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guapka [62]
3 years ago
12

A cold air-standard Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum temperature of t

he air and the rate of heat addition to this cycle when it produces 150 kW of power and the state of the air at the beginning of the compression is 90 kPa and 57 °C. Use constant specific heats at room temperature.
Engineering
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

\dot Q_{in} = 228.659\,kW, T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

Explanation:

The ideal efficiency of the Diesel cycle is given by this expression:

\eta_{th} = \left\{1 - \frac{1}{r^{k-1}} \cdot \left[\frac{r_{c}^{k}-1}{k\cdot (r_{c}-1)} \right]\right\}\times 100\%

Where r and r_{c} are the compression and cutoff ratios, respectively.

\eta_{th} = \left\{1-\frac{1}{18^{0.4}}\cdot \left[\frac{1.5^{1.4}-1}{1.4\cdot (1.5-1)} \right] \right\}\times 100\%

\eta_{th} = 65.648\,\%

The heat addition to the cycle is:

\dot Q_{in} = \frac{\dot W}{\eta_{th}}

\dot Q_{in} = \frac{150\,kW}{0.656}

\dot Q_{in} = 228.659\,kW

The temperature at state 2 is:

T_{2} = T_{1} \cdot r^{k-1}

T_{2} = (330.15\,K)\cdot 18^{0.4}

T_{2} = 1049.108\,K\,(775.958\,^{\textdegree}C)

And the temperature at state 3 is:

T_{3} = T_{2}\cdot r_{c}

T_{3} = (1.5)\cdot (1049.108\,K)

T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

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Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V_s/( 1 - D )

given that; V_s = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

I_L = V_s / ( 1 - D )²R

given that R = 12.5 Ω, V_s = 20 V, D = 0.6

we substitute

I_L = 20 / (( 1 - 0.6 )² × 12.5)

I_L = 20 / (( 0.4)² × 12.5)

I_L = 20 / ( 0.16 × 12.5 )

I_L = 20 / 2

I_L = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmax = [20 / 2 ] + [ 60 / 20 ]    

I_{Lmax = 10 + 3

I_{Lmax = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmin = [20 / 2 ] -[ 60 / 20 ]    

I_{Lmin = 10 - 3

I_{Lmin  = 7 A

Therefore, the maximum inductor current is 7 A

 

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

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Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy =  370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = ln\frac{1}{1-0.2} = 0.223

so

yield strength after cold is

yield strength = K \varepsilon ^n

yield strength =  600*0.223^{0.21)

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = \frac{\sigma u}{1-W}

ultimate strength = \frac{440}{1-0.2}

ultimate strength after cold work = 550 MPa

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Answer:

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Answer:

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Explanation:

NB: kindly check below for the attached picture.

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The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.

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