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3 years ago

21.Why are throttling devices commonly used in refrigeration and air-conditioning applications?

Engineering

1 answer:

Sloan [31]3 years ago

5 0

Answer is given below

Explanation:

we know that some common types of throttling devices are

Hard -throttling devices
Capillary valve
Constant pressure throttling devices
Thermostatic expansion valve
Float expansion valve

so here throttling devices commonly used in refrigeration and air-conditioning because

To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature
To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.

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Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)

Novosadov [1.4K]

Answer:

a) ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

The phase is; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0

3 years ago

Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False

Valentin [98]

Answer:

I would say false but I am not for sure

8 0

2 years ago

How do I do this?<br> Blueprints, complete the missing view.

Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

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2 years ago

A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?

blagie [28]

Answer:

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0

2 years ago

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

3 years ago