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RoseWind [281]
3 years ago
8

Complete the sentence to identify a useful advance in the culinary arts.

Engineering
1 answer:
Archy [21]3 years ago
5 0

Answer:

silos to store grains

Explanation:

I got it from PLATO... hope this helps...

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Explain how the horsepower of a sports car can be improved.
Travka [436]

Answer:

with turbo or nos

Explanation:

6 0
3 years ago
I want a problems and there solutions of The inception of cavitation?​
Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

  • The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
  • Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

    ⇒  A cavitation number is denoted by "σ" .

4 0
3 years ago
For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh
Paul [167]

Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere dp= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction \in =  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas \rho = 0.15 mol/dm ⁻³

viscosity of methane gas \mu = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  0.1 5 *\dfrac{16}{0.1^3}

Density =  2400

Density \rho_f =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

Re = \dfrac{dV \rho}{\mu}

Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}

Re=2276.317705

For Re > 1000

\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}

\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}

\Delta P=8575.755212*2.5

\Delta = 21439.38803 \ Pa

To atm ; we have

\Delta P = \dfrac{21439.38803 }{101325}

\Delta P =0.2115903087  \ atm

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

4 0
3 years ago
An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i
Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object\left ( m\right )=521kg

initial velocity\left ( v_0\right )=90m/s

Final velocity\left ( v\right )=14m/s

kinetic energy of body is given by=\frac{1}{2}mv^{2}

change in kinectic energy is given by substracting  final kinetic energy from initial kinetic energy of body.

Change in kinetic energy=\frac{1}{2}\times m\left ( V_0^{2}-V^2\right )

Change in kinetic energy=\frac{1}{2}\times521\left ( 90^{2}-14^2\right )

Change in kinetic energy=2058.992KJ

7 0
3 years ago
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D
zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

d = \sqrt{4*P / (\pi*\sigma)}

d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm

7 0
3 years ago
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