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2 years ago

What does the air change rate represent?

Engineering

1 answer:

Juli2301 [7.4K]2 years ago

3 0

Answer and Explanation:

The removal or addition of air volume to the space is the air change rate
The rate of air change is positive when air volume is added to the space and the rate of air change is negative when air volume is removed from the space.
The standard built home has a 0.5 to 1 of air change rate.
The rate of air change is dependent on the building (how the building form)

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A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

Alex Ar [27]

Answer:

Throat diameter $d_2$ =28.60 mm

Explanation:

Bore diameter $d_1=63mm$ ⇒ $A_1=3.09\times 10^{-3} m^2$

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004 $\frac{m^3}{s}$

Coefficient of discharge $C_d$ =0.8

We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$h=x(\dfrac{S_m}{S_w}-1)$

$S_m$ =13.6 for Hg and $S_w$ =1 for water.

$h=0.235(\dfrac{13.6}{1}-1)$

h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

⇒ $d_2$ =28.60 mm

So throat diameter $d_2$ =28.60 mm

4 0

3 years ago

simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$