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2 years ago

Argue the importance to society of incorporating green building into an engineer's designs, with at least two examples.

Engineering

2 answers:

Solnce55 [7]2 years ago

7 0

Answer:

Explanation:

How are sustainable design practices used in civil engineering projects? Sustainable designs are made in buildings to reduce heating and cooling costs by using less energy, which in turn helps the environment. Engineering new ways to build infrastructure can help reduce the impact on the environment.

stellarik [79]2 years ago

4 0

Answer:

Green Design is your answer

Explanation:

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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

3 years ago

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

yanalaym [24]

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

$Re=\dfrac{F_i}{F_v}$

For plate can be given as

$Re=\dfrac{\rho\ v\ l}{\mu }$

Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

$Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.$

$Reynolds\ number\is \$

7 0

3 years ago

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of

Alexxandr [17]

Answer:

$P_m_i_n = 442KPA$

Explanation:

We are given:

m = 1.06Kg

$T_H = 1.2T_L$

T = 22kj

Therefore we need to find coefficient performance or the cycle

$COP_R = \frac {1}{(T_R/T_l) -1}$

$= \frac {1 }{1.2-1}$

= 5

For the amount of heat absorbed:

$Q_l = COP_R Wm$

= 5 × 22 = 110KJ

For the amount of heat rejected:

$Q_H = Q_L + W_m$

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= $= \frac{132}{1.06}$

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

$T_L = \frac{T_H}{1.2}$ ;

$T_L = \frac{342.5}{1.2}$

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at $T_L$ = 12.4°c we have 442KPa

6 0

3 years ago

I WILL GIVE 20 POINTS!!

Alex777 [14]

Answer:

Use a resume header

Explanation:

Create a Summary

Research industry, employer keywords

there are some hints okay

5 0

3 years ago

Read 2 more answers

For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with

Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0

3 years ago