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4 years ago

An example of Ferrous alloy is Brass a)-True b)-False

Engineering

1 answer:

djyliett [7]4 years ago

5 0

Answer: False

Explanation: No, brass is not a ferrous alloy.

Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.

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Your company is planning to build a pipeline to transport gasoline from the refinery to a field of storage tanks. The parameters

AleksandrR [38]

Answer:

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

Explanation:

The Reynolds number ( $Re_{D}$ ) is a dimensionless criterion use for flow regime of fluids, which is defined as:

$Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}$ (Eq. 1)

Where:

$\rho$ - Density, measured in kilograms per cubic meter.

$\mu$ - Dynamic viscosity, measured in kilograms per meter-second.

$v$ - Average flow velocity, measured in meters per second.

$D$ - Pipe diameter, measured in meters.

We need to find the equivalent velocity of water used in the prototype system. In this case, we assume that $Re_{D,gas} = Re_{D,w}$ . That is:

$\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}}$ (Eq. 2)

Where subindex $w$ is used for water and $gas$ for gasoline.

If we know that $\rho_{gas} = 690\,\frac{kg}{m^{2}}$ , $\mu_{gas} = 0.006\,\frac{kg}{m\cdot s}$ , $v_{gas} = 0.5\,\frac{m}{s}$ , $D_{gas} = 1\,m$ , $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\mu_{w} = 0.0018\,\frac{kg}{m\cdot s}$ and $D_{w} = 0.05\,m$ , then we get the following formula:

$57500 = 27777.778\cdot v_{w}$

The fluid velocity for the prototype system is:

$v_{w} = 2.07\,\frac{m}{y}$

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

3 0

3 years ago

Need help I’m giving out brainlest whoever get it me correct

ludmilkaskok [199]

Answer:

(C) Prototype Model

Explanation:

I'm sure that is the answer i am very sorry if not :)

7 0

4 years ago

Sensors are used to monitor the pressure and the temperature of a chemical solution stored in a vat. The circuitry for each sens

JulsSmile [24]

Circle because it’s round and we all love round things

5 0

3 years ago

The Environmental Protection Agency is likely to consider the effect of building activity on what aspect of water? (Select all t

Agata [3.3K]

Answer:

Sorry

Explanation:

6 0

3 years ago

Read 2 more answers

If d=0.25m and D=0.40m. Assume headloss from the contraction to the end of the pipe can be found as hų = 0.9 (V is velocity in t

pychu [463]

Answer:

Height=14.25 m

Discharge= $0.8155 m^{3}/s$

Explanation:

Bernoulli’s equation for the two points, let’s say A and B is given by

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}+z_A=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+z_B+h_{L(A-B)}$

But since the elevations are the same then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+h_{L(A-B)}$

Since $h_{L(A-B)}=\frac {0.9v^{2}}{2g}$ then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}$

From continuity equation

$A_AV_A=A_BV_B=0.25\pi D^{2}V_A=0.25\pi d^{2}V_B hence [tex]D^{2}V_A= d^{2}V_B$ and substituting 0.4 for D and 0.25 for d

$V_B=\frac { D^{2}V_A}{d^{2}}=\frac { 0.4^{2}V_A}{0.25^{2}}=2.56V_A$

$\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}$ and substituting $V_B$ with $2.56V_A$ we have