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3 years ago

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A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is

3 m thick. Assume that the data for the sample at 7.0 m are representative of the entire clay profile. Also assume that the clay is heavily over consolidated and that the danse sands at the surface of the profile are so stiff that they do not contribute to the settlement. Find the settlement of the surface due to compression of the clay layer

1 answer:

goblinko [34]3 years ago

8 0

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

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Read 2 more answers

Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to

vredina [299]

Answer: the total chronic daily intake is 0.03296 mg/ kg.d

Explanation:

First of we have to determine the CDI due to ingestion of drinking water

so

CDI = (C x IR x EF x ED) / (BW x AT) let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d

next we determine the CDI due to dermal contact with water.

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) / (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)] × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d

so the total chronic daily intake is 0.03296 mg/ kg.d

4 0

3 years ago

A converging-diverging nozzle with an exit to throat area ratio of 4.0 is designed to expand air isentropically to atmospheric p

34kurt

Answer

0.9, 1172.35kPa

Explanation:

<em>Question (in proper order) Attached below</em>

Air is flowing inside the throat has following inlet conditions

$P_{0}=1000 kPa$

$T_{0}=500 K$

$M=1.8$

$M=\frac{u}{c}=1.8$

'u' is the speed of sound in the air

$\Rightarrow u=1.8\times c$

$=1.8\times 340.29$

$=612.522\frac{m}{sec}$

Therefore volumetric flow rate entering,

$Q=612.522\times 0.0008$

$=0.4900176\frac{m^{3}}{sec}$

Using ideal gas equation

PV=nRT

$n=\frac{PV}{RT}$

$=\frac{1000\times 0.4900176}{8.314\times 500}$

=0.117878 gmoles/sec

Therefore , mass flow rate

$Mass = 0.117878\times 29$

$=3.4184 grams/sec$

Given

$\frac{A}{A_{0}}=2$

$\Rightarrow A=0.0016.m^{2}$

Using continuity equation

$A_{1}V_{1}=A_{2}V_{2}$

$\Rightarrow V_{2}=\frac{A_{1}V_{1}}{A_{2}}$

$=\frac{0.0008\times 612.522}{0.0016}$

$=306.261\frac{m}{sec}$

Hence exit velocity = 306.261 m/sec

Exit Mach number

$M=\frac{u}{c}=\frac{306.261}{340.29}=0.9$

Temperature will remain same as 500 K

Now

Using Bernoulli's equation

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}$

Here

$z_{1} = z_{2}$

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}-\frac{v_{2}^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{\rho g}+\frac{612.522^{2}}{2g}-\frac{306.261^{2}}{2g}=\frac{P_{2}}{\rho g}$

$\Rightarrow \frac{1000000}{1.225}+\frac{612.522^{2}}{2}-\frac{306.261^{2}}{2}=\frac{P_{2}}{1.225}$

$\Rightarrow P_{2}=1172.35kPa$

4 0

3 years ago