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3 years ago

What different tests did the team perform to come up with a workable design?

1 answer:

5 0

They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.

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A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

4 0

3 years ago

Do you understand entropy? Why the concept of entropy is difficult to engineering students?

Darya [45]

Answer:

Entropy:

Entropy is the measure of randomness of system.In other words the entropy is the measurement of tendency of system towards the disorder.

The concept of entropy arise from second law of thermodynamics.It is given as follows

$ds=\int \dfrac{dQ}{T}$

Entropy is a extensive property of system .

Entropy of universe = Entropy of system + Entropy of surrounding.

The entropy of the system can be zero,positive and negative.But entropy of the surrounding can not be negative,but it can be zero or positive.

Actually the concept of entropy is difficult to understand because we can not visualize because it is not like beam and like rods.Only we have to realize that there is entropy.

4 0

2 years ago

Pipelines are a useful means of transporting oil because they: Multiple select question. are fast never fail to deliver are chea

Rufina [12.5K]

Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.

<h3>What is a pipeline?</h3>

A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.

The options are attached

a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.

b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.

c. Pipelines have more locations than water carriers.

d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Learn more about Pipelines

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7 0

1 year ago

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= $4\sqrt{2}$ cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as $P = \dfrac{\pi ^2EI}{L_e^2}$

If area moment of inertia(I) is high then buckling load will be high.

We know that area moment of inertia(I)

For circular tube $I = \dfrac{\pi }{64}(D_o^4-D_i^4)$

For circular bar $I = \dfrac{\pi }{64}D^4$

Now by putting the values

For circular tube $I=62.83 cm^4$

For circular bar $I=50.26 cm^4$

So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.

3 0

2 years ago

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0

3 years ago