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irga5000 [103]
3 years ago
8

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

fluids and thus to reduce the pumping costs. Consider the flow of such a fluid through a 100-m-long pipe of outer diameter 30 cm in calm ambient air at 0°C. The pipe is heated electrically, and a thermostat keeps the outer surface temperature of the pipe constant at 25°C. The emissivity of the outer surface of the pipe is 0.8, and the effective sky temperature is -30°C.
Required:
Determine the power rating of the electric resistance heater, in kW, that needs to be used.
Engineering
1 answer:
Gala2k [10]3 years ago
7 0

Answer:

what id the answer

Explanation:

You might be interested in
-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63
Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0
3 years ago
A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al
marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0
3 years ago
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
Tech a says that the weight of the flywheel smoothest out the engines power pulses. Tech B says that the flexplate and torque co
lakkis [162]

Answer:

both statement is correct

Explanation:

Flywheel engine uses to reduce fluctuations.

And                                                                

FlexPlate is a metal disk that connects the output from the engine to the input of the torque converter. This will replace the flywheel

so that both statement is correct

4 0
3 years ago
Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

                          v = +/- 0.323 m/s

- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

4 0
3 years ago
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