Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can be tripped to a turbulent state by adding roughness to the leading edge of the plate. For a particular situation, experimental results show that the local heat transfer coefficients for laminar and turbulent conditions are

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

Calculate the average heat transfer coefficients for laminar and turbulent conditions for plates of length L = 0.1 m and 1 m.

Answers

Answer 1

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8


Related Questions

Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.

Answers

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

Answer: 0.2025

Explanation: I got it correct

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol

Answers

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

Help this is very hard and I don't get it

Answers

Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.

Answers

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

Answers

Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec

Answers

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement​

Answers

Explanation:

D. B. C. A. E. Is this a good idea

In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.

Answers

Answer:

406.140 KHz

Explanation:

Given data:

Rsig = 100 kΩ

Rin = 100kΩ

Cgs = 1 pF,

Cgd = 0.2 pF,  and   etc.

Determine the expected 3-dB cutoff frequency

first find the CM miller capacitance

CM = ( 1 + gm*ro || RL )( Cgd )

     = ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )

     = ( 11.311 ) pF

now we apply open time constant method to determine the cutoff frequency

Th = 1 / Fh

hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]

                               = [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] =  406.140 KHz

A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.

Answers

Answer: attached below is the power triangles

7.13589 MVAR

Explanation:

Power ( P1 ) = 10 MW

power factor ( cos ∅ ) = 0.6 lagging

New power factor = 0.85

Calculate the reactive power of a capacitor to be connected in parallel

Cos ∅ = 0.6

therefore ∅ = 53.13°

S = P1 / cos ∅ = 16.67 MVA

Q1 = S ( sin ∅ ) = 13.33 MVAR  ( reactive power before capacitor was connected in parallel )

note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2

cos ∅2 = 0.85 ( new power factor )

hence ∅2 =  31.78°

Qsh ( reactive power when power factor is raised to 0.85 )

= P1 ( tan∅1 - tan∅2 )

= 10 ( 1.333 - 0.6197 )

= 7.13589 MVAR

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