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jek_recluse [69]

3 years ago

13

Cual es la definición de la distribución de las instalaciones?

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.

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prove that the heat transfer at the constant pressure is given by the enthalpy change during the process

mihalych1998 [28]

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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Electrical pressure or “force”<br><br> A) current<br> B) resistance <br> C) voltage

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A train travels 650 meters in 25 seconds. What is the train's velocity?

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The train is traveling 26 meters A second .

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2 years ago

Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus

Alina [70]

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4 0

3 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

3 years ago