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Zielflug [23.3K]
3 years ago
13

A __________ is an added note showing additional or more specific information.

Engineering
1 answer:
xxMikexx [17]3 years ago
8 0

Answer:

awnsers should be added to know to show additional

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(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
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Answer:

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Explanation:

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Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mec
olga55 [171]

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

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Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

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When the Moon is in the position shown, how would the Moon look to an observer on the North Pole?
kirill115 [55]

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

3 0
3 years ago
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