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3 years ago

Introduction for site visit

Engineering

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

Site visit is when an external evaluation team goes to an institution to evaluate verbal, written and visual evidence. explanatory context. Site visits, which often last several days, are part of the quality evaluation process.

Explanation:

You might be interested in

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

3 years ago

A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b

mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0

3 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

7 0

3 years ago

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

Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat

attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

$COP_{HP}=\frac{T_H}{T_H-T_L}$

Therefore, the temperature relationship, $T_H=1.15\;T_L$

Then, we should apply the values in the COP.

$=\frac{1.15\;T_L}{1.15-1}$

$=7.67$

The number of heat rejected by the heat pump must then be calculated.

$Q_H=COP_{HP}\times W_{nst}$

$=7.67\times5=38.35$

We must then calculate the refrigerant mass flow rate.

$m=0.264\;kg/s$

$q_H=\frac{Q_H}{m}$

$=\frac{38.35}{0.264}=145.27$

The $h_g$ value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

$T_L=\frac{T_H}{1.15}$

$=\frac{64+273}{1.15}=293.04$

$=19.89\°C$

the lowest reservoir temperature = 258.703 kpa

So, the pressure ratio should be = 7.15

8 0

3 years ago

Introduction for site visit​

Introduction for site visit