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abruzzese [7]
2 years ago
13

There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains

8 ft of oil and is drained at the rate of 0.25 ft/min. After how many minutes will the level of oil in the two tanks be the same?
Engineering
1 answer:
Luba_88 [7]2 years ago
4 0

Answer:

  16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

  (4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

__

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

  y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

  y = 8 -0.25x . . . . . .  height in feet after x minutes (tank B)

These will be equal when ...

  y = y

  12 -0.5x = 8 -0.25x

  4 = 0.25x . . . . . . . . . . add 0.5x -8

  16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

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The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

 ( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0
3 years ago
A 800-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea
Arturiano [62]

Answer:

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes

Explanation:

In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:

Efficiency=\frac{work}{heat transfer to power plant}

Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW

From First law of thermodynamics:

Rate of heat transfer to river=heat transfer to power plant-work done

Rate of heat transfer to river=2000-800

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.

4 0
3 years ago
Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req
Genrish500 [490]

Answer:

Check the explanation

Explanation:

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6 0
3 years ago
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

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3 0
2 years ago
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
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