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gavmur [86]
3 years ago
9

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

Engineering
1 answer:
nikitadnepr [17]3 years ago
3 0

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

μ = dynamic viscosity in (Pa·s) at input temperature T,

\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

T_0 = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

For kinematic viscosity :

\nu = \frac {\mu} {\rho}

kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec

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Answer:

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Explanation:

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In this case, we compute the heat output from coal, given its heating value and the mass flow:

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Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%

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3 years ago
3 MAOP stands for which of the following?
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3 years ago
A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ
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Answer:

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
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The load is 17156 N.

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First compute the flexural strength from:  

σ = FL / πR^{3}

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σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

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4 years 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
Roman55 [17]

Answer:

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

Explanation:

From ideal gas law,

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R is gas constant = 8314.34 J/kgmol.K

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

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

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Heat transfer = -88.13 + 71.24 = -16.89 kJ

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