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Genrish500 [490]
3 years ago
6

What does the supply chain management process involve

Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

It involves the active streamlining of a business's supply-side activities to maximize customer value and gain a competitive advantage in the marketplace

Explanation:

Supply chain management is the management of the flow of goods and services and includes all processes that transform raw materials into final products.

You might be interested in
Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi
PolarNik [594]

Answer:

a)   P₂ = 3219.11 lbf / ft² , b)    P₂ = 721.91 lbf / ft² , c)  P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

     A₁ v₁ = A₂ v₂

     v₂ = v₁ A₁ / A₂

The area of ​​a circle is

    A = π r² = π/4  d²

     v₂ = v₁ (d₁ / d₂)²

Let's calculate

    v₂ = 12 (7/3)²

    v₂ = 65 feet / s

Now let's use Bernoulli's equation

     P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

     P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

      y₁ = y₂

      P₁ - P₂ = ½ ρ  (v₂² –v₁²)

      P₂ = P₁ - ½ ρ (v₂² –v₁²)

     ρ = 62.43 lbf / ft³

     P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

    P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

    P₂ = 7200 - 3980.89

    P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

        y₂ –y₁ = 40 ft

        P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

        7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

        P₂ = 7200 - 2497.2 - 3980.89

         P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

         y₂ –y₁ = -40

         P₂ = 7200 + 2497.2 - 3980.89

         P₂ = 5707.31 lbf / ft²

3 0
3 years ago
¿Qué son alimentos funcionales?
Mnenie [13.5K]
DescripciónAlimentos funcionales son aquellos alimentos que son elaborados no solo por sus características nutricionales sino también para cumplir una función específica como puede ser el mejorar la salud y reducir el riesgo de contraer enfermedades
4 0
3 years ago
A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature
mafiozo [28]

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

6 0
4 years ago
A balanced three-phase 208 V wye-connected source supplies a balanced three-phase wyeconnected load. If the line current IA is m
tatyana61 [14]

Answer:

6.004 Ω

Explanation:

For  a Y- connected system given that :

Line voltage, $V_L = 208 \ V$

Line current , $I_L=20\ A$

and specified that $V_L \ and \ I_L$ are in phase.

Hence the impedance will be pure resistive.

For Y-system

$V_L = \sqrt3 V_{ph}$

$V_{ph}$ = phase voltage

$V_{ph}$ $=\frac{V_L}{\sqrt3} = \frac{208}{\sqrt3}$

    = 120.08 V

Line current = Phase current

$I_L = I_{ph} = 20 \ A$

Now, $z_{ph} = \frac{V_{ph}}{I_{ph}}=\frac{120.08}{20}$

                          = 6.004 Ω

5 0
3 years ago
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