22-Mec-B2 Environmental Control in Buildings · May 2013

Question 3 of 8: Two-stage ammonia refrigeration with a direct-contact intercooler (20 marks)

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

Notes on this paper

Paper format. Professional Engineers of Ontario / EGBC annual examination, 07-Mec-B2 (now 22-Mec-B2) Environmental Control in Buildings, May 2013 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for ammonia (R-717) are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to state any interpretation assumptions with the answer — that latitude is used explicitly below where the printed data are redundant.

All eight problems are worked here. Every psychrometric state has been recomputed from the ASHRAE formulation for saturation vapour pressure rather than scaled off a chart, so the numbers are tighter than a graphical solution would be; chart-quality agreement (about ±0.2 K and ±0.0002 kg/kg) is all that an examiner expects.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry), Ch. 6 (cooling loads), Ch. 10 (cooling towers), Ch. 15 (duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 8 (energy estimating and degree-day methods), Ch. 12–13 (fluid flow and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression and multistage refrigeration).
ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy, and ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality.
Canadian frame: National Building Code of Canada 2020, National Energy Code of Canada for Buildings 2020, and Environment and Climate Change Canada Canadian Climate Normals for degree-day data.

Psychrometric relations used throughout. At barometric pressure $p$, with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation,

$$W = 0.6220\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$

in SI (kJ per kg of dry air), and in the inch-pound system $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air. The thermodynamic wet-bulb temperature is obtained from the adiabatic-saturation equation, which is what a chart's constant-wet-bulb lines represent.

Question 3: Two-stage ammonia refrigeration with a direct-contact intercooler (20 marks)

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. The two-stage ammonia (R-717) plant of the figure printed with the question, with a direct-contact heat exchanger acting as both flash chamber and de-superheater at the intermediate pressure.

Given data
Quantity	Value
Refrigeration capacity	30 tons = 360,000 Btu/h = 6000 Btu/min
Evaporator exit, state 1	saturated vapour at −20 °F (18.3 psia)
Intermediate pressure (heat exchanger)	80 psia
Condenser pressure	250 psia
Compressor-2 inlet, state 3	saturated vapour at 80 psia
Isentropic efficiency, both stages	85%
Expansion-valve inlets, states 5 and 7	saturated liquid

Find. the mass-flow ratio $\dot{m}_2/\dot{m}_1$, the power input to each stage in horsepower, and the coefficient of performance.

The plant, with the state numbering of the figure supplied in the question. Compressor 1 handles the evaporator flow $\dot{m}_1$; compressor 2 handles the whole condenser flow $\dot{m}_2$. The direct-contact heat exchanger receives superheated vapour at 2 and the throttled liquid at 6, and delivers saturated vapour at 3 to compressor 2 and saturated liquid at 7 to the low-stage expansion valve.

Approach. Read the eight enthalpies from the R-717 tables and the supplied p–h diagram, correct both compressions for 85% isentropic efficiency, then get the flow ratio from a combined mass and energy balance on the direct-contact heat exchanger and scale to the 30-ton duty.

Fix the states. All enthalpies are on the usual R-717 datum ($h_f = 0$ at −20 °F is not used; the datum is $h_f = 0$ at −40 °F), read from the saturation table and the appended pressure–enthalpy chart. Saturation temperatures are 44.4 °F at 80 psia and 110.8 °F at 250 psia.

Ammonia states
State	Condition	$h$ (Btu/lb)	$s$ (Btu/lb·R)
1	sat. vapour, −20 °F	604.5	1.4041
2s	80 psia, $s = s_1$	691.9	1.4041
2	80 psia, actual	707.4	—
3	sat. vapour, 80 psia	623.3	1.2934
4s	250 psia, $s = s_3$	693.3	1.2934
4	250 psia, actual	705.7	—
5 = 6	sat. liquid, 250 psia (throttled)	167.8	—
7 = 8	sat. liquid, 80 psia (throttled)	91.6	—

