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olchik [2.2K]
3 years ago
5

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin

g its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

Engineering
1 answer:
lord [1]3 years ago
5 0

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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Answer:

Ф = 0.02838 ft

F  = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

                                               = 90- (-20)

                                               =  110 F

Ф =  6.45 ×10^(-6) × 110 × 40

Ф =  28,380 × 10^(-6) ft

Ф = 0.02838 ft     .... total gape required for expansion of steel rails

Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

  σ      =  1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

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3 years ago
Who is father of Engineer?
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3 years ago
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Answer:

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Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i
IceJOKER [234]

Answer:

Gc(s) = \frac{0.1s + 0.28727}{s}

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

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3 years ago
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com
Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

P1 = P4  ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

<u>From pressure table of R 134a refrigerant</u>

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance  

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

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