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2 years ago

What would be the structure for the body points for a persuasive presentation?.

1 answer:

5 0

A persuasive speech is structured like an informative speech. It has an introduction with an attention-getter and a clear thesis statement. It also has a body where the speaker presents their main points and it ends with a conclusion that sums up the main point of the speech.

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A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

Molodets [167]

Answer:

Explanation:

Mean temperature is given by

$T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}$

Tmean = (Ti + T∞)/2

$T_mean = 107.5^{0}$

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

$\mu$ = 221.6 × 10⁻⁷N.s/m²

$\kappa$ = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/ $\mu$

Substituting for Pv

Re = 4m/(πD $\mu$ )

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0

2 years ago

Joey has a car that uses the hand crank to open the windows. Joey is wondering where the energy comes from to open the windows.T

Sedaia [141]

Explanation:

Joey has a car that uses the hand crank to open the windows. Joey is wondering where the energy comes from to open the windows.The sunHuman-powered energy from JoeyThe hand crankThe moving car

5 0

2 years ago

In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t

Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0

3 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

3 years ago

Which group might be classified as a gang?

Furkat [3]

Answer:

The following criteria are commonly used for classifying groups as gangs: The group has three or more members, generally aged 12–24. Members share an identity, typically linked to a name, and often other symbols. Members view themselves as a gang, and they are recognized by others as a gang.

Explanation:

8 0

2 years ago

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