9

You receive the bill for servicing your diesel engined car. As a good engineer you immediately notice that one of the items on b

vichka [17]

Answer:

b. spark plugs

Explanation:

Diesel engines are characterized in that the mechanism that activates the explosion of fuel is high pressures, so when the piston reaches the top of the cylinder, the pressure of the air-diessel mixture is so high that it causes the explosion, this It is what generates the power in a diesel engine.

3 0

3 years ago

As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

katen-ka-za [31]

yes it will

Explanation:

5 0

4 years ago

Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e

marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

$\Delta x=\frac{PL^3}{3EI}$

Hence we can write

$P=\frac{3EI\cdot \Delta x}{L^3}$

Comparing with the standard spring equation $F=kx$ we find the cantilever analogous to spring with $k=\frac{3EI}{L^3}$

Now the angular frequency of a spring is given by

$\omega =\sqrt{\frac{k}{m}}$

where

'm' is the mass of the load

Thus applying values we get

$\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}$

$\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec$

8 0

3 years ago

Read 2 more answers

The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 700°C is 2 × 10-6. Calculate the number of vacancies

True [87]

Answer:

$1.16*10^{23} vacancies/m^3$

Explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

$\frac{N_v}{N}=2*10^{-6}\\N_v=2*10^{-6}N\\$

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

$N=\frac{p_{Ag} * N_A}{A_{AG}} \\N=\frac{6.022*10^{23} * 10.35*10^6g/m}{107.87g/mol} \\N=5.78*10^{28}atom/m^3$

if we substitute the value of the number of lattice into the first equation, we arrive at

$N_v=2*10^{-6} *5.78*10^28\\N_v=1.16*10^{23} vacancies/m^3$

4 0

4 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