Question

Oration 3 The problems refer to the ideal vapor compression cycke schematically shown below. This ideal vapor compression cycle is used to air condition acar. The air conditioner uses R134a refrigerant and the compressor power input is 1.5KW. Saturatie vapor enters the compresa 200 LPa and Saturated liquid cuits the condenser at 1.2 MPa. The evaporator is used to cool 30 "Cabintai from outside the car down to 10°C for delivery by a fan into the car's cabin 1. The phase of R-134a State 4 is which of the following C a. 2.phase, -0.39 b. saturated vapor, approximately superheated vapor d. sub-cooled or compressed liquid 2. The mass flow rate of the R-134a is most nearly a 1.33 kg's 0.52 kg's 004ks 23.4 kg's d. 3. If the capacity of the air conditioner is 2 Toms (The evaporator removes heat at this rate from the cabin supply air). the mass flow rate of the air being delivered to the cabin is most nearly! The air may be treated as an ideal gas with a constant specific heat of 1.004 - a 17.1 kg's b. 0.11 kg's c. 0.35 kg's d 467 kg's 4. For 2 Tons of cooling capacity (70337 kW) and the given compressor power (15 kW), the Coefficient of Performance (COP) of this cycle is most nearly a 2.77 b 14.2 c. 0.21 d468

Answer 1

Solution :

Given data : -

W; = 1.5 kᎳ

P1 = 200 kPa

x1 = 1 (Saturated vapor

P = P3 = 1.2 MPa = 1200 kPa

(Tali = 30°C

(T.). = 10°C

.

The T-s diagram of ideal vapor compression cycle is given as :

Water 400 - 300 W in T[°C] 1200 kPa 200 kPa -1.0 0.1 1.2 2.3 3.4 6.7 7.8 8.9 10.0 4.5 5.6 s [kJ/kg-K]

.

The properties of refrigerant R-134 at various states:

State -1: P1 = 200 kPa and 21 = 1

..h1 = h 0200 kPa = 244.50 kJ/kg

.$1 = SgQ200 kPa = 0.93788 kJ/kg. K

.

State - 2 : P2 = 1200 kPa and s2 = $1

.. h2 = 281.914 kJ/kg

.

State - 3: P3 = 1200 kPa and x3 = 0 (saturated liquid)

.. h3 = hf91200kPa = 117.79 kJ/kg

.

State - 4: h4 = hz (isenthalpic process and P4 = 200 kPa

.h4 = 117.79 kJ/kg

h = 38.41 kJ/kg and hq = 244.50 kJ/kg

Here , h, <h4 < hq Hence the state - 4 is 2-Phase.

.

Part - 1: The phase of the R-134a at state 4 is determined as :

1. h4 - hf .. 14 hen

...14 = 117.79 – 38.41 206.09

A = 0.3852 39 (Answer - a)

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Part-2 : The mass flow rate of the R-134a is determined as :

As we know the work input by compressor is given as

:: Win = m(h2 - h1)

:: 1.5 = m(281.914 - 244.50)

:: m = 0.04 kg/s (Answer - c)

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Part-3: The mass flow rate of air is determined as:

Given : Q = 2 Tons = 2 x 3.517 kW = 7.0337 kW

(pa = 1.004 kJ/kg · K

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Here, as we know that the heat gained by evaporator is equal to

Heat lost by the air.

.: Q2 = Qa

... QL = m(g) (T – T)

.: 7.0337 = m (1.004)30 – 10)

...ma = 0.3502 kg/s 0.35 kg/s (Answer - c)

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Part - 4: The COP of the cycle is determined as:

::COP = QL Win

::COP = 7.0337 1.5

· COP = 4.689 (Answer - d)