A 2-stage dcv that has an internal pilot does not work well (if at all) on

Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air con

PSYCHO15rus [73]

Answer:

See explaination

Explanation:

The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.

Kindly check attachment for the step by step solution of the given problem.

4 0

3 years ago

A 0.4-W cylindrical electronic component with diameter 0.3 cm and length 1.8 cm and mounted on a circuit board is cooled by air

Katyanochek1 [597]

Answer:

The surface temperature of the component 54.6 degrees celsius.

Explanation:

Please see attachment.

7 0

3 years ago

Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential

Deffense [45]

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/( $A^{*}$ ₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃ in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403 $e^{j51.34 }$

c) value of A₃ in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937 $e^{j81.34 }$

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁ = 5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( $\frac{5}{4}$ )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A $e^{j\beta }$

so, A₂ in exponential form will be;

A₂ = 6.403 $e^{j51.34 }$

exponential form of A₂ = 6.403 $e^{j51.34 }$

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( $\frac{8 }{9.196}$ )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°

A₃ = 12.19∠41.02°

Therefore, value of A₃ in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/( $A^{*}$ ₂)

giving your answer in exponential form

we know that $A^{*}$ ₂ is the complex conjugate of A₂

so

$A^{*}$ ₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A $e^{j\beta }$

$A^{*}$ ₂ = 6.403 $e^{-j51.34 }$

also

A₁ = 6∠30

we convert to polar form

A₁ = 6 $e^{j30 }$

so A₅ = A₁/( $A^{*}$ ₂)

A₅ = 6 $e^{j30 }$ / 6.403 $e^{-j51.34 }$

A₅ = (6/6.403) $e^{j(30+51.34) }$

A₅ = 0.937 $e^{j81.34 }$

Therefore A₅ in exponential form is 0.937 $e^{j81.34 }$

6 0

3 years ago

Write a program to control the operation of the RED/GREEN/BLUE LED (LED2) as follows: 1. If no button is pressed, the LED should

aalyn [17]

Answer:

See explaination

Explanation:

int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }

4 0

3 years ago

What is engine knock? What cause the engine knock problem?

antiseptic1488 [7]

Answer:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

Explanation:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

The engine knock problem can be caused due to the following reason

a) When the octane rating of the fuel used is low.

b) The deposition of the carbon around the cylinder walls takes place.

c) The spark plug used in the vehicle is not correct.

3 0

3 years ago