Lecture 6 worked example 3 - R134a capillary-tube throttlingTutorialCurrent specThird6 min

ENGR5003 Lecture 6 worked example 3 (slides 16-18)

A refrigerant (R134a) enters a capillary tube of a refrigerator as a saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the temperature of the initial state and the temperature and dryness fraction of the final state.

R134a saturation data (pressure table):

p (kPa)T (C)vfv_f (m3/kg)vgv_g (m3/kg)hfh_f (kJ/kg)hfgh_{fg} (kJ/kg)hgh_g (kJ/kg)
120-22.30.00073240.162122.5214.5237.0
80031.30.00084590.025695.5171.8267.3
Formulas you may need
  • SFEE with q=w=0q = w = 0 and negligible KE, PE: h1=h2h_1 = h_2 (throttling is isenthalpic) (on the formula sheet as the SFEE)
  • Dryness fraction: h=(1−x)hf+xhgh = (1 - x)h_f + xh_g, so x=h−hfhg−hfx = \dfrac{h - h_f}{h_g - h_f} (on the formula sheet)