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2 years ago

Define volume flow rate of air flowing in a duct of area A with average velocity V.

Engineering

1 answer:

KATRIN_1 [288]2 years ago

6 0

Explanation:

Step1

Volume flow rate is the rate of change of volume of fluid that is flowing in the duct of pipe per unit time. It is measure in m³/s or l/s. Volume flow rate is very important parameter in fluid analysis.

Step2

For the given duct, the volume flow rate is the product of average velocity to the cross section area of duct.

Expression for volume flow rate is given as follows:

Q=AV

Here, Q is the flow rate, A is area of the duct and V is the average velocity of flowing fluid.

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The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is

Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2. load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

for example, in matlab,

x=t{1}

y=t{2}

4. plot the wind velocity and electricity generated.

plot(x, y)

5. Label the individual axis and name the graph title.

title('Graph of wind velocity vs approximate electricity generated for the year')

xlabel('wind velocity')

ylabel('approximate electricity generated for the year')

5 0

3 years ago

Read 2 more answers

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

2 years ago

This question is 100 points<br> I NEED HELP!!!

Mamont248 [21]

Answer:

hey if u repost this i can answer it u and u dont have to waste this much points but its super blury and not even able to read a single word

8 0

2 years ago

Read 2 more answers

The B-pillar may also be called the:

slega [8]

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up

6 0

2 years ago

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

2 years ago