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german
3 years ago
5

Can a heat engine operating with only one thermal reservoir?

Engineering
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

No

Explanation:

Heat engine is a device which operates between two thermal reservoir.One thermal reservoir supply the thermal energy to the heat engine and another thermal reservoir absorb heat rejected by heat engine.These thermal reservoir are infinite amount of heat capacity system.Temperature of thermal reservoir remain constant always.

So we can say that heat engine can not operate by using only one thermal reservoir.

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A car is about to start but it blows up. what is the problem with the car<br> ?
ratelena [41]

Answer:

because there is a bomb

6 0
3 years ago
Read 2 more answers
Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mec
olga55 [171]

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

6 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
3 years ago
What does the supply chain management process involve
VMariaS [17]

Answer:

It involves the active streamlining of a business's supply-side activities to maximize customer value and gain a competitive advantage in the marketplace

Explanation:

Supply chain management is the management of the flow of goods and services and includes all processes that transform raw materials into final products.

5 0
3 years ago
From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% ar
egoroff_w [7]

Answer:

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

= 0.75*50*60 + 0.25*50*40

=2750 vehicles / hour

(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(c) Percentage of truck by aerial photo = 25%

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

4 0
3 years ago
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