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2 years ago

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Air at 27°C, 1 atm flows parallel to a flat plate, which is electronically heated. The plate is 0.5 m long in the direction of f

low. The uniform heat flux from the plate to the air is 3,000 W/m². A trip wire placed at the leading edge of the plate ensures that the boundary layer is turbulent over the entire plate. A thermocouple mounted at the trailing edge of the plate records a temperature of 127°C. Consider the following:
a) Determine the local Nusselt number Nuy at the trailing edge of the plate.
b) Calculate the Reynolds number Rex at the trailing edge of the plate.
c) Evaluate the air free stream velocity (m/s).

1 answer:

dmitriy555 [2]2 years ago

8 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno

Ivan

Answer:

Technician A

Explanation:

Ohms law: I= E/R so rest resistance must be present along with E/potential difference. Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law. Current flow is directly proportional to the voltage and inversely proportional to R (resistance or impedance).

8 0

2 years ago

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

3 years ago

Read 2 more answers

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

3 years ago

In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

kipiarov [429]

Answer:

Moisture/ water content w = 26%

Void ratio , e = 0.73

Explanation:

Initial mass of saturated soil w1 = mass of soil - weight of container

= 113.27 g - 49.31 g = 63.96 g

Final mass of soil after oven w2 = mass of soil - weight of container

= 100.06 g - 49.31 g = 50.75

Moisture /water content, w = $\frac{w1-w2}{w2}$ = $\frac{63.96-50.75}{50.75}$ = 0.26 = 26%

Void ratio = water content X specific gravity of solid

= 0.26 X 2.80 =0.728

5 0

2 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

2 years ago