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3 years ago

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How does a 2.5 MW wind turbine costing $ 4 million compare to a 5-kw wind turbine $3 /W? a) Same $/w b) Smaller $/w c) Larger $/

w

1 answer:

My name is Ann [436]3 years ago

4 0

MW means megawatt, and one megawatt is a million Watts.
The 2.5 MW turbine is 4/2.5=1.6 $/w
Answer B

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Consider a fluid with mean inlet temperature Ti flowing through a tube of diameter D and length L, at a mass flow rate m'. The t

EleoNora [17]

Answer:

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$

Explanation:

Our data given are:

$T_i =$ Mean temperature (inlet)

D = Diameter

L = Length

$\dot{m}=$ Mass flow rate

Equation to surface flow as,

$q(x) = a+bsin(x\pi/L)$

We need to consider the perimeter of tube (p) to apply the steady flow energy balance to a tube , that is

$q(x)pdx=\dot{m}c_p dT_m$

Where $p=\pi D$

Re-arrange for $dT_m,$

$dT_m = \frac{q(x)pdx}{\dot{m}c_p}$

Integrating from 0 to x (the distance intelt of pipe) we have,

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x q(x)dx$

Replacing the value of q(x)

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x (a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}\int_0^x(a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}(ax-\frac{bL}{\pi}cos(\frac{x\pi}{L}))^x_0$

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$

8 0

3 years ago

The inlet for a high-bypass-ratio turbofan engine has an area A1 of 6.0 m 2 and is designed to have an inlet Mach number M~ of 0

Lelu [443]

Answer:

The additive drag at flight condition will be found by the following equation

Area = A1 = 6m2

Da = Additive drag

Cda = Additive drag coefficient

P = prassure at altitude of 12Km

Po = Prassure at sea level

\gamma = Ratio of specific heat capacity

The formula of additive drag is given below

D = Cda q A

q = Dynamic prassure , A = cross sectional area

q =( \gamma/ 2) P0 M02

D = Cda ( \gamma/2) po Mo2 A

Cda=0.32

\gamma=1.4

M=0.8

p0= 101325pa

D = 0.32 (1.4/2)(101325pa)(0.6)2 6

D = 49025N

Explanation:

The additive drag at the flight conditions will be D= 49025N

8 0

3 years ago

A slug travels 3 centimeters in 3 seconds. A snail travels 6 centimeters in 6 seconds. Both travel at constant speeds. Mai says,

Irina-Kira [14]

Answer:

i dont agree with mai because they were both going 1cm per second

Explanation:

3÷3=1

6÷6=1

they both are difrent numbers but equal the same thing

8 0

3 years ago

Which of the following activities could be considered unethical?

Zigmanuir [339]

Answer: u slap birds

Explanation: u order cheeseburgers with n cheese

8 0

3 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago