Lab1 Calc.ns Pt3

# Lab1 Calc.ns Pt3 - The temperature at entrance of the...

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PART 3 3. What is the saturation temperature for the low pressure? (Remember that gauge pressure is given, not absolute.) How does the measured refrigerant temperature at the evaporator exit compare? Is this logical based on the T-s diagram? If your pressure gauge read 20% higher than the actual pressure, what would be the effect on the value of h that you determined at the exit of the evaporator? Low pressure was determined after the expansion valve, before the evaporator, and it was averaged at 242Kpa (gauge). The conversion to absolute pressure was ( Pgauge + Prelative = Pabsolute) For this case; 242Kpa + 101Kpa = 343Kpa. Using that value, the saturation temperature was determined to be 277K, 4 ° C from the gas property tables.
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Unformatted text preview: The temperature at entrance of the evaporator was 275.85K and at the exit was 278.59K. It was a small increase and that implies it doesn’t absorb that much heat but the mass flow was high enough to cause effective cooling. If the pressure was 20% higher, actual values show in figure XXXXX, then the h f would increase from 66.6 kj/kg to 72.10 kj/kg. That would make increase the cooling rate of the evaporator because that force the change in enthalpy to be greater. Q = m(h e - h i ) 3. Pressure Error Check absolute(measured) absolute(+20%) Pmin 343.6 412.36 T(K) *C T(K) *C *C Temp 275.22 2.22 281 8.00 2.94 Figure XXXXX T shows absolute pressure and as well as 20% increase and how it affects the Temperature....
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