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4 years ago

1. Under what conditions can soils be chemically stabilized?

Engineering

1 answer:

marshall27 [118]4 years ago

7 0

Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

a). Plasticity Index :A soil with a high value of plasticity Index is not good for various engineering projects. The introduction of line helps in reducing plasticity due cation exchange reaction.Pozzolanic reaction over time reduces plasticity and increase index strength due to the formation of calcium - silicate hydrate.

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A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago

Consider an open loop 1-degree-of-freedom mass-spring damper system. The system has mass 4.2 kg, and spring stiffness of 85.9 N/

Marat540 [252]

Answer:

Damping ratio $\zeta =0.0342$

Explanation:

Given that

m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m

We need to find damping ratio

We know that critical damping co-efficient

$C_c=2\sqrt {mk}$

$C_c=2\sqrt {4.2\times 85.9}$

$C_c=37.98$ N.s/m

Damping ratio( $\zeta$ ) is the ratio of damping co-efficient to the critical damping co-efficient

So $\zeta =\dfrac{C}{C_c}$

$\zeta =\dfrac{1.3}{37.98}$

$\zeta =0.0342$

So damping ratio $\zeta =0.0342$

3 0

4 years ago

Answer every question of this quiz

Reil [10]

I'd say number 4, number 3 looks like an exhaust valve

5 0

4 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was

shepuryov [24]

Answer:

Time of submersion in years = 7.71 years

Explanation:

Area of plate (A)= 16in²

Mass corroded away = Weight Loss (W) = 3.2 kg = 3.2 x 106

Corrosion Penetration Rate (CPR) = 200mpy

Density of steel (D) = 7.9g/cm³

Constant = 534

The expression for the corrosion penetration rate is

Corrosion Penetration Rate = Constant x Total Weight Loss/Time taken for Weight Loss x Exposed Surface Area x Density of the Metal

Re- arrange the equation for time taken

T = k x W/ A x CPR x D

T = (534 x 3.2 x 106)/(16 x 7.9 x 200)

T = 67594.93 hours

Convert hours into years by

T = 67594.93 x (1year/365 days x 24 hours x 1 day)

T = 7.71 years

3 0

4 years ago