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adoni [48]
3 years ago
13

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

Engineering
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

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Answer:

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2 years ago
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic
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Answer:

The answer is below

Explanation:

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n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm

2) The speed of the rotor is the motor speed. The slip is given by:

Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm

3) The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz

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6 0
3 years ago
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea
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Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

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