Actual first-stage discharge. The isentropic compression from 18.3 to 80 psia raises the enthalpy by $691.9 - 604.5 = 87.4$ Btu/lb, so $$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_c} = 604.5 + \frac{87.4}{0.85} = \boxed{707.4\ \text{Btu/lb}}$$ which on the chart is about 186 °F — strongly superheated, and exactly why intercooling is worth the extra hardware.
Actual second-stage discharge. From saturated vapour at 80 psia to 250 psia, $$h_4 = h_3 + \frac{h_{4s} - h_3}{\eta_c} = 623.3 + \frac{693.3 - 623.3}{0.85} = 705.7\ \text{Btu/lb}$$ about 212 °F. Note that de-superheating in the direct-contact exchanger has brought the second stage back to the saturation line, so its discharge is cooler than the first stage's despite the higher pressure.
Mass-flow ratio from the direct-contact heat exchanger. The exchanger is adiabatic and receives $\dot{m}_1$ of superheated vapour at 2 plus $\dot{m}_2$ of throttled two-phase refrigerant at 6; it delivers $\dot{m}_2$ of saturated vapour at 3 and $\dot{m}_1$ of saturated liquid at 7. Energy balance: $$\dot{m}_1 h_2 + \dot{m}_2 h_6 = \dot{m}_2 h_3 + \dot{m}_1 h_7 \;\Rightarrow\; \frac{\dot{m}_2}{\dot{m}_1} = \frac{h_2 - h_7}{h_3 - h_5}$$ Substituting, $$\frac{\dot{m}_2}{\dot{m}_1} = \frac{707.4 - 91.6}{623.3 - 167.8} = \frac{615.8}{455.5} = \boxed{1.352}$$ The high stage carries 35% more refrigerant than the low stage, the extra being the vapour flashed off in the intercooler.
Low-stage mass flow from the 30-ton duty. The refrigerating effect per pound is $h_1 - h_8 = 604.5 - 91.6 = 512.9$ Btu/lb, so $$\dot{m}_1 = \frac{6000}{512.9} = 11.70\ \text{lb/min},\qquad \dot{m}_2 = 1.352 \times 11.70 = 15.82\ \text{lb/min}$$
Power input to each stage. With 1 hp = 42.41 Btu/min, $$\dot{W}_1 = 11.70\,(707.4 - 604.5) = 1204\ \text{Btu/min} = \boxed{28.4\ \text{hp}}$$ and for the high stage $$\dot{W}_2 = 15.82\,(705.7 - 623.3) = 1302\ \text{Btu/min} = \boxed{30.7\ \text{hp}}$$ The two stages are almost equally loaded, which is what a well-chosen intermediate pressure achieves.
Coefficient of performance. $$\text{COP} = \frac{\dot{Q}_{evap}}{\dot{W}_1 + \dot{W}_2} = \frac{6000}{1204 + 1302} = \boxed{2.39}$$ For comparison, a single-stage machine between the same −20 °F and 110.8 °F limits at 85% efficiency lands near 2.0, so the second stage buys roughly a 20% improvement as well as a much cooler discharge.

Final results
Quantity	Result
Mass-flow ratio $\dot{m}_2/\dot{m}_1$	1.352
Low-stage mass flow	11.70 lb/min (0.0884 kg/s)
High-stage mass flow	15.82 lb/min (0.1195 kg/s)
Power, compressor 1	28.4 hp (21.2 kW)
Power, compressor 2	30.7 hp (22.9 kW)
Total compressor power	59.1 hp (44.1 kW)
Coefficient of performance	2.39

Topic. Two-stage vapour compression with a flash-type direct-contact intercooler.

Key relations. Isentropic efficiency: $\eta_c = (h_{2s} - h_1)/(h_2 - h_1)$. Throttling: $h$ constant. Flash-chamber balance: $\dot{m}_2/\dot{m}_1 = (h_2 - h_7)/(h_3 - h_5)$. Duty: $\dot{Q} = \dot{m}_1(h_1 - h_8)$, one ton of refrigeration = 200 Btu/min = 3.517 kW.

Why this works. Splitting the compression does two independent good things. First, it cuts the work by returning the vapour to saturation between stages instead of compressing progressively hotter, more voluminous gas — ammonia has a steep saturation line and a large superheat penalty, so this matters. Second, throttling the condensate in two steps and separating the flash gas at the intermediate pressure means that flash vapour is never expanded down to the evaporator pressure and re-compressed from there. The direct-contact exchanger does both jobs in one vessel: the liquid sprayed in from state 6 de-superheats the low-stage discharge, and the vapour boiled off joins the high-stage suction.

Common pitfall. Writing the ratio as $(h_2 - h_7)/(h_3 - h_6)$ with $h_6$ taken as a two-phase enthalpy computed from a quality — it is simply $h_5$, because the valve is isenthalpic. The other common error is applying the 30-ton duty to $\dot{m}_2$; the evaporator only sees $\dot{m}_1$.

Source. Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley, Ch. 10 §10.5 (multistage compression with intercooling) and Fig. 10.16; ASHRAE Handbook – Refrigeration (2018), Ch. 1 and the R-717 property tables.

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