All categories

3 years ago

Cual es la definición de la distribución de las instalaciones?

Engineering

2 answers:

Nat2105 [25]3 years ago

7 0

Answer:

Nao ablo espanhol

Explanation:

jeka57 [31]3 years ago

6 0

La distribución de las instalaciones consiste en determinar dónde se colocarán los departamentos, los grupos de trabajo de los departamentos, las estaciones de trabajo y los puntos donde se guardan los materiales de trabajo dentro de una instalación productiva. El objetivo es ordenar estos elementos de manera que se garantice el flujo continuo del trabajo en una fábrica o un patrón de tránsito dado en una organización de servicios.

You might be interested in

A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur

SOVA2 [1]

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W $m^{-1}K^{-1}$

diameter (d1) = 10 cm

so radius (r1) = 10 /2 = 5 cm = 5 × $10^{-2}$

radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × $10^{-2}$

temperature = 25°C

surface heat transfer coefficient = 6 6 W $m^{-2}K^{-1}$

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to 15°C

first pass through through resistance R1 i.e.

R1 = ( r2 - r1 ) / 4 $\pi$ × r1 × r2 × K

R1 = ( 5.2 - 5 ) $10^{-2}$ / 4 $\pi$ × 5 × 5.2 × 16 × $10^{-4}$

R1 = 3.825 × $10^{-3}$

same like we calculate for resistance R2 we know i.e.

R2 = 1 / ( h × area )

here area = 4 $\pi$ r2²

area = 4 $\pi$ (5.2 × $10^{-2}$ )² = 0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 × $10^{-3}$ + 4.90499

heat loss = 2.037152 W

8 0

3 years ago

If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 6 V, while the current through th

Trava [24]

Given Information:

Current of Ideal source = Is = 100 mA = 0.100 A

Open circuit voltage = Voc = 6 V

Short circuit current = Isc = 200 mA = 0.200 A

Required Information:

Power absorbed by idea current source = ?

Answer:

Power absorbed by idea current source = 0.3 Watts

Explanation:

Ideal Current Source:

An ideal current source doesn't have internal resistance and provides constant current regardless of the voltage supplied to the circuit.

The power absorbed by the ideal source can be found using

P = Is²R

Where Is is the ideal source current and R can be found using

R = Voc/Isc

Where Voc is the open circuit voltage and Isc is the short circuit current.

R = 6/0.200

R = 30 Ω

Therefore, the power absorbed by the ideal current source is

P = Is²R

P = (0.100)²*30

P = 0.3 Watts

7 0

3 years ago

4. The outer end of a control arm is attached to the steering knuckle through a

arlik [135]

Is attached to the spring

3 0

3 years ago

Read 2 more answers

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation $N_{o} = N_{i} * \frac{E_{f}-E_{i} }{KT}$

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, $N_{d} = 1.65*10^{16} - 10^{16} \\ = 6.5 * 10^{15} per cm cube$

5 0

3 years ago

Read 2 more answers

The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher

DedPeter [7]

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows

ε = $\frac{Roughness _. value}{ Diameter}$ Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the $h_{l}$ equation we get $h_{l}$ = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

6 0

3 years ago