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Sergio039 [100]
2 years ago
13

A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a volumetric flow rate of 0.1 lit

er/s through a 2.5-cm-diameter hose. A jet of water exits at 23 C, 1 atm, with a velocity of 50 m/s at an elevation of 5 m. At steady state, the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input. Determine the power input to the motor, kW.

Physics
1 answer:
VladimirAG [237]2 years ago
5 0

Answer:

W=1259W=1.2Kw

Explanation:

Hello!

The first step to solve is to find the enthalpies at the entrance (state 1) and the exit of the washer, for this we use the thermodynamic tables.

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

h1(T=20C,P=1atm)=83.93kJ/kg

state 2

h1(T=23C,P=1atm)=96.48kJ/kg

now we find the mass flow remembering that it is the product of the flow rate by the density=

m= mass flow

M=\alpha q

q=flow=0.1l/S=0.0001M^3/s

α=Density=1000kg/m^3

m=0.1kg/s

Now we draw the energy flows in the washer (see attached image) and propose the first law of thermodynamics that states that the energy that enters must be equal to the one that comes out

mh1+W+\frac{v1^2}{2g} =0.1W+mh2+\frac{v2^2}{2g}+mgH

where

m=mass flow

h=entalpy

v=speed

W=power input

g=gravity

H=height

SOLVING FOR W

W=\frac{m2h2-m1h1+m\frac{v2^2-v1^2}{g}+mgH}{1.1}

W=\frac{(0.1)(96480)-(0.1)(83930)+\frac{0.1(50^2-0.2^2)}{2}+0.1(9.81)(5)}{1.1}

W=1259W=1.2Kw

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Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa
Aleksandr-060686 [28]

Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

 

 

Given:

 

T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

P1 = 260 kPa

T1 = 747 degrees Celsius

V1 = 80 m/s ->nN = 92% -> P2 = 85 kPa

Solution:

From the isentropic relation,

Pr2<span> = (P2 / P1)PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>

 

There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

 

h2a = 1068.89 kJ/kg – (((728.2 m/s)­2 – (80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg\

From the air table, we read T2a  = 786.3 K

5 0
1 year ago
A skateboarder travels on a horizontal surface with an initial velocity of 3.2 m/s toward the south and a constant acceleration
Naddika [18.5K]

Answer:

A. 0.432

B. -1.92

C. 1.44 units/second

D. -3.2 units/second

Explanation:

A. To calculate her x position, we just use the following equation of motion to find the distance traveled:

    s=u*t+\frac{1}{2} (a*t^2)

here s = displacement

t = time (in seconds)

a = acceleration

Solving for the distance, we get:

s = 0 * 0.6 + \frac{1}{2}(2.4 * 0.6^2)

s = 0.432 m

Since 0.432 meters east is equals to 0.432 meter in the positive x-direction, the x position is also 0.432.

B. Since the skater has a constant v - velocity of -3.2 m/s, (south means negative y axis), the total distance traveled is:

Distance = speed * time = -3.2 * 0.6 = -1.92 m

The answer is -1.92 units in the y-axis.

C. The x velocity component is the final speed in the east direction, which is going to be:

v^2 - u^2=2*a*s

v^2 = 2*2.4*0.432

v = 1.44 units/second (in positive x direction)

D. Her y velocity component does not change, since the velocity towards the south is a constant 3.2 m/s

Thus the answer is -3.2 units/second in the y-axis.

8 0
2 years ago
You have found a treasure map that directs you to start at a hollow tree, walk 300 meters directly north, turn and walk 500 mete
Dmitry_Shevchenko [17]

Answer:633 m

Explanation:

First we have moved 300 m in North

let say it as point a and its vector is 300\hat{j}

after that we have moved 500 m northeast

let say it as point b

therefore position of b with respect to a is

r_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}

Therefore position of b w.r.t to origin is

r_b=r_a+r_{ba}

r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}

r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}

after this we moved 400 m 60^{\circ} south of east i.e. 60^{\circ} below from positive x axis

let say it as c

r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=r_{b}+r_{cb}

r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}

magnitude is \sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}

=633.052

for directiontan\theta =\frac{250\sqrt{2}+300-200\sqrt{3}}{250\sqrt{2}+200}

tan\theta =\frac{307.139}{553.553}

\theta =29.021^{\circ} with x -axis

7 0
2 years ago
The magnitude of the force associated with the gravitational field is constant and has a value F. A particle is launched from po
Goryan [66]

Answer:

Energy gained by the second particle = 12Uo

Explanation:

Given Data;

Resistant force = 12F

Initial kinetic energy = Uo

Calculating the kinetic energy gained, we have;

u = f *r

where f= resistant force = 20F

r = initial kinetic energy = Uo

Therefore,

U = 12 * uo

  = 12 Uo

Therefore, energy gained by the second particle = 12Uo

4 0
2 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Thepotemich [5.8K]

Answer:

T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

5 0
1 year ago
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