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Sauron [17]
1 year ago
5

Sometimes we need to create heat, such as in circuit breakers and rear window

Engineering
2 answers:
Verizon [17]1 year ago
8 0

Answer: True

Explanation:

Circuit breakers are used to stop the current by using heat to detect if it has a critically large current. Defoggers use heat to stop condensation on the windows.

mrs_skeptik [129]1 year ago
4 0
The answer to your question is true
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An orchestra is having a recording done of 2 performances in the same concert hall. The first show is sold out. They struggled t
konstantin123 [22]

Answer:

yes, the recordings sound is same

Explanation:

given data

recording done = 2 performances

1st  show = sold out

2nd show =  lightly attended

to find out

recordings sound the same and why

solution

as per given in

  • 1st show is sold out it mean in this case concert hall is full so that recording sound should be high here
  • 2nd case only few people are attended and struggle for ticket  and orchestra

it mean it sound performance so in both case recording sound will be same

because we do not other all are sitting at front row or they sit as they want

4 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
Why do engineers play a variety of roles in the engineering process?
katrin2010 [14]

Answer:

b

Explanation:

7 0
2 years ago
X-Ray Inspection
Softa [21]
The answer is the test is being tested towards the lungs the test is done by scanning your body the tools are called “the x rat visional lock space” and the rubber tool is called a deeldo it’s purple with a pencil looking shape perfect for the body.
5 0
3 years ago
R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its
attashe74 [19]

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

5 0
2 years ago
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