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3 years ago

There are four distinct events that occur in the Carnot cycle. Name the events and describe each one.

Engineering

1 answer:

OLEGan [10]3 years ago

8 0

Answer and Explanation:

There four events or stages of reversible nature in a Carnot cycle are:

1. Isothermal expansion of gas

2. Adiabatic expansion

3. Isothermal compression of gas

4.Adiabatic compression of gas

All these 4 processes are reversible processes.

1. A reversible Isothermal expansion of gas takes place in Carnot cycle where an ideal gas absorbs or intake certain quantity of heat from a heat reservoir or source at an elevated temperature which results in the expansion of gas and work is done on the surroundings.

2. A reversible expansion of gas adiabatically takes place in Carnot cycle in an environment with thermal insulation where the gas expand and work is done on the surrounding which results in the lowering of the temperature of the system.

3. A reversible compression of gas isothermally takes place such that the work is done on the system by the surrounding resulting in heat loss.

4. A reversible compression of gas adiabatically, takes place in an environment with thermal insulation and the work is done continuously on the system by the surroundings resulting in the rise in temperature.

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A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur

SOVA2 [1]

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W $m^{-1}K^{-1}$

diameter (d1) = 10 cm

so radius (r1) = 10 /2 = 5 cm = 5 × $10^{-2}$

radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × $10^{-2}$

temperature = 25°C

surface heat transfer coefficient = 6 6 W $m^{-2}K^{-1}$

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to 15°C

first pass through through resistance R1 i.e.

R1 = ( r2 - r1 ) / 4 $\pi$ × r1 × r2 × K

R1 = ( 5.2 - 5 ) $10^{-2}$ / 4 $\pi$ × 5 × 5.2 × 16 × $10^{-4}$

R1 = 3.825 × $10^{-3}$

same like we calculate for resistance R2 we know i.e.

R2 = 1 / ( h × area )

here area = 4 $\pi$ r2²

area = 4 $\pi$ (5.2 × $10^{-2}$ )² = 0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 × $10^{-3}$ + 4.90499

heat loss = 2.037152 W

8 0

3 years ago

List irreversibilities

natali 33 [55]

Answer:

Some of the irreversibilities are listed below:

Plastic deformation of solids
Transfer of heat over finite difference of temperature
When two fluids are mixed together the process is irreversible
Combustion of a gas
Current flowing through a finite resistor
Diffusion and free compression or expansion of gas
Relative motion of body with force of friction
Processes involving chemical reactions(spontaneous)

6 0

3 years ago

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?

IceJOKER [234]

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,