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2 years ago

6

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th

e flow. If the differential height between the water columns connected to the two outlets of the probe is 2.4 cm, determine
(a) the flow velocity and
(b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 45°C and 98 kPa, respectively.

1 answer:

Korolek [52]2 years ago

4 0

Answer:

20.94 m/s, 235.44 Pa

Explanation:

Acceleration due to gravity g = $9.81 m/s^2$

height h = 0.024 m

From density of air = P/RT

= (98000)/(287 * 318.14) = $1.073 kg/m3$

Using Bernoulli equation

$(P/density*g) + (V^2/2g) + z = constant$

$(P1/density*g) + (V1^2/2g) + z1 = (P2/density*g) + (V2^2/2g) + z2$

Here z1 = z2 (since the outlets have the same differential height) and V2 = 0 (no velocity at the tip)

Solving and making V1 subject of the formula

$V1 = \sqrt{(P2 - P1)/density of air}$

$V1 = \sqrt{(2*density of water* g*h)/density of air}$

$V1 = \sqrt{(2*1000*9.81*0.024/1.073}$

= 20.94 m/s

Change in pressure P2 - P1= density of water * g * height

= 1000*9.81*0.024

=235.44 Pa

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Functions to create a median and mode of a set of numbers

Explanation:

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2 years ago

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The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air

Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

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=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

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Answer:

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b)Mass in lb = 376.8 lb

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d) $w=1276 \ lb/ft.s^2$

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d= 2 ft

r= 1 ft

h= 3 ft

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We know that volume V given as

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$V=9.42\ ft^3$

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Mass = Density x volume

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1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

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Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

$v=\dfrac{1}{\rho}\ ft^3/lb$

$v=\dfrac{1}{40}\ ft^3/lb$

$v=0.025\ ft^3/lb$

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

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2 years ago

Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b

grandymaker [24]

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

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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

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