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Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

$COP_{i} = \frac{250\,K}{300\,K-250\,K}$

$COP_{i} = 5$

The real Coefficient of Performance is:

$COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}$

$COP_{r} = 12.571$

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0

3 years ago

What is need for using fins?

antiseptic1488 [7]

Answer: It is a term of heat transfer process in which fins are surface that are the extension of the object to work for the heat exchangers to increase the heat exchanging rate.

Explanation: Fins are considered to help the heat exchanger surface to lead the process of heat transfer by increasing the are of the surface which is exposed to the surroundings. Fins work really well with materials having high thermal conductivity and will be more effective. They are preferred because they increase the rate of exchange of heat by increment in the convection.

7 0

3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

3 years ago

A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan

hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

$R = \dfrac{u^2}{15(e+f_s)}$

where;

R= radius

$f_s$ = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

$f_s$ = 0.20

So;

$R = \dfrac{30^2}{15(0.08+0.20)}$

$R = \dfrac{900}{15(0.28)}$

$R = \dfrac{900}{4.2}$

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0

3 years ago

6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4

S_A_V [24]

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

$\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb$

Mass of cement used is 62.5 lb

$\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb$

Mass of gravel used is 225 lb

7 0

3 years ago

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