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2 years ago

Select three ways an engineer can create a view of a design.

Engineering

1 answer:

Angelina_Jolie [31]2 years ago

5 0

Sketching projection

Sketching will give the brief planning of the engineering work to be done so it's important

Isometric projection

Yes doing a work in 3D factor out the mistakes which can be optimised

Third angle projection:-

Viewing angles matter a lot
So it will help the engineer to make it perfect from all angles

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What type of oil pressure gauge should be used when

liq [111]

Answer:

The mechanical gauge would be the one for the job

Explanation:

6 0

3 years ago

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

Vilka [71]

Answer:

<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

6 0

3 years ago

thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat

eduard

Answer:

$q_{in} = 528.6\,\frac{kJ}{kg}$

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

$q_{in} = h_{g} - h_{mix}$

The specific enthalpies are:

Liquid-Vapor Mixture:

$h_{mix} = 2217.2\,\frac{kJ}{kg}$

Saturated Vapor:

$h_{g} = 2745.8\,\frac{kJ}{kg}$

The thermal energy per unit mass required to heat the steam is:

$q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}$

$q_{in} = 528.6\,\frac{kJ}{kg}$

7 0

3 years ago

Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid

AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for this case we can define the following equation for mass flow using the first law of thermodynamics

$m=\frac{Q}{h1-h2}$

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

$Q=\frac{5000}{1123-184.5} =5.328Ibm/hr$

3 0

3 years ago

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa

mote1985 [20]

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

4 0

3 years ago