The following vessel has an example of which type of engine attached to it?

Under certain conditions, wind blowing past a rectangular speed limit Sign can cause the sing to oscillate with a frequency omeg

aniked [119]

Answer:

Explanation:

Given that,

Omega is a function of the following

ω = f(b, h, v, ρ, k)

Where, all unit have a dimension of

ω = T^-1

b = L

h = L

V = LT^-1

ρ = FL^-4T²

k = FL

Then,

From the pie theorem

The required pi term is 6—3 = 3 terms,

So, we use V, p and b as a repeating term.

For first pi

π1 = ω•b^a•v^b•ρ^c.

Since

ω= T^-1, b = L, v = LT^-1 and

ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(T^-1)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b-1)• L^(a+b-4c)

Comparing coefficient

c = 0

2c - b - 1 = 0

b = 2c - 1 = 0 - 1 = -1

a + b - 4c = 0

a = 4c - b = 0 - -1 = 0+1

a = 1

Then, a = 1, b = -1 and c = 0

So, π1 = ω•b^a•v^b•ρ^c.

π1 = ω•b^1•v^-1•ρ^0

π1 = ω•b / v

Check dimensions

ωb/v = (T^-1)L / LT^-1 = L^0•T^0 = 1

Then, π1 is dimensionless

For second pi

π2 = h•b^a•v^b•ρ^c.

Since

h = L, b = L, v = LT^-1 and ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c = 0

2c - b = 0

b = 2c = 0

1 + a + b - 4c = 0

a = 4c - b - 1 = 0 -0 - 1 = -1

a = -1

Then, a = -1, b = 0 and c = 0

So, π2 = h•b^a•v^b•ρ^c.

π2 = h•b^-1•v^0•ρ^0

π2 = h / b

Check dimensions

h / b = L / L = 1

Then, π2 is dimensionless

For third pi

π3 = k•b^a•v^b•ρ^c.

Since

k= FL, b = L, v = LT^-1 and ρ=FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(FL)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^(c+1)•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c + 1= 0

Then, c = -1

2c - b = 0

b = 2c = -2

1 + a + b - 4c = 0

a = 4c - b - 1 = -4 +2 - 1 = -3

a = -3

Then, a = -3, b = -2 and c = -1

So, π3 = k•b^a•v^b•ρ^c.

π3 = k•b^-3•v^-2•ρ^-1

Therefore,

π3 = k / b³•v²•ρ

Let check for dimension

π3 = FL / (L³• L²T^-2 • FL^-4T²)

π3 = FL / (L^(3+2-4) • T^(-2+2) •F)

π3 = FL / (L• T^(0) •F)

π3 = FL / LF = 1

π3 is also dimensions less

So.

I. There are three none dimensional pi

II. The none dimensional group are

π1 = ω•b / v

π2 = h / b

π3 = k / b³•v²•ρ

III. Reynolds Number. The Reynolds number is the ratio of inertial forces to viscous forces and it is dimensionless

So, the π3 can be considered as a Reynolds number

6 0

3 years ago

Suppose a steam locomotive is rated at 7500 horsepower. If its efficiency is 6%, how much wood must be burned in a 3-hour trip?

Sati [7]

Answer:

50,625kg of wood

Explanation:

Power input = power output ÷ efficiency

Power output = 7500 horsepower = 7500×750W = 5625000W, efficiency = 6% = 0.06

Power input = 5625000/0.06 = 93750000W

Energy input = power input × time

time = 3hour = 3×3600 = 10800sec

Energy input = 93750000 × 10800 = 1.0125×10^12J

Mass = Energy ÷ Energy density = (1.0125×10^12J) ÷ (20×10^6J/kg) = 50,625kg of wood

3 0

3 years ago

The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is discovered the lower deck is

leva [86]

Answer:

It will be around 146,27 min since the pump is turned on until the deck is clear of the water.

Explanation:

When the leak is discovered and the pump is turned on, the lower deck is already submerged and the leak is not fixed; then, in order to have the deck clear of water, the bilge pump has to remove the accumulated water ( $V_{0}$ ) and the water that is taking on ( $r_{in}*t$ ) through the leak. We can represent this mathematically as follow:

$V_{0} +r_{in} *t-r_{out}*t=0$ Equation 1

Where:

$V_{0}$ : is the accumulated water when the leak was discovered

$r_{in}$ : is the takes on rate through the leak = 57.5 gal/min

$r_{out}$ : is the removing rate of the bilge pump = 73.8 gal/min

t= is the time since the pump is turned on until the deck is clear of water.

To calculate the accumulated water ( $V_{0}$ ), we will model the lower deck as a flat-bottomed container with a bottom surface area of 510 $ft^{2}$ and straight vertical sides. Knowing that the level submerged is 7.5 inches, and performing the corresponding unit conversions, we obtain: