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3 years ago

Used oil is not always considered hazardous if managed in a prescribed manner.

Engineering

2 answers:

Lana71 [14]3 years ago

4 0

A) True because the oil has already ran through it’s cycle (oil change)

saw5 [17]3 years ago

3 0

Answer:

A. True

Explanation:

common sense.... I think

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What is the definition of a struck by injury

dmitriy555 [2]

Produced by forcible contact or impact between the injured person and an object or a piece of equipment

Explanation:

it mean you get injured after you've already been injured

3 0

3 years ago

Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a

defon

Answer:

Explanation:

Wow

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3 years ago

In a steam power plant, 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW.

Free_Kalibri [48]

Answer:

a) $\eta = 42\,\%$ , b) $COP_{R} = 29$

Explanation:

a) The thermal efficiency is:

$\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%$

$\eta = \frac{1\,MW-0.58\,MW}{1\,MW} \,\times 100\,\%$

$\eta = 42\,\%$

b) The coefficient of performance is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{0.58\,MW}{0.02\,MW}$

$COP_{R} = 29$

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3 years ago

You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your

olchik [2.2K]

Solution:

The given formula,

$x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}$

$y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}$

$\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}$

From the table,

$1) $\quad x=2 \cdot 87, \quad y=0.96$\\$\frac{x}{y}=\frac{2187}{0.96}$22895\\\\2) $x=3 \cdot 466 \cdot y=6 \cdot 998$\\$\frac{x}{y}=\frac{3 \cdot 466}{0.898}$$=3 \cdot 4729$$

$3$x=3 \cdot 229, y=1 \cdot 08$\\$\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}$\\=2.9898\)\\\\4) $x=6.525, y=1.094$\\$\frac{x}{y}=\frac{5.625}{1.094}$\\=5.0502$

8 0

4 years ago

The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (

mr Goodwill [35]

Answer:

a) the output voltage is 50 V

- 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

- 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

7 0

3 years ago