22-Mec-B2 Environmental Control in Buildings · December 2017

Question 4 of 8: Ammonia cold-storage plant for a milk chilling duty

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

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

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 and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 4: Ammonia cold-storage plant for a milk chilling duty (20 points)

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.

Quantity	Symbol	Value
Milk throughput	$\dot V_m$	$9000\ \text{L/h}$
Milk cooled from / to	$t_{in},\ t_{out}$	$27\,{}^{\circ}\text{C}$ to $4\,{}^{\circ}\text{C}$
Milk specific heat / density	$c_{p,m},\ \rho_m$	$3.77\ \text{kJ/(kg}\cdot\text{K)}$, $1030\ \text{kg/m}^3$
Plant heat gain	$\dot Q_{gain}$	$3600\ \text{kJ/min}=60\ \text{kW}$
Evaporating temperature	$t_1$	$-6\,{}^{\circ}\text{C}$, dry saturated at exit
Compressor delivery pressure	$p_2$	$10.34\ \text{bar}$
Condenser liquid undercooled to	$t_3$	$24\,{}^{\circ}\text{C}$
Compression	$-$	isentropic
Mechanical / volumetric efficiency	$\eta_m,\ \eta_v$	$0.90$ / $0.85$
Compressor	$-$	twin cylinder, single acting, $200\ \text{rpm}$
Brine rise, specific heat, density	$\Delta t_b,\ c_{p,b},\ \rho_b$	$3\ \text{K}$, $2.93\ \text{kJ/(kg}\cdot\text{K)}$, $1190\ \text{kg/m}^3$

Find. The plant sketch and the cycle on the attached R-717 p–h diagram, then the compressor shaft power, the swept volume of each cylinder, and the volumetric rate of brine circulation.

Part (a) — the plant. Ammonia circulates 1→2→3→4 between the evaporator and the condenser; a secondary brine circuit carries the duty from the evaporator to the milk coolers, its temperature rise limited to 3 K.

Approach. Add the two components of the evaporator duty, fix the four cycle states from the ammonia tables, take the refrigerating effect per unit mass to get the refrigerant flow, and then take the compressor work, the induced volume and the brine flow in turn.

Evaporator duty. The milk mass flow follows from the volumetric throughput and density: $$\dot m_m=\frac{9000}{1000}\times\frac{1030}{3600}=2.575\ \text{kg/s}$$ $$\dot Q_{milk}=\dot m_m\,c_{p,m}\,\Delta t=2.575\times3.77\times(27-4) =223.28\ \text{kW}$$ Adding the plant heat gain of $3600/60=60\ \text{kW}$, $$\dot Q_e=223.28+60=\boxed{283.28\ \text{kW}}$$ which is $80.6$ tons of refrigeration — a substantial industrial plant, and consistent with a slow-speed reciprocating ammonia machine.

Part (b) — fix the four cycle states. State 1 is dry saturated vapour at $-6\,{}^{\circ}\text{C}$; state 2 is the isentropic end point at the delivery pressure; state 3 is liquid undercooled to $24\,{}^{\circ}\text{C}$ at that same pressure; and state 4 follows from throttling at constant enthalpy.

Point	Description	$p$ (bar)	$t$ (°C)	$h$ (kJ/kg)	$s$ (kJ/kg·K)	$v$ (m³/kg)
1	dry saturated vapour, compressor suction	3.41	−6.0	1455.1	5.700	0.3597
2	after isentropic compression	10.34	71.0	1609.6	5.700	—
3	undercooled liquid leaving the condenser	10.34	24.0	312.7	—	—
4	after throttling, wet vapour	3.41	−6.0	312.7	—	—

At $10.34\ \text{bar}$ ammonia condenses at $26.0\,{}^{\circ}\text{C}$, so the stated $24\,{}^{\circ}\text{C}$ liquid represents $2\ \text{K}$ of undercooling — a modest and entirely realistic figure. Compression ends at $71\,{}^{\circ}\text{C}$, well within the $150\,{}^{\circ}\text{C}$ limit that governs ammonia discharge temperatures.

