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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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Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Maru [420]

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

5 0
3 years ago
Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.
Ann [662]

The number of tubs that each of them sold is; 24 tubs each

The number of days it will take for both of them to sell same amount of tubs is; 4 days

Number of cookies that Nicole had already sold = 4 tubs

Number of cookies sold by Josie before counting = 0 cookies

Nicole now sells 5 tubs per day and

Josie now sells 6 tubs per day.

Let the number of days it will take for them to have sold the same amount of cookies be d.

Thus;

5d + 4 = 6d + 0

6d - 5d = 4

d = 4 days

Thus, total number of cookies for both are;

Total for Nicole = 4 + 5(4) = 24 cookies

Total for Josie = 6(4) = 24 cookies

Read more about proportion at; brainly.com/question/870035

6 0
3 years ago
(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
olya-2409 [2.1K]

Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

equation is:

75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

5 0
3 years ago
Eight energy corporations made plans to increase their combined spending on efficiency programs to $50 million per year for the
tangare [24]

Answer:

F=531831381

Explanation:

There are two  ways of doing this question:

1) By Formula

2) By Using  Compound interest Table

By Formula:

F=A*\frac{(1+i)^{n}-1}{i}

Where:

F is future value

A is annual amount per year

i is interest rate

n is number of years

F=50 million*\frac{(1+0.08)^{8}-1}{0.08}

F=531831381

By Using  Compound interest Table:

F=A(F/A,i,n)

From Table F/A at i=8% and n=8 is 10.6366

F=50000000*(10.6366)

F=531830000≅531831381

The difference in two answers is due to decimal points if you take value from table to greater decimal points you will get the exact answer as by using formula.

7 0
3 years ago
You recall an algorithm from elementary school for factoring a number N: Divide out all factors of 2, then of 3, then of 4, then
Contact [7]

Answer:

let number = 0

while number < 1

  begin

     print "Enter a positive integer: "

     read number

  end

end_while

find and print number's factors:

let prime = TRUE

let currentFactor = 2

let lastFactor = the square root of number truncated

  to an integer value

while currentFactor <= lastFactor

  begin

     if number is evenly divisible by currentFactor

        begin

           print currentFactor

           let number = number / currentFactor

        end

     else

        let currentFactor = currentFactor + 1

     end_if

  end

end_while

print a message if number is prime:

if prime == TRUE

  print "Your number is prime"

end_if

Explanation:

4 0
3 years ago
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