Question

System Model: Automotive Air Conditioning System

Outside Air Temp: 45 oC Car Interior Air Temp: 22 oC

Heat Load = 4.2 kJ/sec  rate at which heat must go in to the air conditioner to be removed from the car’s interior

Refrigerant: R134a

HINT: Heat enters an air conditioning system where?

Refrigerant 134a …

- leaves the evaporator coil and enters the compressor as a saturated vapor at the evaporator pressure of 2.8 bars.

- It leaves the compressor at the condenser pressure of 14 bars

.- It leaves the condenser at the same pressure as a saturated liquid. Finally, after passing through the expansion valve (a throttling process), …

- the refrigerant enters the evaporator coil as a saturated liquid-vapor mixture at the evaporator pressure (2.8 bars).

a) Draw a schematic sketch of the machine labeling pertinent parts. Sketch the line between the low and high pressure sides of the system. Also draw a temperature-entropy plot for this unit. You may start with the ideal model and modify this later if the data shows that this is more realistic. (5 pts)

b) Based on the description of the ideal vapor compression model given above, find …

i. the pressure at each point (states 1, 2, 3, and 4). Give your answers in bars, kPa, and in psi (pounds per square inch). Show the pressures in bars on your schematic (part a).

ii. What property from the tables (v, u, h, s) is assumed to be constant between the compressor inlet and the compressor outlet? Show this on your schematic.

iii. What property from the tables (v, u, h, s) is assumed to be constant between the expansion (throttling) valve inlet and outlet? Show this on your schematic. (5 pts)

c) At each of the four states, list the pressure, the temperature, the quality (where appliable), enthalpy (kJ/kg) and entropy (kJ/kg K) of the R134a. (40 pts)

d) Find the mass flow rate of refrigerant necessary for an air conditioning system heat input rate (Heat In) equal to the heat load. (10 pts)

e) Convert the heat load from kJ/sec (= kW) into BTU/hr. Also give this value in tons of refrigeration capacity (1 ton = 12,000 BTU/hr). (2 pts)

f) Using the mass flow rate found in d), determine the compressor power demand. Give the power demand in kW (=kJ/sec) and horsepower. (10 points)

g) What are the theoretical and the actual coefficients of performance for this machine? (8 pts)

h) Using the mass flow rate found in d), determine how much heat is rejected in the condenser coil in kW. Also give this answer in BTU/hr. (5 points)

i) A more realistic model of vapor compression refrigeration is given in Fig 10.5.

i. For a compressor isentropic efficiency of 80% (see sec. 6.12), find the power needed to run the compressor for a more realistic process. (5 pts)

ii. Your model was based on saturated liquid leaving the condenser and entering the expansion valve. How much additional heat would be rejected if this was a subcooled (compressed) liquid at a temperature of 30 oC entering the expansion valve? (5 pts)

Now – move from the refrigerant inside the machine to the air inside the car.

The conditioned air inside the car is moved by the fan (a squirrel cage blower – do a quick Google search if you don’t know what this looks like) across the surfaces of the evaporator coil. As it passes through the evaporator the air is cooled and dehumidified.

The air enters the evaporator at a pressure of 1 bar and a temperature of 22 oC. The design ΔT across the coil (the drop in temperature) is 10 oC.

j) For the air passing through the coil, find ..

i. The temperature as the air leaves the coil (3 pts)

ii. The change in enthalpy and the mass flow rate of air necessary for the rate of heat going out of the air to equal the design heat input to the evaporator coil. (10 pts)

iii. As the air enters the coil, find the density of the air and the volume flow rate of air through the coil. Give the volume flow rate in m3/sec, liters, and cubic feet per minute (CFM). (12 pts)

I am having trouble understanding the tables on whether I use superheated or liquid or vapor properties of the stages C) I know E-H has been answered somewhere else in Chegg but I am still lost at the whole system

Answer 1

e.

Heat Load = 4.2 kW

1 kW = 3412.142 BTU/hr

Heat Load = 4.2 * 3412.142

Heat Load = 14330.9964‬ BTU/hr

Heat Load = 14330.9964‬ / 12000 = 1.1942497‬ TR

f.

W1-2 = m (h2 - h1)

= 0.03463 * (279.8 - 246.52)

W1-2 = 1.1524864‬ kW  

1 kW = 1.34102 hp

W1-2 = 1.5455 hp  

g.

COP = RE / W1-2

COP = 4.2 / 1.1524864‬

COP = 3.6443

h.

Q2-3 = m (h2-h3)

= 0.03463 * (279.8 - 125.26)

Q2-3 = 5.3517202 kW

1kW = 3412.142 BTU/hr

Q2-3 = 18260.8292 BTU/hr

Please rate my answer. Thank You.