------------2
Where x measured from mid position.
We know that velocity in simple harmonic given as
Where A is the amplitude and ω is the natural frequency of simple harmonic motion.
From equation 1 and 2
------3
--------4
Now by dividing equation 3 by 4
So A=0.2826 m
A=282.6 mm
Now by putting the values of A in the equation 3
ω=14.609 rad/s
Frequency
ω= 2πf
14.609= 2 x π x f
f= 2.24 Hz
Maximum acceleration
Time period T
T=0.42 sec