An air conditioner with refrigerant 134-a as the working fluid is used to keep a

Question

An air conditioner with refrigerant 134-a as the working fluid is used to keep a room at 23 defy by rejecting the waste heat to the outdoor air at 37 deg. When the Xbox is being used to Play mw3 in confirmed kill
Matches the room gains heat through the walls and windows at a rate of 250kj/min while heat generated by the Xbox , lights, and tv amount to 900W the refrigerant enters the compressor at 400 kpa as a saturated vapor at a rate of 100l/min and leaves at 1200 kpa and 70 deg

Determine
Actual cop
Max cop
Minimum flow rate of the refrigerant at the compressor inlet
For the compressor inlet ans exit conditions

Explanation / Answer

Given:
Chemical R-134a
T room: 23 C
T outside: 37 C
Q walls: 250 KJ/min
Q elec: 900 W

Assumption:
The air-conditioner operates at steady-state
The compressor is adiabatic
Changes in kinetic and potential energies are negligible int he compressor

Equation 1.......... COP ref = Q(dot) c / W(dot) ref

Equaiton 2.......... Q(dot) c = Q walls + Q elec

Q c 5.06

W ref = -W s,comp

Equation 3......... Q(dot) comp - W(dot) s,comp = m(dot)(H^ + E^ kin + E^ pot)

Assuming that the compressor is adiabatic

Equation 4......... W(dot) s,comp = m(dot) (H^ 1 - H^2) < 0

Equation 5......... m(dot) = V(dot) 1/ V^ 1

R-134 Saturated vapor at 400 kPa

V1 0.051207 m^3/kg

H1 403.72 KJ/kg

Plugging values into equation 5 gives you: m 0.0325 kg/s

We need to determine H2 so we can use equation 4

Looking up the specific enthalpy in the thermo tables.

H2 = 448.7 kJ/kg

Now plug into equation 4 to evaluate W s,comp since W ref = -W s,comp

W s,comp = -1.466 kW

W ref = 1.466 kW

Going back to equation 1 to evaluate the COP, we have COP = 3.46 is your final answer

For COP max

Carnot Cycle tells us that it will yield the max efficiency of any cycle between two reservoirs.

Equation 6............... COP R, rev = 1 / ((Th/ Tc)-1)

The cold reservoir is the air inside the room being air-conditioned, so Tc = 23 C = 296.15 K

The hot reservoir is the outside air, so Th= 37C = 310.15 K

Plugging into equation 6 gives you.... COPmax = 21.15

For COP min

equation 7.............. W(dot) ref,min = Q(dot) c / COP max

equation 8.............. W(dot) ref,min = -W(dot) s,comp = m(dot) (H^2 - H^1) > 0

Rearranging equation 5

Equation 9.............. V(dot)1 = m(dot) V^1

Equation 10............. m(dot) = W(dot) ref,min / H^2 - H^1

Plugging in values in for eqautions 7, 9, 10

You get: W ref,min = 0.2395 kW

m = 0.00532 kg/s

V1, min = 0.0002723 m^3/s

So your final answer is V1, min = 16.33 L/min

My neck hurts. Hope this helps

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