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4 years ago

Identify each statement as referring to a series or parallel circuit.

Engineering

1 answer:

jolli1 [7]4 years ago

7 0

Answer:

1. Parallel circuit

2. Parallel circuit

3. Series circuit

4. Series circuit

5. Parallel circuit

6. Parallel circuit

Explanation:

1. In a parallel circuit, there are multiple paths for current to flow. The path each current takes depends on the resistance of the resistors on that path.

2. In a parallel circuit, current splits up into various paths to get the total current through the circuit, the current flow through each resistor is added.

3. In a series circuit the voltage across each resistor is not the same. to get back the total voltage, the voltages across each resistor needs to be added.

4. Series circuits have voltage drops across each resistor. this makes the voltage across each resistor depend on the resistance of the resistor.

5. In parallel circuits voltage is the same across each resistor because they are all connected directly to the same source.

6. In parallel circuits, the power is the same in each resistor of equal resistance since the voltage across each resistor is the same

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Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago

What is the focus of 7th grade civics

Jobisdone [24]

C. seems like the best answer. i may be wrong so don’t quote me on that

7 0

3 years ago

Read 2 more answers

Select the correct answer. Jennifer, a construction manager at a construction firm, has to hire several contractors and subcontr

Ymorist [56]

A. Quality management

7 0

3 years ago

Read 2 more answers

Distinguish between systems analysis and systems design?

Zielflug [23.3K]

Answer:

System analysis can be defined as a deep analysis of a part of the structure of a module that has been designed before. System design means to make any module or a part of the structure from scratch and build it completely without estimation.

Explanation:

3 0

3 years ago

Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m

Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0

3 years ago