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kherson [118]
3 years ago
8

Ask a question about your assignment

Engineering
2 answers:
lesantik [10]3 years ago
6 0

Answer:

Why is school so hard lol

Explanation:

Shalnov [3]3 years ago
6 0

Answer:

look at each of the section headings below and make a prediction about the main idea of each section.

Explanation:

Can you help me with this please?

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A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,
Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

10

−

11

N

6 0
3 years ago
Read 2 more answers
Is the COP of a heat pump always larger than 1?
Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump=  1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.  

3 0
3 years ago
Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Maru [420]

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

5 0
3 years ago
If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0
3 years ago
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