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Aleonysh [2.5K]
3 years ago
6

Consider a fully developed laminar flow in a circular pipe. The velocity at R/2 (midway between the wall surface and the centerl

ine) is given by _____ provided that Vmax is the maximum velocity in the tube.
Engineering
1 answer:
ryzh [129]3 years ago
4 0

Answer:

The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

Explanation:

Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as

v = (ΔPR²/4μL) [1 - (r²/R²)]

where v = fluid velocity at any point in the radial direction

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

But the maximum velocity of the fluid occurs at the middle of the pipe when r=0

Hence, maximum veloxity is

v(max) = (ΔPR²/4μL)

So, velocity at any point in the radial direction is

v = v(max) [1 - (r²/R²)]

At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

So,

1 - (r²/R²) = 1 - (1/4) = (3/4)

Hence, v at r = (R/2) is given as

v = v(max) × (3/4)

Hope this Helps!!!

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aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

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Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

3 0
3 years ago
When its 100-hp engine is generating full power, a small airplane with mass 700 kggains altitude at a rate of 2.5 m/s. What frac
snow_lady [41]

For the given problem it is necessary to recap the concepts about Power, that is, is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time.

The equation for power can be written as

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Where,

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By the second newton law, force can be:

F = mg

Where m means the mass and g the gravity acceleration.

We can also write the equation as,

P = mgv

Replacing the values

P = 700*9.8*25

P = 17.15kW

Tenemos unidades en dos sistemas diferentes, por lo que convertimos los HP a Sistema internacional y tenemos que

1hp = 0.746kW

Then the fraction \etais,

\eta = \frac{17.15kW}{0.746kW}

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Therefore the fraction of the engine power is being used to make the airplane climb is 22.984%

5 0
3 years ago
Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state
lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

P_1=1\ bar,V_1=4.25 {m^3}

P_2=6.8\ bar

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

Now by putting the values                (1.4 bar =140 KPa)

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=140\times 4.25 \ln \dfrac{1.4}{6.8}

W=-940.36 KJ

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7 0
4 years ago
A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

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          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0
3 years ago
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Finger [1]

Answer:

specific heat

Explanation:

7 0
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