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3 years ago

5

The Reynolds number, rhoVD/μ is a very important parameter in fluid mechanics. Determine its value for ethyl alcohol flowing at

a velocity of 4 m/s through a 2-in.-diameter pipe.

1 answer:

ElenaW [278]3 years ago

8 0

Answer:

The Reynold Number is 146415.34

Explanation:

The density of ethyl alcohol = ρ = 789 kg/m^3

The dynamic viscosity of ethyl alcohol = μ = 0.001095 kg/m.s

Velocity = V = 4 m/s

Diameter of pipe = D = 2 in = 0.0508 m

The Reynold's number (Re), will be:

Re = ρVD/μ

Re = (789 kg/m^3)(4 m/s)(0.0508 m)/(0.001095 kg/ms)

<u>Re = 146415.34</u>

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A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (

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Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

The flux = volume of water per unit time = flow rate of water through the hoop.

The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.

This means that

Flow rate = AV

where A is the area of the hoop

V is the velocity of the water through the hoop

This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

Q = AV

From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2

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3 years ago

Why will screws never replace nails

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Answer:

because people have different opinions on nails and screws

Explanation:

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3 years ago

Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$

$\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}$

$\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}$

$G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}$

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

$P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))$

The rate of heat:

$Q=G_{m}(h_{3}-h_{2})$

The thermal efficiency:

$\mu_{ther}=\frac{P_{net}}{Q}$

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2 years ago

Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch

Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 $\frac{ft^{3}}{sec}$

$D_{1}$ = 6 inch=0.5 ft

$D_{2}$ =2 inch=0.1667 ft

As we know that Q=AV

$A_{1}\times V_{1}=A_{2}\times V_{2}$

So $V_{2}=\frac{Q}{A_2}$

$V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}$

$V_{2$ =0.687 ft/sec

We know that Head loss due to sudden contraction

$h_{l}=K\frac{V_{2}^2}{2g}$

If nothing is given then take K=0.5

So head loss $h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}$

=0.00366 ft

So head loss=0.00366 ft

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2 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

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$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

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2 years ago