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3 years ago

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You are an engineer in an electric-generation station. You know that the flames in the boiler reach a temperature of 1200 K and

that cooling water at 300 K is available from a nearby river. You also know that the metallurgical temperature limit for the blades in the turbine is 1000 K before they will incur excessive creep. Now what is the maximum efficiency for this plant

1 answer:

andre [41]3 years ago

6 0

Answer:

...wat

Explanation:

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A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where

Arturiano [62]

Answer:

$\mathbf{F_D \approx 1.071 \ lbf}$

Explanation:

Given that:

The height of a triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = $1.2075 \times 10^{-5} \ ft^2/s$

the density of water ρ = 62.36 lb /ft³

$Re_{max} = \dfrac{Ux}{v}$

$Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}$

$Re_{max} = 414078.6749$

$Re_{max} = 4.14 \times 10^5$ which is less than < 5.0 × 10⁵

Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

$dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2$

where;

$(2-x) dy$ = strip area

$Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}$

Therefore;

$dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})$

$dF_D = 1.136 \times(2-x)^{1/2} \ dy$

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy$

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy$

Let U = (2-2y)

-2dy = du

dy = -du/2

$F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}$

$F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du$

$F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2$

$F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2$

$F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ]$

$F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ]$

$F_D = 1.071031071 \ lbf$

$\mathbf{F_D \approx 1.071 \ lbf}$

8 0

3 years ago

Two point charges of values +2.0 and +5.5 μC, respectively, are separated by 0.50 m. What is the potential energy of this 2-char

IgorC [24]

Answer:

U=0.198J

Explanation:

The potential energy associated with two point charges is given as

U=(Kq1q2)/r

Where k=8.99*10⁹N.m²/C²

r=0.5m,

q₁=2μc,

q₂=5.5μc

If we substitute values into the equation we arrive at

U=(8.99*10⁹N.m²/C²* 2*10⁻⁶*5.5*10⁻⁶)/0.5

U=197.78*10⁻³

U=0.198J

From the calculation above, we can conclude that the potential energy between the two system of charges is 0.198J

8 0

4 years ago

Read 2 more answers

The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

5x−2y=30 Complete the missing value in the solution to the equation. (8, )

givi [52]

Answer:

Value of y = 5

(8,5)

Explanation:

Given:

5x−2y=30

Value of x = 8

Find:

Value of y

Computation:

⇒ 5x−2y=30

⇒ 5(8)−2y=30

⇒ 40−2y=30

⇒ -2y = -10

y = 5

Value of y = 5

(8,5)

4 0

3 years ago

In fuel oil piping systems, all exterior above-grade fill piping shall be __________ when tanks are abandoned or removed.

castortr0y [4]

Answer:

Removed

Hope this Helps!

6 0

3 years ago