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fgiga [73]
2 years ago
14

Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra

gms can deteriorate over time. Who is correct
Engineering
1 answer:
diamong [38]2 years ago
8 0

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

Learn more about technicians on:

brainly.com/question/1548867

#SPJ12

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Answer:

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2 years ago
. A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa a
Delvig [45]

Answer:

Qin = 1857 kJ

Explanation:

Given

m = 0.5 Kg

T₁ = 25°C = (25 + 273) K = 298 K

P₁ = 100 kPa

P₂ = 500 kPa

First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:

T₂ = T₁*P₂/P₁

⇒  T₂ = 298 K*(500 kPa/100 kPa) = 1490 K

Until the piston starts rising no work is done so the heat transfer is the change in internal energy

Qin = ΔU = m*cv*(T₂-T₁)

⇒   Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ

8 0
3 years ago
a refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evapora
timurjin [86]

Answer:

I. 3.316 kW

II. 1.218 kW

III. 2.72

Explanation:

At state 1, the enthalpy and entropy are determined using the given data from A-13.

At P1 = 200kpa and T1 = 0,

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus

h2(s) = 295.95 kJ/kg

The actual enthalpy is then gotten by

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

h3 = h4 = 120.43 kJ/kg

Heating load is determined from energy balance, thus,

Q'l = m'(h1 - h4)

Q'l = 0.025(253.07 - 120.43)

Q'l = 0.025 * 132.64

Q'l = 3.316 kW

Power is determined by using

W' = m'(h2 - h1)

W'= 0.025(301.8 - 253.07)

W'= 0.025 * 48.73

W'= 1.218 kW

The Coefficient Of Performance is Q'l / W'

COP = 3.316/1.218

COP = 2.72

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Explanation:

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