Arefrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evaporator and condenser pressures are 200 kpa and 1400 kpa, respectively. the isentropic efficiency of the compressor is 88 percent. the refrigerant enters the compressor at a rate of 0.025 kg/s superheated by 10.1°c and leaves the condenser subcooled by 4.4°c. determine

(a) the rate of cooling provided by the evaporator, the power input, and the cop. determine
(b) the same parameters if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits.

a. The COP of the cycle is 2.725 b. The COP of the cycle is 2.886

Explanation:

Working fluid = R-134a

Evaporator pressure P1 = P4 = 200 kPa

Condenser presser P2 = P3 = 1400 kPa

Isentropic efficiency of the compressor ηc = 0.88

Mass flow rate to compressor m = 0.025kg/s

Sub cooled temperature T3’ = 4.4 C

a.State 1

Obtain the saturation temperature at evaporator pressure. Since, the refrigerant enters the compressor in super heated state

Obtain the saturation temperature from the super heated refrigerant R-134a table at P1 = 200kPa and T(sat) = -10.1 C

Calculate the temperature at state 1. As the refrigerant super heated by 10.1 C when it leaves the evaporator.

T1 = (-10.1) + 10.1 = 0 C

Obtain the specific enthalpy and specific entropy at state 1 from the table at T1 = 0 C and P1 = 200 kPa, which is, h1 = 253.05 kJ/kg and s1 = 0.9698 kJ/kg.K

State 2

Obtain the ideal specific enthalpy and saturation temperature at state 1 from refrigerant R-134a table at P2 = 1400 kPa and s1 = s2 = 0.9698kj/kg.K

Using the interpolation

h(2s) = 285.47 + (0.09698 – 0.9389) (297.10 – 285.47)/(0.9733 – 0.9389)

h(2s) = 295.91 kJ/kg

T([email protected]) = 52.40 C

State 3 and State 4

Calculate the temperature at state 3

T3 = T([email protected]) – T3

= 52.40 – 4.4 = 48 C

Obtain the specific enthalpy from the saturated refrigerant R -134a temperature table at T3 = 48 C, which is, h3 = hf = 120.39 kJ/kg

Since state 3 to state 4 is the throttling process so enthalpy remains constant

h4 = h3 = 120.39 kJ/kg

Calculate the actual enthalpy at state 2. Consider the Isentropic efficacy of the compressor

ηc = (h(2s) – h1)/(h2 – h1)

0.88 = (295.91) – (253.05)/h2 – (253.05)

h2 = 301.75 kJ/kg

Calculate the cooling effect or the amount of heat removed in evaporator

Q(L) = m (h1 – h4)

= (0.0025) (253.05 – 120.39)

= 3.317 kW

Therefore, the rate of cooling provided by the evaporator is 3.317 kW

Calculate the power input

W(in) = m (h2 – h1)

= (0.025) (301.75 – 253.05)

= 1.217 kW

Therefore, the power input to the compressor is 1.21 kW

Calculate the Coefficient of Performance

COP = Q(L)/W(in)

= 3.317/1.217

= 2.725

Therefore, the COP of the cycle is 2.725

b.Ideal vapor compression refrigeration cycle

State 1

Since the refrigerant enters the compressor is superheated state. So, obtain the following properties from the superheated refrigerant R-134a at P1 = 200 kPa

X1 = 1, h1 = 244.46kJ/kg, s1 = 0.9377 kJ/kg.K

State 2

Obtain the following properties from the superheated R-134a table at P2 = 1400kPa, which is s1 = s2 = 0.9377kJ/kg.K

Using the interpolation

h2 = 276.12 + (0.9377 – 0.9105) (285.47 – 276.12)/(0.9389 – 0.9105)

= 285.08kJ/kg

State 3

From the saturated refrigerant R-134a, pressure table, at p3 = 1400kPa and x3 = 0

H3 = hg = 127.22 kJ/kg

Since state 3 to state 4 is the throttling process so enthalpy remains constant

H4 = h3 = 127.22 kJ/kg

(hg should be hf because in ideal case it is a should exist as a liquid in state 3)

Calculate the amount of heat removed in evaporator

Q(L) = m (h1 – h4)

= (0.025) (244.46 – 127.22)

= 2.931 kW

Therefore, the rate of cooling provided by the evaporator is 2.931 kW

Calculate the power input to the compressor

W(H) = m (h2 – h1)

= (0.025) (285.08 – 244.46)

= 1.016 kW

Therefore, the power input to the compressor is 1.016 kW

Calculate the COP of the ideal refrigeration cycle

COP = Q(L)/W(in)

= 2.931/1.016 = 2.886

Therefore, the COP of the cycle is 2.886

the warning light flashes for a second or two when the engine starts. if it stays on, there's must be something wrong with the coolant. you should never dare to continue driving when the light is on. pull over as soon as possible and shut the engine off.

hope this is what you're talking about

answer:

walmart

explanation: