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22-Mec-A1 Applied Thermodynamics and Heat Transfer · Undated paper

Question 2 of 8: Throttling of saturated liquid ammonia

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer — 3 hours, open book. Two parts: Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8). The rubric asks for five questions (three from one part, two from the other), each of equal value; all eight are solved.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Moran & Shapiro, Fundamentals of Engineering Thermodynamics. Property data: saturated ammonia and R-134a tables printed with the paper.

Question 2 — Throttling of saturated liquid ammonia (Part A, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantitySymbolValue
Inlet: saturated liquid$T_1,\,x_1$20 °C, $x_1=0$
Inlet enthalpy (table)$h_1=h_f(20°\text{C})$409.3 kJ/kg
Exit pressure$P_2$0.43 MPa = 430 kPa
Table anchors at 0 °C / 4 °C$P_\text{sat}$411.3 / 468.6 kPa

Find. (a) exit quality $x_2$ and $\Delta x$; (b) exit temperature $T_2$; comment on valve work.

Approach. A throttling valve is adiabatic with negligible work and kinetic change, so it is isenthalpic ($h_2=h_1$); locate the saturation state at $P_2$ by interpolation and read the quality off $x_2=(h_2-h_{f2})/h_{fg2}$.

  1. Interpolate the saturation state at 430 kPa. Between the table rows 0 °C (411.3 kPa) and 4 °C (468.6 kPa), the interpolation fraction is $$f=\frac{430-411.3}{468.6-411.3}=0.326\ \Rightarrow\ T_2=0+4(0.326)=\boxed{1.31\ ^\circ\text{C}}$$ Interpolated saturation enthalpies: $h_{f2}=286.6+(310.6-286.6)(0.326)=294.4$ kJ/kg and $h_{fg2}=1210.8+(1198.2-1210.8)(0.326)=1206.7$ kJ/kg.
  2. Apply the isenthalpic condition. The valve does no shaft work and is adiabatic, so $h_2=h_1=409.3$ kJ/kg. The exit lies in the two-phase dome, and $$x_2=\frac{h_2-h_{f2}}{h_{fg2}}=\frac{409.3-294.4}{1206.7}=\boxed{0.0952}$$
  3. Change in quality and temperature drop. The inlet is a saturated liquid ($x_1=0$), so $$\Delta x=x_2-x_1=0.095,\qquad \Delta T=20-1.31=\boxed{18.7\ ^\circ\text{C (drop)}}$$ About 9.5 % of the ammonia flashes to vapour; the temperature falls by the Joule–Thomson effect.
QuantityResult
Exit temperature $T_2$1.31 °C
Exit quality $x_2$0.095
Change in quality $\Delta x$+0.095 (flash fraction)
Valve work $w$0 (throttle, isenthalpic)
Check — the "work done" sub-part.
The question also includes a "determine the work done by the gas" sub-part. A throttling valve produces no work ($w=0$) — the enthalpy is conserved and the pressure energy is dissipated internally. The physically meaningful companion result is the temperature drop (Joule–Thomson cooling), reported above.