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Travka [436]
3 years ago
9

Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa

ter and the other is saturated with alcohol. The unit weight of water is approximately 1g/cm3 while that of alcohol is about 0.8 g/cm3. Which soil sample has the larger water content? Why?
Engineering
1 answer:
mariarad [96]3 years ago
5 0

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with  water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol  = 0.8 g/cm³

solution

we get here water content that is express as

water content = S_s \times \gamma _w   ....................1

here soil is full saturated so S_s is 100% in both case

so put here value for water

water content = 100 % ×  1

water content = 1 g

and

now we get for alcohol that is

water content  = 100 % × 0.8

water content  = 0.8 g

so here water sample have more water content

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A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of
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Answer:

second-law efficiency  = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C  =  293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same  temp reservoir

so here

efficiency ( reversible ) η1 = 1 - \frac{T2}{T1}      ............1

efficiency ( reversible ) η1  = 1 - \frac{293}{1473}  

so efficiency ( reversible ) η1  = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = \frac{thernal\ efficiency}{0.801}

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2 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
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