8.5C (H)

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

Which example describes how the water cycle carries on energy transfer? a flooding river depositing silt on a floodplain a warm

Irina18 [472]

Answer:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

5 0

2 years ago

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Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

grigory [225]

Answer : The equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$

Explanation : Given,

Equilibrium constant = 14.5

Concentration of $CO$ at equilibrium = 0.15 M

Concentration of $H_2$ at equilibrium = 0.36 M

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get:

$14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}$

$[CH_3OH]=2.82\times 10^{-1}M$

Therefore, the equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$

5 0

3 years ago

A block of copper has dimensions of 32cm x 21cm x 11cm. What is the mass of the block of copper ?

MatroZZZ [7]

Explanation:

The volume of the copper block is:

$32 \times 21 \times 21 = 14112 \: {cm}^{3}$

The density of copper is 8.96 g/cm³

So the mass of the copper block is

$14112 \times 8.96 = 126443.52 \: g$

or 126.44352 kg.

3 0

2 years ago

Which formula represents an alkene?

lisov135 [29]

Alkenes are organic compounds which contains double bonds between two carbon atoms. The formula C2H4 can be classified as an alkene since it most likely contains a double bond between the carbon-carbon atoms. An alkene has the general formula of CnH2n.

6 0

3 years ago

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