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2 years ago

7

A trap in the condensate line prevents ____.A) air and foreign material from being drawn up into the drain line. B.) foreign par

ticles

1 answer:

BartSMP [9]2 years ago

7 0

A trap in the condensate line prevents D. All of the above.

<h3>What is the idea of the trap?</h3>

It should be noted that the idea of the condensate drain trap is to use the weight of the water to stop the flow of air.

Based on the information given, all the options are correct. Therefore, the correct option is D.

a. air and foreign material from being drawn up into the drain line

b. additional cooling load from warm air drawn up into the air handler

c. insect invasion of the system

d. all of the above

Learn more about cooling on:

brainly.com/question/13748261

#SPJ12

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Which of the following best describes a central idea of the text?

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During an office party, an office worker claims that a can of cold beer on his table warmed up to 20oC by picking up energy from

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Answer:

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Explanation:

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3 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

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

7 0

3 years ago

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago