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A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?

Mazyrski [523]

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

4 0

4 years ago

Neon is compressed from 100 kPa and 20◦C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

PIT_PIT [208]

Answer:

The specific volume is reduced in 80 per cent due to isothermal compression.

Specific enthalpy remains constant.

Explanation:

Let suppose that neon behaves ideally, the equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in cubic meters.

$n$ - Molar quantity, measured in kilomoles,

$T$ - Temperature, measured in kelvins.

$R_{u}$ - Ideal gas constant, measured in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

On the other hand, the molar quantity ( $n$ ) and specific volume ( $\nu$ ), measured in cubic meter per kilogram, are defined as:

$n = \frac{m}{M}$ and $\nu = \frac{V}{m}$

Where:

$m$ - Mass of neon, measured in kilograms.

$M$ - Molar mass of neon, measured in kilograms per kilomoles.

After replacing in the equation of state, the resulting expression is therefore simplified in term of specific volume:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$P\cdot \nu = \frac{R_{u}\cdot T}{M}$

Since the neon is compressed isothermally, the following relation is constructed herein:

$P_{1}\cdot \nu_{1} = P_{2}\cdot \nu_{2}$

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$\nu_{1}$ , $\nu_{2}$ - Initial and final specific volume, measured in cubic meters per kilogram.

The change in specific volume is given by the following expression:

$\frac{\nu_{2}}{\nu_{1}} = \frac{P_{1}}{P_{2}}$

Given that $P_{1} = 100\,kPa$ and $P_{2} = 500\,kPa$ , the change in specific volume is:

$\frac{\nu_{2}}{\nu_{1}} = \frac{100\,kPa}{500\,kPa}$

$\frac{\nu_{2}}{\nu_{1}} = \frac{1}{5}$

The specific volume is reduced in 80 per cent due to isothermal compression.

Under the ideal gas supposition, specific enthalpy is only function of temperature, as neon experiments an isothermal process, temperature remains constant and, hence, there is no change in specific enthalpy.

Specific enthalpy remains constant.

8 0

3 years ago

A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature

mafiozo [28]

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

6 0

4 years ago

University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, r

abruzzese [7]

Solution :

The percentage of the students who have a chance of repeating their current year = 3%

The drop out students for the first year and the sophomores = 6%

Drop out rate of first year and the seniors = 4%

Now for the state space :

S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }

Therefore

the first year students are indicated as '1'

Sophomores are indicated as '2'

Juniors are indicated as '3'

Seniors are indicated as '4

Graduates are indicated as 'G'

Dropouts are indicated as 'D'

The transition diagram is attached below.

The probability of the students who have the chance of repeating their current year = 3/100 = 0.03

Probability of first year dropouts and sophomores = 6/100 = 0.06

Probability of dropout rate of juniors and seniors = 4/100 = 0.04

Therefore, the probability matrix can be made as :

1 2 3 4 G D

$\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}$ $\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}$

Here, G represents 'graduates' and D represents 'Dropouts.'

5 0

3 years ago

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

Anit [1.1K]

Answer:

i dont know

Explanation:

4 0

3 years ago