(35 points) This is a legit question that I have for a device FOR my homework.

6. Question

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

3 years ago

For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum

Digiron [165]

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ $_y$ = 690 Mpa

plane strain fracture toughness K $_{Ic$ = 32 MPa- $m^{1/2$

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K $_{Ic$ / σ $_y$ )²

so we substitute our values into the formula

t ≥ 2.5( 32 / 690 )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

7 0

3 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW

c)COP=6.01

d) $Q_R=17.99 KW$

Explanation:

Given

$T_L$ = -12°C, $T_H$ =40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

$COP=\dfrac{T_L}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

So

5.01= $\dfrac{15}{W_{in}}$

b) $W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

$Q_R=Q_A+W_{in}$

Given that $Q_A=15$ KW

We know that $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d) $Q_R=17.99 KW$

5 0

4 years ago

Given V1 = 8 vpp, R1 = 6 kΩ, V2 = 5 vpp, R2 = 3 kΩ, V3 = 10 vpp, R3 = 3 kΩ, and Rf = 14 kΩ, find Vout. Vout = vpp. (Round your a

iragen [17]

Answer:

Vout= 93.3V

Explanation:

For this question, consider circuit in the attachment 1.

This is the circuit of an inverting amplifier. In an inverting amplifier

Vout/Vin= -Rf/Rin

To calculate the Vout, we must find Rin and Vin. For this we must solve the input circuit (attachment 2) using Thevinine theorem. Thevnine theorem states that all voltage sources in a circuit can be replaced by an equivalent voltage source Veq and and all resistances can be replaced by an equivalent resistance Req. To find out Req all voltage sources must be short circuited (attachment 3)

1/Req= 1/R1+1/R2+1/R3

1/Req=1/6+1/3+1/3

Req=6/5

To find out Veq consider circuit in attachment 4. We will solve this circuit using nodal analysis. In nodal analysis, we use the concept that sum of currents entering a node is equal to the sum of currents leaving a node. So,

I1= I2+I3

(10-Veq)/6= (Veq-5)/3+(Veq-10)/3

Veq=8V

Now the input circuit can be simplified as shown in attachment 5. Solve for Vout using equation

Vout/Veq= -Rf/Req

Vout/8= -14/(6/5)

Vout= - 93.3

It is at an angle of 180° from Veq

7 0

3 years ago

Construct a Mohr circle for the stress element at A in problem 2. Using ruler and compass, draw the Mohr circle to the scale. Dr

serious [3.7K]

Answer:

hello your question is incomplete attached below is the missing diagram to the question and the detailed solution

Answer : principal stresses : 0.82 MPa, -33.492 MPa

shear stress = 17.157 MPa

∅ = 9.09 ≈ 10°

Explanation:

The principal stress ( б1 ) = 0.82 MPa

( б2 ) = -33.492 MPa

The shear stress = 17.157 MPa

∅ = 9.09 ≈ 10°

attached below is the detailed solution and the Mohr's circle

7 0

3 years ago