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3 years ago

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3. Suppose that a class named Bicycle contains a private nonstatic integer named height, a public nonstatic String named model,

and a public static integer named wheels. Which of the following are legal statements in a class named BicycleDemo that has instantiated an object as Bicycle myBike new Bicycle C);? f. Bicycle. model Hurricane a. myBike height 26; b. my Bike model Cyclone g. Bicycle. int 3 3; c. myBike Wheels 3 d. my Bike .model 108; i. Bicycle wheels 2 e. Bicycle height 24; j. Bicycle yourBike myBike

1 answer:

Vilka [71]3 years ago

7 0

Answer:

The solution to the given problem is provided below.

Explanation:

a.) myBike.height = 26; Not Legal statement

b.) myBike.model = “Cyclone”: Legal statement

c.) myBike.wheels = 3; Legal statement

d.) myBike.model = 108; Not legal statement

e.) Bicycle.height = 24; Not Legal statement

f.) Bicycle.model = “Hurricane”; Not legal statement

g.) Bicycle.int = 3; Not Legal statement

h.) Bicycle.model = 108; Not Legal Statement

i.) Bicycle.wheels = 2; Legal Statement

j.) Bicycle yourBike = myBike; Legal Statement

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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

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4 years ago

Explain three (3) modes of heat transfer in air conditioning system.

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Answer:

1. Conduction

2. Convection

3. Radiation

Explanation:

The 3 modes of heat transfer i an air conditioning system:

1. Conduction:

The transfer of heat by conduction takes place in solid and is when the conduction takes place as a result of direct contact in between the interacting material which transfer the heat energy from particle to particle thus conducting the heat through out the system.

2. Convection:

The other mode for the transfer of heat which takes place especially in fluids - gases and liquids is through the technique of convection in which the transfer of heat takes place by the circular motion of the atoms and molecules of the fluid which carries the heat energy and results in the distribution of the heated fluid in the entire system thus transferring all the heat energy in the entire system.

3. Radiation:

The third mode of heat transfer in the air conditioning system is through radiation. This method transfers the heat by making use of the electro-magnetic radiation in the infra red spectrum where the waves of the spectrum transfers the heat energy with the help of a medium or without any medium at all.

Thus making the radiation method of heat transfer as the only method out of the three methods which does not require the material medium for the transfer of heat energy.

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3 years ago

Tools for forming aluminum tray

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Answer:

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3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

<em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

.<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

<em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

<em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition

zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

$V_{f}=0.61$

$V_{m}=1-V_{f}$

$=1-0.61$

$=0.39$

$E_{f}=10 \ Mpa$

$\sigma_{f}=0.35 \ Mpa$

$E_{m}=0.45 \ Mpa$ , $\sigma_{m}=9\times 10^{-3} \ Mpa$

As we know,

⇒ $E_{e}=fE_{f}+mE_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 10+0.39\times 0.95$

⇒ $=6.27 \ Mpa$

Now,

⇒ $\sigma_{c}=f\sigma_{f}+m\sigma_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 0.35+0.39\times 0.009$

⇒ $=0.217 \ Mpa$

Therefore,

The load carried by fiber,

$=\frac{f\sigma_{f}}{\sigma_{c}}$

$=\frac{0.35\times 0.61}{0.217}$

$=0.98$ i.e., 98%

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4 years ago