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3 years ago

La viscosidad de un liquido es igual a 0.04 N s/m^2 . Este valor en Dinas s/cm^2 se encuentra en el literal

Engineering

1 answer:

nadezda [96]3 years ago

5 0

Answer:

0.4 Dinas*s/cm^2

Explanation:

Tenemos una viscosidad:

V = 0.04 N*s/m^2

Y queremos reescribir esto en Dinas*s/cm^2

Primero transformemos la unidad del denominador, es decir, tenemos que pasar de 1/m^2 a 1/cm^2

Para ello, usamos que:

1m = 100cm

entonces:

(1m/100cm) = 1

Si elevamos ambos lados al cuadrado, obtenemos:

(1m/100cm)^2 = 1

Ahora podemos multiplicar el valor de la viscosidad por esto (que es igual a 1)

V = 0.04 N*s/m^2*((1m/100cm)^2 = 0.00004 N*s/cm^2

Ahora debemos convertir de Newtons a Dinas

Sabemos que:

1 N = 100,000 dinas

1 = (100,000 dinas/1N)

Entonces, de vuelta podemos multiplicar nuestra viscosidad por (100,000 dinas/1N), que es igual a 1 (asi que no cambia el valor, solo sirve para cambiar las unidades)

0.00004 N*s/cm^2 = (100,000 dinas/1N)*(0.00004 N*s/cm^2)

= (100,000 dinas)*(0.00004 s/cm^2)

= 0.4 Dinas*s/cm^2

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Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

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Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

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$v_{i} = 3.986 m/s$

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$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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1 year ago

A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic

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Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

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3 years ago