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3 years ago

The figure below appeared three heat treatments processes of steel (A, B and C),

Engineering

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Contact [7]3 years ago

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Answer:

Explanation:

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0

3 years ago

7–53 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at

qwelly [4]

Answer:

a. The coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is 1.96368kW

Explanation:

Given

First we need to get the enthalpy of R-34a.

When T = 35°C and P = 800kPa;

h1 = 271.24kj/kg

When x2 = 0 and P = 800kPa;

h1 = 95.48kj/kg

To calculate the COP, first we need to calculate the energy balance.

This is given as

Q = m(h1 - h2)

Where m = 0.018kg

Q = 0.018(271.24 - 95.48)

Q = 3.16368Kw

COP is then calculated as Q/W

Where W = Power consumption of the compressor = 1.2kW

COP = 3.16368Kw/1.2Kw

COP = 2.6364

Hence, the coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is calculated as ∆Heat Rate

∆Heat Rate = Q - W

Where Q = Energy Balance = 3.16368Kw

W = Power consumption of the compressor = 1.2kW

∆Heat Rate = 3.16368Kw - 1.2kW

∆Heat Rate = 1.96368kW

Hence, The rate of heat absorption from the outside air is 1.96368kW

6 0

3 years ago

Read 2 more answers

A. Imagine the white car in the left lane is moving more slowly than the surrounding traffic. How is this a violation of a state

asambeis [7]

The roads have pre-described lanes that should follow a particular speed and rules. When a driver drives slowly than he makes the other drivers impatient.

<h3>What are traffic rules?</h3>

Traffic rules are the regulations that are defined to regulate the vehicles and other conveyances that walk on the road. They are different in various places and have separate punishments for not following them.

The left lane of the road is generally for the drivers who want to pass by the vehicle ahead by following the mph rule. If the driver drives slowly then they make the other people impatient and result in honking.

Therefore, the driver driving slowly in the left lane makes others impatient.

Learn more about traffic rules here:

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2 years ago

Read 2 more answers

A three-phase, 318.75kVA, 2300-Volt, alternator has an armature resistance of 0.35/phase and a synchronous reactance of 1.2/phas

Darya [45]

Answer:

attached below

Explanation: