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4 years ago

Given in the following v(t) signal.

Engineering

1 answer:

tatuchka [14]4 years ago

3 0

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

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The water level in a tank z1, is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end

Alika [10]

Answer:

the maximum height h to which the the water steam could rise is 40.65m

Explanation:

Given;

$Z_{1}$ = 20m,

$P_{1gage}$ = 2atm ≈ 20265 N/m

Density of water ρ = 1000 kg/ $m^{2}$

Note: we take point 1 at the free surface of the water in the tank and point 2 at the top of the water trajectory. we also take reference level at the bottom of the tank.

fluid velocity at the free surface of tank is very low ( $V_{1}$ ≅ 0) and at the top of the water trajectory $V_{2}$ = 0

Step 1: Applying Bernoulli equation between poin 1 and point 2

$P_{1}$ /ρg + $V_{1} ^{2}$ /2g + $Z_{1}$ = $P_{2}$ /ρg + $V_{2} ^{2}$ /2g + $Z_{2}$

$P_{1}$ /ρg + $Z_{1}$ = $P_{atm}$ /ρg + $Z_{2}$

$Z_{2}$ = ( $P_{1}$ - $P_{atm}$ )/ρg + $Z_{1}$

Substituting values into $Z_{2}$ we have,

$Z_{2}$ = $\frac{2atm}{(1000 kg/m^{3} )(9.81 m/s^{2} )} (\frac{101325N/m^{2} }{1atm} )(\frac{1 kg.m/s^{2} }{1N} ) + 20 = 40.65 m$

6 0

3 years ago

The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be

lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = $\frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}$ .....................1

Q = $\frac{T \infty - Tso}{\frac{1}{hA}}$ .....................2

now we compare both equation 1 and 2 and put here value

$\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}$

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0

3 years ago

Explain 3 types of lubrication occurs in bearings?

Angelina_Jolie [31]

Answer:

The three types of lubrication occurred in the bearing are:

When the lubrication occur in the bearing it play an important role in the rolling life of bearing element. Basically, lubricant task is to reduced the friction.

1) Grease lubricant: Grease contain various type of additives which help in enhance the performance. It consist of oil where thickness are added as, this thickness improved the characteristics of grease.

2) Oil lubricant: This is used to reduced or minimized the friction and the lubricant oil is used to in those vehicles when they are motorized. Bearing lubricant oil is used to higher the speed capability.

3) Solid films: These are non fluid coating surface that are applied in the friction surface for prevention. It is used in very extreme situation where oil and grease types of lubricant does not work.

8 0

3 years ago

Chairs and bolsters are used to:

weeeeeb [17]

Answer:e. support reinforcing bars in beams and slabs, prior to concrete placement.

Explanation: Chairs are used for construction of foundations,large steel supports,deck constructions and for underground works. Chairs can be for rebar support ( rebar support chairs),post tension chairs.

Bolsters are usually long made of metallic materials mainly used as support for construction of different infrastructures like roads it helps to ensure the concretes and other construction materials stay firmly connected. Image 1 is a metal be chair

Image 2 is a metal bolster

3 0

3 years ago

A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4

Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

gd = Y/(1 + w)

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0

3 years ago