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alexandr1967 [171]
2 years ago
9

Only answer this if your name is riley

Engineering
2 answers:
Fittoniya [83]2 years ago
8 0

Explanation: im not a reiley but thx for points BiG bRaIn PlaYz

Sati [7]2 years ago
3 0

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

You might be interested in
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz
sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0
2 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
2 years ago
An op-amp is connected in an inverting configuration with R1 = 1kW and R2 = 10kW, and a load resistor connected at the output, R
Svetllana [295]

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

  1. no current will go in the inverting(-) and noninverting(+) side of the opamp
  2. V₊ = V₋  , so whatever voltage is at the noninverting side will also be the voltage at the inverting side

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋  = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

8 0
3 years ago
Create a project named CarDealer that contains a Form for an automobile dealer. Include options for at least three car models. A
PIT_PIT [208]

Answer:

/****************** The code for a form application that uses radioButtons * *and displays new form based on the user selection *****************/

Using system;

Using system windows forms;

namespace CarDealer

{

Public partial class Form1 : Form

{

public Form1()

{

//default constructor to initialize components

InitializeComponent();

}

//when user clicks on the details button

private void detail_Click

(Object sender, EventArgs e)

{

if (model1,Checked)

{

Forms2 f = new Form2()

f.ShowDialog()

model1.Checked = false

}

if (model2.Checked)

{

Form3 f = new Form3();

f.ShowDialog();

model2.Checked = false;

}

if (model3.Checked)

{

Form4 f = new Form4();

f.ShowDialog();

model3.Checked = false;

}

}

}

}

Explanation:

Program plan

- Design form: Place label controls with text as select car model and one empty label control to display total price. Change front type and size from each other labels properties window.

- Add three radio buttons with text Renault Kwit, Tata Tiago, Mahindra KUV 100.

- Add button with text View Details.

- Add 3 new windows form from project menu for each car model.

- Change form name and respective car models.

- Place picturebox control to each form by using image property add .jpeg image and place label control containing price of each model to each form

- When user select car model and click on view details, new form containing car model details is displayed.

Form Design is attached below

7 0
2 years ago
Serves as a protective barrier to prevent contact with engergized ("hot") parts<br> within the unit
erik [133]

Answer:

thanks hot hot

Explanation:

4 0
3 years ago
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