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kicyunya [14]
3 years ago
5

1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con

stant heat addition from 100 kPa and 27∘C to 700 kPa 2-3 Isothermal expansion until V3 = 7v23-1 P = constant heat rejection to the initial state2. Assume air has constant properties with cv = 0.718 kJ/kg K, cp = 1.005 kJ/kg K, R = 0.287 kJ/kg K, and k = 1.4.(a) Sketch the P- and T-s diagrams for the cycle.(b) Determine the ratio of the compression work to the expansion work (the back work ratio).(c) Determine the cycle thermal efficiency.
Engineering
1 answer:
QveST [7]3 years ago
5 0

Answer:

Explanation: Here it is: 67 Hope that helps! :)

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A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min

4 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
3 years ago
A debugging process where you, the programmer, pretend you are a computer and step through each statement while recording the va
Ipatiy [6.2K]

Answer:

hand tracing

Explanation:

as a programmer when we pretend  computer in the  debugging process by the step of each statement in recording    

then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method      

so while recording value of each variable each step is hand tracing

5 0
3 years ago
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = \frac{q^2Nd^2*Xn^2}{6Eo} * f

also note that ; Pactive ∝ Nd2 (

tD = K . \frac{Vdd}{(Vdd - Vt )^2}  since K = constant

Hence : Nd ∝ rt

5 0
3 years ago
Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti
erik [133]

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.

7 0
3 years ago
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