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2 years ago

2) The switch in the circuit below has been closed a long time. At t=0, it is opened.

Engineering

1 answer:

saul85 [17]2 years ago

8 0

Answer:

il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

il(t) = e^(-t/.01)

il(t) = e^(-100t) . . . amperes

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Refrigerant 134a enters a horizontal pipe operating at steady state at 40°C, 300 kPa, and a velocity of 25 m/s. At the exit, the

salantis [7]

Answer:

a) 2.42 $kg/s$

b) 37.20 m/s

c) 120.56 kW

Explanation:

Given that:

The fluid in the Refrigerant = R-134a

Diameter (d) = 0.1 m

In the Inlet:

Temperature $T_1 = 40^0C$

Pressure $P_1= 300kPa$

Velocity $V_1$ = 25 m/s

At the exit:

Temperature $T_2 = 90^0C$

Pressure $P_2 = 240 kPa$

From the Table A-12 for Refrigerant R-134a at $T_1 = 40^0C$ and $P_1= 300kPa$

Specific Volume $v_1 = 0.0809 m^3/kg$

From the Table A-12 for Refrigerant R-134a at $T_2 = 90^0C$ and $P_2 = 240 kPa$

Specific Volume $v_2 = 0.12038 kJ/kg$

Their corresponding Enthalpy $h_1$ and $h_2$ are as follows:

Enthalpy $h_1$ =284.05 kJ/kg

Enthalpy $h_2$ = 333 kJ/kg

a) The mass flow rate of the refrigerant can be calculated as :

$m_1 = \frac{AV_1}{v_1}$

$m_1 = \frac{\frac{\pi (0.1)^2}{4}*25}{0.08089}$

$m_1 = 2.42 kg/s$

b) The velocity at the exit point:

we knew that:

$m=m_1 =m_2$

∴

$\frac{AV_1}{v_1} =\frac{AV_2}{v_2}$

$V_2 = \frac{v_2}{v_1} V_1$

$V_2 = \frac{0.12038}{0.08089} *25$

$V_2 = 37.20 m/s$

c) Expression for calculating heat transfer (as long as there is no work that is said to be done and the pipe is horizontal) can be represented as:

$Q_{cv}= m[(h_2-h_1)+\frac{1}{2}(V_2^2-V_1^2)]$

$Q_{cv}= 2.42*[(333.49-284.05)+\frac{1}{2}(37.20^2-25^2)]$

$Q_{cv}= 2.42*[49.44+379.42]$

$Q_{cv}= 119.6448kW+918.19W(\frac{1kW}{1000W} )$

$Q_{cv}= 119.6448kW+0.92 kW$

$Q_{cv} = 120.56 kW$

3 0

3 years ago

According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca

Ilia_Sergeevich [38]

Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

Explanation:eNotes was founded in 1998 by Brad Satoris and Alexander Bloomingdale, to enable and enhance the ability of students to solve assignments as given and other home works,eNote is also used by students to prepare effectively and efficiently for examinations.

Since its launch eNote has been a great source of educational materials for students who have gained a lot in the use of its materials.

bullwhip effect is a concept in supply chain, where forcasts causes the supply chain to be inefficient, it usually starts from retailers who raise concern of high demands.

CRMWRONGEDLP, is the program that minimizes the bullwhip effects according to eNote.

7 0

3 years ago

What’s the answer???

Zanzabum

D IS THE ANSWER ❤︎❤︎❤︎spread love and share knowledge

3 0

3 years ago

Describe what V1-V4 is

photoshop1234 [79]

Answer:

it is wave inversion

Explanation:

8 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

3 years ago