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FromTheMoon [43]
3 years ago
12

During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem

ains at 11°C. An inventor claims to have devised a power cycle working between these temperatures having a thermal efficiency of 10%.
Investigate this claim.

(a) What is the maximum thermal efficiency for these conditions?

(b) Is the inventor's claim possible?
Engineering
1 answer:
maw [93]3 years ago
4 0

Answer:

a) \eta_{th} = 10.910\,\%, b) Yes.

Explanation:

a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K}  \right) \times 100\,\%

\eta_{th} = 10.910\,\%

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

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What flight patterns do groups of birds utilize and why?
lara31 [8.8K]
Birds mostly fly in a V because the lead bird cuts al of the wind and has the hardest work and then after a while the lead bird goes to the back and another bird takes his place cutting all the wind for the other birds
7 0
3 years ago
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
2 years ago
A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force
svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required  hydraulic flow rate liquid if working pressure and  if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time  = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is  =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

5 0
3 years ago
Plzzzz helppp design process in order
MA_775_DIABLO [31]

Answer:

generate

define

present

evaluate

develop

construct and test

7 0
2 years ago
What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit
jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

v=\dfrac{K}{n_e}\dfrac{dh}{dl}

v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s

v=3.22\times 10^{-6}\ m/s

The velocity in m/d      ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0
2 years ago
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