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SIZIF [17.4K]
3 years ago
5

In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The ar

tery bifurcates (splits) into two identical blood vessels that are each 3mm in diameter. What are the average velocity and the mass flow rate upstream and downstream of the bifurcation? The density of blood is 1.06g·cm−3
Engineering
1 answer:
antiseptic1488 [7]3 years ago
7 0

9514 1404 393

Answer:

  see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

  (5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

  (5.3 g/s)/2 = 2.65 g/s

__

The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

  (5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

  (2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s

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harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0
4 years ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
weqwewe [10]

Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

 cout << "Too small";

 cout << endl;

}

else if(bagOunces > 10){

 cout << "Too large";

 cout << endl;

}

else{

 cout << (6 * bagOunces) << " seconds" << endl;

}

}

int main() {

  PrintPopcornTime(7);

  return 0;

}

Explanation:

Using C++ to write the program. In line 1 we define the header "#include <iostream>"  that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

8 0
3 years ago
I need help with this question
Ad libitum [116K]

Answer:

LOL where is the question, that u need help with?

Explanation:

5 0
3 years ago
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic
DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm

2) The speed of the rotor is the motor speed. The slip is given by:

Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm

3) The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=1*60=60\ Hz

6 0
3 years ago
A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of
Sonbull [250]

Answer:

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

First, we calculate the torque while the load is being lifted in terms of 'F'.

This is calculated by

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

Efficiency = 16.67%

8 0
4 years ago
Read 2 more answers
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