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3 years ago

9

How much force is required if the ramp is 15 ft long

1 answer:

lord [1]3 years ago

6 0

Answer:

do yah stiil need help

Explanation:

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Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5

pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$

$Pb = 70000\ Pa-27303\ Pa - 14715\ Pa$

$Pb = 27,996\ Pa = 28\ kPa$

6 0

3 years ago

Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a

Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0

3 years ago

4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is gener

Dmitrij [34]

Answer:

The rate of irreversible loss will be "55.22 MW".

Explanation:

The given values are:

Elevation,

h = 120 m

Flow of water,

Q = 100 m³/s

Efficiency,

= 80%

i.e,

= 0.8

Efficiency turbine,

= 50 MW

Now,

Without any loss,

The power generated by turbine will be:

⇒ $P=\delta gQh$

On substituting the values, we get

⇒ $=1000\times 9.8\times 100\times 120$

⇒ $=117.72 \ MW$

Power generated in actual will be:

= $\frac{50}{0.8}$

= $62.5 \ MW$

Hence,

Throughout the piping system,

The rate of irreversible loss is:

= $Power \ generated \ by \ turbine-Power \ generated \ in \ actual$

= $117.72-62.5$

= $55.22 \ MW$

4 0

3 years ago

Which of the following statements about pitot-static systems is FALSE? a). A pitot probe measures the Total Pressure of the free

Pavlova-9 [17]

Answer:

C

Explanation:

Pitot tube:

Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.

As we know that stagnation pressure is the summation of dynamic and static pressure.

Stagnation pressure = Static pressure + Dynamic pressure

So

Dynamic pressure = Stagnation pressure - Static pressure

We know that dynamic pressure

$P_{dynamic}=\dfrac{\rho V^2}{2}$

On the other hand Pitot tube measure the dynamic pressure.

So option C is correct.

5 0

3 years ago

A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL

prohojiy [21]

Answer:

i don't think i understand the question

Explanation:

7 0

2 years ago

Read 2 more answers