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2 years ago

Question

Engineering

1 answer:

Leto [7]2 years ago

4 0

Answer:

True

Explanation:

The CNC is the primary interface between the machine operator and the machine.

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A block of copper (SL model) is subjected to a change of state. The process is isentropic (that is, entropy does not change. Whi

Fantom [35]

To develop this problem it is necessary to apply the definitions of entropy change within the bodies

The change of entropy in copper would be defined as

$\Delta S= \frac{\delta Q}{T}$

Where,

Q= Heat exchange

T = Temperature

For an incompressible substance, the change in the heat exchange is defined as

$\Delta Q =mcdT$

Where,

m = Mass

c = Specific heat

Replacing in our equation we have that

$\Delta S = \frac{mcdT}{T}$

$\Delta S = mc ln \frac{T_f}{T_i}$

Since $\Delta S = 0$ , then

$mcln\frac{T_f}{T_i} = 0$

$T_f = T_i$

In this way for the change of enthalpy and internal energy you have to

$\Delta U = \Delta h = mc\Delta T = mc(T_f-T_i)$

As $Tf_= T_i$ , then

$\Delta U = 0, \Delta H = 0$

Therefore the correct option is A. No change at All

5 0

2 years ago

A 2 nominal schedule 40 PVC pipe is 75 ft long. It is to convey water. The available pump can provide a pressure drop of 1.84 ps

nexus9112 [7]

Answer:

10.22 m^3/s

Explanation:

To estimate the pressure drop in a pipe we use the Darcy-Weisbach equation

$\frac{(deltaP)}{L} =\alpha *(\frac{density}{2})*(\frac{v^{2}}{D})$ (equation 1)

With:

$\alpha$ = Darcy-Weisbach friction coefficient

L = length of duct or pipe

v = velocity of fluid

D= hydraulic diameter

Also flow rate is:

$Q=v*A$

Where v is:

$v=\frac{Q}{A}$

Area as a function of the diameter is:

$A=\pi *\frac{D^{2}}{4}$

$v=\frac{4*Q}{\pi*D^{2}}$ (equation 2)

For a laminar regime the the Darcy-Weisbach friction coefficient is function of the Reynolds number (Re) as:

$\alpha=\frac{64}{Re}$

$Re=\frac{density*v*D}{u}$

With: v =velocity, D= diameter if the pipe and u= viscosity.

With this information alpha would be:

$\alpha=\frac{64*u}{density*v*D}$ (equation 3)

Replacing equation 3 in equation 1 we have:

$\frac{(deltaP)}{L} =\frac{32*u*v}{D^{2}}$

And finally replacing the value for v in this equation we have:

$\frac{(deltaP)}{L} =\frac{128}{\pi}*\frac{u*Q}{D^{4}}$

Clearing for Q we get an expression to estimate the expected flow rate in the pipe.

$Q=\frac{deltaP}{L}*\frac{\pi}{128}*\frac{D^{4}}{u}$

We know

Delta P = 1.84 psi or $\frac{lb}{in^{2}}$

L= 75 ft or 900 in

D for a 2 nominal schedule 40 PVC is 2.047 in. In tables you find External diameter and internal diameter. For calculations you use internal diameter (ID)

U for water at 20°C is $2.034*10^{-5}\frac{lb*s}{ft^{2}}$ or $1.4113*10^{-7}\frac{lb*s}{in^{2}}$

$Q=\frac{1.84\frac{lb}{in^{2}}}{900in}*\frac{\pi}{128}*\frac{(2.047in)^{4}}{1.4113*10^{-7}\frac{lb*s}{in^{2}}} = 6.24*10^{3}\frac{in^{3}}{s}$

So the flow expected for this pipe is $6.24*10^{3}\frac{in^{3}}{s}$ or $10.22 \frac{m^{3}}{s}$

7 0

3 years ago

Diodes are used in DC circuits but not AC circuits.<br> A. True<br> B. False

Juliette [100K]

Answer:

True diodes are not allowed at AC circuits

6 0

1 year ago

Which one of these gases needs the most amount of energy to ignite? Select one: 1. Butane 2. Propane 3. Ethylene 4. Hydrogen 5.

Anvisha [2.4K]

Answer:

The correct option is: 2. Propane

Explanation:

Minimum ignition energy, abbreviated as MIE, is the minimum energy required to ignite the flammable mixture of air at atmospheric pressure and room temperature. The unit used to measure this energy is millijoule, mJ

The values of the minimum ignition energy for the following gases are:

Butane = 0.25 mJ

Propane = 0.26 mJ

Ethylene = 0.07 mJ

Hydrogen = 0.017 mJ

Cyclohexane = 0.22 mJ

<u>Therefore,among the following gases, Propane requires the most ignition energy, 0.26 mJ. </u>

3 0

2 years ago

As you increase the amount of pulleys in a pulley system, the mechanical

hammer [34]

Answer:

The correct option is;

Easier

Explanation:

The force required to lift a given load reduces when the number of pulleys in a pulley system is increased because the number of ropes sharing the applies pulling force is increased, however, the distance by which the rope is to be pulled is doubled when the required force to lift a given load is half the weight of the load.

7 0

2 years ago