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vekshin1
3 years ago
8

A 10-kW cooling load is to be served by operating an ideal vapor-compression refrigeration cycle with its evaporator at 400 kPa

and its condenser at 800 kPa. Calculate the refrigerant mass flow rate and the compressor power requirement when refrigerant-134a is used.
Engineering
1 answer:
Leni [432]3 years ago
5 0

Answer:

0.0625kg/s

0.9031kW

Explanation:

At state 1, enthalpy and entropy are determined from table R-134a.

P1 = 400kpa

h1 = 255.55kj

S1 = 0.92691

h2 = 270kj/kg

Enthalpy at states 3 and 4 can be gotten from the pressure and state with data from a-12

h3 = h4 = 95.48kj/kg

1.)

The mass flow rate

= QL/h1 - h4

= 10/255.55-95.48

= 10/160.07

= 0.0625kg/s

2.)

The power requirement

= 0.0625(270-255.55)

= 0.9031 kW

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A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume
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Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

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What are the two reasons for a clear cut
Inessa [10]

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A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a
8090 [49]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

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