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2 years ago

13

g A plane stress element has components sigma x = 160 MPa, tau xy = 100 MPa (CW). Determine the two values pf sigma y for which

the maximum shear stress is 73 MPa.

1 answer:

Volgvan2 years ago

6 0

Answer:

The question is mentioned in the attachment.

Explanation:

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A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20

viktelen [127]

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

$h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\$

$x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\$

The power produced and consumed by turbine and pump respectively are:

$W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW$

7 0

3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

so

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

2 years ago

Why is data driven analytics of interest to companies?

Svetlanka [38]

Answer:

Advantages of data analysis

Ability to make faster and more informed business decisions, backed by facts. Helps companies identify performance issues that require action. ... can be seen visually, allowing for faster and better decisions.

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2 years ago

A video inspection snake is use

LekaFEV [45]

Answer:

very good thx

Explanation:

5 0

2 years ago

Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer

Vilka [71]

Answer:

thrust washer

can be used to eliminate rubbing friction of wheel touching frame

5 0

2 years ago