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Alex73 [517]
3 years ago
11

Sơ đồ nguyên lý của một hệ thống lạnh

Engineering
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

A = 5h (B + b); solve for B.

2

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I want a problems and there solutions of The inception of cavitation?​
Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

  • The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
  • Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

    ⇒  A cavitation number is denoted by "σ" .

4 0
3 years ago
Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r
Evgesh-ka [11]

Answer:

Explanation:

In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.

For the step by step solution to the question you asked, go through the attached documents.

4 0
3 years ago
Whats a meat trimmer
baherus [9]

Answer:

trime the meat

Explanation:

4 0
3 years ago
What type of fire extinguisher is used for electrical fires.
AysviL [449]
The all purpose one will work or you could use type C I believe.
8 0
3 years ago
Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is
ryzh [129]

Answer:

a) \dot W = 0.978\,kW, b) I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}

COP_{HP} = 14.198

The reversible work input is:

\dot W = \frac{\dot Q_{H}}{COP_{HP}}

\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

\dot W = 0.978\,kW

b) The irreversibility is given by the difference between real work and ideal work inputs:

I = \dot W_{real} - \dot W_{ideal}

I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

7 0
4 years ago
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