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hram777 [196]
3 years ago
6

Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft

of pipe length required to maintain this flow at the specified rate.
The density and dynamic viscosity of water at 70°F are rho = 62.30 lbm/ft^3 and μ = 6.556 x 10^-4 lbm/ft*s. The roughness of copper tubing is 5 x 10^-6 ft.
The pumping power per ft of pipe length required to maintain this flow at the specified rate = _________ W (per ft length)

Engineering
1 answer:
Lelu [443]3 years ago
8 0

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

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Aleks [24]

Answer:

Technician A is corect

Explanation:

Because he said the correct reason

6 0
3 years ago
Write a do-while loop that continues to prompt a user to enter a number less than 100, until the entered number is actually less
chubhunter [2.5K]

Answer:

#include <iostream>//including iostream library to use functions such as cout and cin

using namespace std;

int main() {

int userInput = 0;

do

{

 cout << "Enter a number < 100: " ;

 cin >> userInput;

 if (userInput < 100)//condition if number is less than 100

 {

  cout << "Your number < 100 is: " << userInput << endl;

 }

} while (userInput > 100);//do while loop condition

return 0;

}

Explanation:

A do-while loop executes regardless in the first iteration. Once it has run through the first iteration, it checks if the condition is being met. If, the condition is TRUE, the loop begins the second iteration. If FALSE, the loop exits. In this case, the condition is the userInput. after the first iteration, lets say the userInput is 120, the condition userInput > 100 is true.Therefore, the loop will run again until eventually the number is less than hundred, lets say 25. In that case the condition would be 25 > 100, which would be false, so the loops will break.

8 0
4 years ago
Write IEEE floating point representation of the following decimal number. Show your work.<br> 1.25
lisabon 2012 [21]

Answer:

00111111101000000000000000000000

Explanation:

View Image

0   01111111   01000000000000000000000

The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.

The next 8-bits are for the exponents.

The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.

And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.

You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰

The exponent bit value is 127+0=127 which is 01111111 in binary.

The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:

010...00000000000000000000

6 0
3 years ago
A dead-man system should shut off fuel flow within ____ of the system’s maximum flow rate.
AysviL [449]

Answer:

I think it is 10 percent because

3 0
4 years ago
Read 2 more answers
Find the magnitude of the steady-state response of the system whose system model is given by dx(t)/dt+ x(t)-f(t), where f(t) 2co
NISA [10]

This question is incomplete, the complete question is;

Find the magnitude of the steady-state response of the system whose system model is given by

dx(t)/dt + x(t) = f(t)

where f(t) = 2cos8t.  Keep 3 significant figures

Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

Given that;

dx(t)/dt + x(t) = f(t)  where f(t) = 2cos8t

dx(t)/dt + x(t) = f(t)

we apply Laplace transformation on both sides

SX(s) + x(s) = f(s)

(S + 1)x(s) = f(s)

f(s) / x(s) = S + 1

x(s) / f(s) = 1 / (S + 1)

Therefore

transfer function = H(s) = x(s)/f(s) = 1/(S+1)

f(t) = 2cos8t →   [ 1 / ( S + 1 ) ]   →  x(t) = Acos(8t - ∅ )

A = Magnitude of steady state output

S = jw

S = j8

so

A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )

A = 2/√65 = 0.2481

∅ = tan⁻¹( 1/1) = 45°

therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )

7 0
3 years ago
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