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3 years ago

Hằng số phổ biến chất khí

Engineering

1 answer:

drek231 [11]3 years ago

7 0

Answer:

Business activities may broadly be classified into two categories namely (A) Industry and (B) Commerce. Industry involves production of goods and services whereas commerce is concerned with the distribution of goods and services.

Explanation:

hope helps

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The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

V = 0.5 m/s

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3 years ago

How do I do this?<br> Blueprints, complete the missing view.

Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

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3 years ago

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following

torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches $5\times 10^5$

$(a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\$

$(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e$ $(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}$

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3 years ago

A cold storage room is used to keep the temperature inside the room maintain at low temperature.

lianna [129]

Answer:

The power of the brick wall it may be how the soiled ness of the wall too keep in the cold

Explanation:

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3 years ago

Read 2 more answers

The uniform slender rod has a mass m.

Nikolay [14]

Answer: 3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

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3 years ago