ditferential height h of the mercury colunm Refrigerant 134-a enters the evapora

Question

ditferential height h of the mercury colunm Refrigerant 134-a enters the evaporator coils placed at the back of the freezet section of a houschold retrigerator at 100kPa with a quality of 20 percent and leaves at 100kPa and -26 C as shown in Figure 2. If the compressor consumes 600 W of power and the COP of the refrigerator is 1.2, determine (a) the mass flow rate of the refrigerant (b) the tate of heat rejected to the kitchen air Stcam enters an adiabatic turbine at 3MPa and 400 C and expands to 30 kPa with an isentropic efficiency of 90 percent. Determine the power output of thrs turbine when the mass flow rate is 2kg/s An Otto cycle with a compression ratio of 8 has the initiai pressure of 9SkPa and temperature of IS c. find the heat rejected and power produced by this cycle. Air enters the heating section of an air-conditioning at I atm, 10 C, and 70 percent reiative humidity at a rate of 35 m/min, and it leaves the humidifying section at 20 C with a relative humidity of 90 percent. If the humidifier supplies wet steam at 100 C, determine (a) The temperature and relative humidity of the air leaving the heating section (b) The rate of heat transfer in the heating section Ol Condenser so 1Pa Expansion valve Compressor 75 cm

Explanation / Answer

Part a) There is sufficient information at the inlet and the exit to determine the state points on a psychrometric chart.

h1 = 23.5 kJ/kgdry air

?1 = 0.0053 kg H2O/kgdry air

v1 = 0.808 m3/kgdry air h? 3 = 42.3 kJ/kgdry air

?3 = 0.0087 kg H2O/kgdry air

We note that m? a,1 = ?ma,2 = ?ma,3 = ?ma and

m = V1/ v1 = 35 m3/min /0.809 m3/kg = 43.3 kg/min

Performing an energy balance between state point 2 and 3

( ?mh)a,2 +( ?mh)w,2 + ?mwhw = (?mh)a,3 +( ?mh)w,3 m? ah? 2 + ?mwhw = ?mah? 3

But mw ma = ?1 ? ?2

Therefore

m h2 = ?mh3 + (?2 ? ?3)hw h2 = h3 + (?2 ? ?3)hg at100 ?C

= 42.3 kJ/kg + (0.0053 ? 0087) kg H2O/kgdry air × 2675.6 kJ/kg

= 33.2 kJ/kgdry air

Therefore at state 2 we now have the enthalpy and ?2 = ?1 = 0.0053 kg H2O/kgdry air.

From the psychrometric chart T2 = 19.5 C = Temperature

?2 = 0.378 = 37.8%= Relative humidity.

Part b)

In the heating section

Q?= ?m (h2 ?h1) = (43.3 kg/min)(33.2 kJ/kg ?23.5 kJ/kg) = 420 kJ/min

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