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kenny6666 [7]
3 years ago
10

A thin‐walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory

in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.006 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 2.5 m/s. Assume the thermophysical properties of the exhaust gas are those of air
Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube.

Engineering
1 answer:
Serhud [2]3 years ago
5 0

Answer:

The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K

Explanation:

The detailed solution is attached as files below.

However, the steps followed are highlighted:

1) The average temperature was calculated as 380.5 K

2) The properties of air at 380.5 K was highlighted

3) The Prandti number was calculated. Pr = 0.693

4) The Reynold number was calculated, Re = 28716.77

5) The Nusselt umber was calculated, Nu = 75.94

6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.

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Give two causes that can result in surface cracking on extruded products.
Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

           Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

          Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0
3 years ago
La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
liq [111]

Answer:

La probabilidad pedida es 0.820196

Explanation:

Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

X: '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''

Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}    con x=0,1,2,...,n

Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x} con x=0,1,2,...,10

Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)

Calculando P(X=8), P(X=9) y P(X=10) por separado y sumando, obtenemos que P(X\geq 8)=0.820196

7 0
3 years ago
Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig
marin [14]

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of  a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

5 0
3 years ago
A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c
monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of  the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0
3 years ago
A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract
Hitman42 [59]

Answer:

radius = 9.1 × 10^{-3} m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support =  45 mm

solution

we apply here minimum radius formula that is

radius = \sqrt[3]{\frac{FL}{\sigma \pi}}      .................1

here F is applied load and  is length

put here value and we get

radius =  \sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}  

solve it we get

radius = 9.1 × 10^{-3} m

8 0
3 years ago
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