Because of the need for increased toughness, shackles for most overhead lifting are made from

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

Visual interpretations when driving are two things: first, what the driver sees around him or her passively, so it is what the e

Phoenix [80]

Answer:

A. How the driver understands and processes the things his or her eyes receive.

Explanation:

When driving, visual interpretation occurs in two things;

What the driver sees around him or her passively, mainly what is received by the eye
How the driver understands and processes the things his or her eyes receive

Proper vision is vital while driving. There is the central vison and the peripheral vision. The central vision is what the driver sees in front through the windshield straight ahead. The Peripheral vision is what is seen at the corners of the eyes.

8 0

3 years ago

Which dimensionless parameter tells whether flow disturbances will be attenuated or amplified? a. Pr b. Re C G d. St e. Fe

Aleonysh [2.5K]

Answer:

b. Re

Explanation:

Reynolds number describe the type of flow of fluid. If Reynolds number has a high value then it is called turbulent flow and if Reynolds number is low then it is called laminar flow. Reynolds number given as follows:

$Re=\dfrac{\rho VD}{\mu }$

For internal pipe flow, if Reynolds number greater than 4000 then, it is called turbulent flow and if Reynolds number less than 2000 then it is called laminar flow. The Reynolds number between 2000 to 4000 the flow is called transition flow.

7 0

3 years ago

A copper-nickel alloy of composition 60 wt% Ni-40 wt% Cu is slowly heated from a temperature of 1250°C (2280 °F). (a) At what te

makkiz [27]

Answer:

a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about 1310 degree Celsius. 

b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; 47wt %of Ni is of a tie line formed across the (a+ L) phase area at 1310 degrees.

c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around 1350 degrees.

c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, being 72wt % of Ni.

4 0

3 years ago

8.26 Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of

Elis [28]

Answer:

Fatigue lifetimes will be ranked as B>A>C

Explanation:

Start by calculating the mean stress for all samples

σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa

σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa

σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa

Now calculate the stress amplitudes of all three samples

σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa

σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa

σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa

The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.

Attached picture shows the justification using and S- N plot.

6 0

4 years ago