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damaskus [11]
3 years ago
15

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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What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP
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Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

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Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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4 years ago
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A mixing basin in a sewage filtration plant is stirred by a mechanical agitator with a power input/WF L T=. Other parameters de
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Answer: π= G[√(u.V/W)]

STEP 1

Given parameters:

Power Input W= FL/T,

Absolute Viscosity u= FT/L²

Basin volume V= V/L³

Velocity gradient G= V/L³

STEP 2

We start by expressing the velocity gradient G as a function of W, u, V

G= G(W,u,V)

To get the pii terms, we use the dimension number formula n=k - r

where n and k are natural numbers representing number of fundamental dimensions and variable present respectively.

n= 4-3=1

STEP 3:

We expressed the pii terms as

π= G.W^a.u^b.V^c

The three fundamental F L T

We can write as

Fⁿ.Lⁿ.Tⁿ= 1/T. (FL/T)^a.(FT/L²)^b.(L³)

Using the exponential rule and by comparing coefficient on both sides;

Fⁿ.Lⁿ.Tⁿ= F^a+b. L^a-2b+3c. T^-a+b-1

Fⁿ= F^a+b = a+b= 0..............I

Lⁿ= L^a-2b+3c=0 = a-2b+3c=0...........ii

Tⁿ=L^-a+b-1=0. -a+b-1=0............iii

From the above equations we have,

a+b =0: b=-a...........iv

putting eq. iv into iii , we have

-a-a-1=0: -2a-1=0: a= -1/2

substituting the above value of a into eq iv, we have

b= 1/2

substituting the value of b above into eq 2, we have,

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Lastly, from the pii terms given above we can obtain dimensionless relationship,

π=G(W^-1/2.u^1/2.V^1/2)

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5 0
4 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
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<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

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