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3 years ago

7

Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui

d columns is 3.2 cm and the density of oil is 860 kg/m3 .

1 answer:

Burka [1]3 years ago

6 0

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

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Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti

12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

4 0

2 years ago

Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

CAo = 16.1 / 2 mol/dm^3

Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = $\frac{Xai}{1-Xai}$ ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm --- ( 2 )

<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>

XA2 = 0.90

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4 0

3 years ago

Find the total present worth of a series of cash flows with an annual interest rate of 2% per year. Round your answer to the nea

prisoha [69]

The total present worth is $19,783.01

The present worth of a series of cash flow is the value of the cash flows in year 0 (today)

Cash flow in year 0 = 5330

Cash flow in year 1 = 0

Cash flow in year 2 = 0

Cash flow in year 3 = 13075 / (1.02)^3 = 12,320.86

Cash flow in year 4 = 2308 / (1.02)^4 = 2,132.24

Present worth = $19,783.01

A similar question was solved here: brainly.com/question/9641711?referrer=searchResults

5 0

2 years ago

Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67

defon

Answer:

#include <iostream>

#include <string>

using namespace std;

bool isPalindrome(string str)

{

int length = str.length();

for (int i = 0; i < length / 2; i++)

{

if (tolower(str[i]) != tolower(str[length - 1 - i]))

return false;

}

return true;

}

int main()

{

string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};

int i;

for(i=0; i<6; i++)

{

//Testing function

if(isPalindrome(s[i]))

{

cout << "\n " << s[i] << " is a palindrome... \n";

}

else

{

cout << "\n " << s[i] << " is not a palindrome... \n";

}

}

return 0;

}

5 0

3 years ago

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

Δ $U = Q + W$ ⇒ $Q =$ Δ $U -W$

$Q = 79 - (-63) = 142 kJ$

Therefore, the gas gained heat by an amount of 142 kJ.

3 0

3 years ago