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3 years ago

Fill in the blank to correctly complete the statement below.

Engineering

1 answer:

lapo4ka [179]3 years ago

6 0

Explanation:

The invention of the pendulums driver ____ ao in the 1600s paved the way for a new industrial era. Add answer.

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.

Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e = $\frac{V_v}{V_s}$ = $\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)$

Specific gravity $G_{s}$ = $\frac{P_s}{P_w} =$ $\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)$

% Saturation S = $\frac{V_w}{Vv}$ × $\frac{100}{1}$ = $\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}$ × $\frac{100}{1}--------(3)$

water content w = $\frac{M_w}{M_s}$ = $\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)$

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w = $\frac{M_w}{Ms}$

Recall; mass = Density × volume

w = $\frac{V_wP_w}{V_sP_s} ------(5)$

From eqn. (2) $G_{s}$ = $\frac{P_s}{P_w}$

∴ $\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)$

putting eqn. (6) into (5)

w = $\frac{V_w}{V_sG_s} -----(7)$

Again, from eqn (1)

$V_s = \frac{V_v}{e}$

substituting into eqn. (7)

$w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S$

∴ $w = \frac{Se}{G_s} -----(8)$

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, $G_{s} = 2.7, e= 0.9$

∴ $w= \frac{0.9*1}{2.7} = 0.333$

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

$w= \frac{0.01*0.9 }{2.7} = 0.0033$

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil ( $P_w=\frac{1_g}{cm^{3} }$ )

Given, $V_{s} = 100 cm^{3 }$ , S = 50% = 0.5

%S = $\frac{V_w}{V_v}$ × $\frac{100}{1}$ = $\frac{V_w}{eV_s}$ × $\frac{100}{1}$

0.50 = $\frac{V_w}{0.9* 100} = 45cm^{3}$

mass of water = $P_wV_w= 1 * 45 = 45_{g}$

7 0

4 years ago

Tech A says never use a water hose to clean up dust after a repair. Tech B says never use a floor scrubber to clean up dust afte

DENIUS [597]

Answer:

Tech B

Explanation:

You ruin your mop that way. it will clump up and than you'll have a mud mess.

5 0

4 years ago

A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of

Art [367]

Answer:

Explanation:

a) Given C [ cal pro fat calc sod]

[140 27 3 13 64]

P = [cal pro fat calc sod]

[180 4 11 24 662]

B = [cal pro fat calc sod]

[50 5 1 82 20]

To find C+2P+3B = [140 27 3 13 64] + 2[180 4 11 24 662] + 3[50 5 1 82 20]

= [650 54 28 307 1448]

The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.

7 0

3 years ago

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Fofino [41]

Answer:

DRAWING LOAD IS $3.67 A_{O}\sigma$

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = $A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})$

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

$A_{f} = A_{O}- 0.6A_{O}$

= $0.4 A_{O}$

Due to change in area ,drawing load p is

p = $0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})$

p = $3.67 A_{O}\sigma$

4 0

3 years ago

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

4 0

3 years ago