Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.

g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression:

$f = \frac{64}{Re}$

If we get that $Re = 211.230$ , then the friction factor is:

$f = \frac{64}{211.230}$

$f = 0.303$

The friction factor is 0.303.

4 0

2 years ago

Create a Relational Schema for the following scenario. Include all primary and foreign keys and list any assumptions you make. 

NARA [144]

Answer:

See explaination

Explanation:

Some of the Important Observation used in Entity Relational Diagram.

1. A Chemist can be involved one project or many projects

2. A project can hire just one or many chemists.

3. A chemist can lead over one or many projects

4. A project can have only one project leader chemist.

5. A project can be done in 1 or many laboratories.

6. A laboratory can host 0 or 1 projects.

7. A laboratory can have 1 or many equipments.

8. An Equipment can be in 1 or 0 laboratories.

9. A chemist can have 0 or many types of equipments.

10. An Equipment can have 0 or 1 chemist at a time

Notations:

PK=Primary Key:

A primary key, also called a primary keyword, is a key in a relational database that is unique for each record. It is a unique identifier, such as a driver license number, telephone number

FK=Foreign Key:

A foreign key is a column or group of columns in a relational database table that provides a link between data in two tables. It acts as a cross-reference between tables because it references the primary key of another table, thereby establishing a link between them

Composite key:

A composite key is a combination of two or more columns in a table that can be used to uniquely identify each row in the table when the columns are combined uniqueness is guaranteed, but when it is taken individually it does not guarantee uniqueness

4 0

3 years ago

Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over

sashaice [31]

The reason why giant stars become planetary nebulas is Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.

<h3>Why do giant stars become planetary nebulae?</h3>

A planetary nebula is known to be formed or created by a dying star. A red giant is known to be unstable and thus emit pulses of gas that is said to form a sphere around the dying star and thus they are said to be ionized by the ultraviolet radiation that the star is known to releases.

Learn more about giant stars from

brainly.com/question/27111741

#SPJ1

3 0

2 years ago

A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t

Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

$tan\phi=\frac{\bigtriangleup l}{h}$

$tan\phi=\frac{0.08}{1.38}$

$tan\phi= 0.05797$

For small angle of $\phi$ , $tan\phi$ can take as $\phi$ .

$\phi = 0.05797 rad.$

Thus, the shear strain is 0.05797 rad.

7 0

2 years ago

What are the four categories of engineering materials used in manufacturing?

alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They have low weight and are poor conductors of electricity and heat

8 0

2 years ago