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kupik [55]
2 years ago
11

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

e origin and the mirror/mistboundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate
Engineering
1 answer:
anyanavicka [17]2 years ago
3 0

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     <em>=  191 mPa </em>

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Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr
omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

                                   W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                         c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

                  c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

                             q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

                             q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

                            q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                          Q_in = Δ u_2,3

                                         q_2,3 = u_3 - u_2

                                         q_2,3 = c_v*(T_H - T_2)  

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

                                         q_2,3 = c_v*(T_H - T_L*r^(n-1) )    

                                         q_2,3 = 0.718*(1173-288*8(1.3-1) )

                                        q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                     q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                           c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n +  q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

                  c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

                             q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

                             q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

                            q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

                                          Q_out = Δ u_4,1

                                         q_4,1 = u_4 - u_1

                                         q_4,1 = c_v*(T_4 - T_L)  

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

                                         q_4,1 = c_v*(T_H*r^(1-n) - T_L )    

                                         q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

                                        q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

                                         q_net,in = q_3,4+q_2,3

                                         q_net,in = 129.8+456

                                         q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

                                         q_net,out = q_4,1+q_1,2

                                         q_net,out = 244+60

                                         q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

                                         n = 1 - q_net,out / q_net,in

                                         n = 1 - 304/585.8

                                         n = 0.481

6 0
3 years ago
1. Which of the following will cause a spark knock?
zlopas [31]

Answer:

I couldn't find options for your question online, but I can give you an explanation so you can choose the correct option.

Explanation:

A spark knock is a form of unpredictable behavior that occurs in combustion, that is, in the chemical reaction that occurs between oxygen and an oxidizable material. Such combustion is usually manifested by incandescence or flame.

The spark knock is a detonation that occurs when there is a lot of pressure in the fuel.

<u>Some situations in which this can happen are: </u>

  • Engine overloaded.
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5 0
3 years ago
A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator a
SashulF [63]

Answer:

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Explanation:

Coefficient of Performance=11

T_H=\text {Absolute Temperature of high temperature reservoir=300 K}

T_L=\text {Absolute Temperature of low temperature reservoir}

\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}

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