Refrigerating effect and refrigerant mass flow. The evaporator takes the fluid from state 4 to state 1, so $$q_e=h_1-h_4=1455.1-312.7=1142.3\ \text{kJ/kg}$$ $$\dot m_r=\frac{\dot Q_e}{q_e}=\frac{283.28}{1142.3} =\boxed{0.2480\ \text{kg/s}}\quad(893\ \text{kg/h})$$ The undercooling is worth having: without it the liquid would enter the throttle at $26\,{}^{\circ}\text{C}$ and the refrigerating effect would fall by roughly $10\ \text{kJ/kg}$, requiring about $1\%$ more refrigerant for the same duty.
Part (c)(i) — compressor power. Isentropic compression work per unit mass, then the indicated power, then the shaft power through the mechanical efficiency: $$w=h_2-h_1=1609.6-1455.1=154.6\ \text{kJ/kg}$$ $$\dot W_{ind}=\dot m_r\,w=0.2480\times154.6=38.33\ \text{kW}$$ $$\dot W_{shaft}=\frac{\dot W_{ind}}{\eta_m}=\frac{38.33}{0.90} =\boxed{42.59\ \text{kW}}$$ The plant coefficient of performance on shaft power is $\text{COP}=283.28/42.59=6.65$ (ideal, on indicated power, $7.39$), which is what one expects of ammonia working across only $32\ \text{K}$ of temperature lift. The condenser must reject $\dot m_r(h_2-h_3)=321.6\ \text{kW}$.
Part (c)(ii) — swept volume per cylinder. The volume the compressor actually induces is the mass flow times the suction specific volume; the swept volume is larger by the reciprocal of the volumetric efficiency, which accounts for re-expansion of the clearance gas: $$\dot V_{ind}=\dot m_r\,v_1=0.2480\times0.3597=0.08921\ \text{m}^3/\text{s}$$ $$\dot V_{swept}=\frac{\dot V_{ind}}{\eta_v}=\frac{0.08921}{0.85} =0.10495\ \text{m}^3/\text{s}$$ Single acting means one induction stroke per revolution per cylinder, so at $200\ \text{rpm}$ with two cylinders the machine completes $2\times200/60=6.667$ induction strokes per second: $$V_{cyl}=\frac{0.10495}{6.667}=\boxed{0.01574\ \text{m}^3=15.74\ \text{litres}}$$ For a square engine ($\text{bore}=\text{stroke}$) that is a bore of $272\ \text{mm}$ — a large, slow, industrial ammonia compressor, exactly the machine the stated $200\ \text{rpm}$ implies.
Part (c)(iii) — brine circulation rate. The brine carries the whole evaporator duty from the chiller to the milk coolers, and its temperature rise is limited to $3\ \text{K}$: $$\dot m_b=\frac{\dot Q_e}{c_{p,b}\,\Delta t_b}=\frac{283.28}{2.93\times3} =32.23\ \text{kg/s}$$ $$\dot V_b=\frac{\dot m_b}{\rho_b}=\frac{32.23}{1190}\times1000 =\boxed{27.08\ \text{L/s}}$$ which is $97.5\ \text{m}^3/\text{h}$. The narrow $3\ \text{K}$ rise is what makes the flow so large; it is specified to keep the brine temperature nearly uniform through the milk coolers, and it is the price paid for using a secondary refrigerant rather than expanding ammonia directly in a food-contact heat exchanger.

Part (b) — the cycle on the attached R-717 pressure–enthalpy diagram. 1→2 isentropic compression, 2→3 desuperheating, condensation and 2 K of undercooling, 3→4 throttling at constant enthalpy, 4→1 evaporation to dry saturated vapour.

Final results.

Quantity	Result
Milk cooling load	$223.28\ \text{kW}$
Evaporator duty (with heat gain)	$283.28\ \text{kW}$ ($80.6\ \text{TR}$)
Refrigerating effect	$1142.3\ \text{kJ/kg}$
Refrigerant mass flow	$0.2480\ \text{kg/s}$ ($893\ \text{kg/h}$)
Isentropic work	$154.6\ \text{kJ/kg}$; indicated power $38.33\ \text{kW}$
(c-i) Power input at $\eta_m=0.90$	$\mathbf{42.59\ \text{kW}}$
Plant COP (on shaft power)	$6.65$
(c-ii) Swept volume per cylinder	$\mathbf{15.74\ \text{litres}}$ ($0.01574\ \text{m}^3$)
(c-iii) Brine circulation	$\mathbf{27.08\ \text{L/s}}$ ($97.5\ \text{m}^3/\text{h}$)
Condenser duty	$321.6\ \text{kW}$

Check: assumptions declared under cover-page instruction 1. (i) Ammonia properties are quoted on the datum of the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ at $0\,{}^{\circ}\text{C}$); only the differences $h_1-h_4$ and $h_2-h_1$ enter the answers, so any consistent set of tables gives the same duties. (ii) The stated plant heat gain of $3600\ \text{kJ/min}$ is taken as an additional load on the evaporator, which is where a cold-store cabinet gain must appear. (iii) Pressure drops in the suction and discharge lines are neglected, so the suction state is the saturated state at $-6\,{}^{\circ}\text{C}$ as stated. (iv) "Single acting" is read literally — one induction stroke per revolution per cylinder; a double-acting machine of the same duty would need half this swept volume. (v) The two densities printed at the foot of the question are matched to the fluids by physical sense: $1030\ \text{kg/m}^3$ is whole milk and $1190\ \text{kg/m}^3$ a calcium-chloride brine.

Topic. The vapour-compression cycle with undercooling, and secondary-refrigerant plant.

Key relations. Refrigerating effect $q_e=h_1-h_4$ with $h_4=h_3$ (isenthalpic throttling); work $w=h_2-h_1$ at $s_2=s_1$; $\text{COP}=q_e/w$. Shaft power $\dot W=\dot m_r w/\eta_m$. Compressor displacement: $\dot V_{swept}=\dot m_r v_1/\eta_v$ and $V_{cyl}=\dot V_{swept}/(N n_{cyl})$ for a single-acting machine at $N$ rev/s. Secondary circuit: $\dot m_b=\dot Q_e/(c_{p,b}\Delta t_b)$.

Why this works. The throttling valve is the one irreversible element that the cycle cannot do without: expanding the liquid through a turbine would recover work, but the recovery is small and the machinery impossible in a two-phase stream, so the industry accepts $h_4=h_3$. That is precisely why undercooling pays — every kelvin of liquid subcooling moves state 4 to the left and lengthens the evaporator line at no cost in compressor work. The volumetric efficiency matters for a different reason: it does not change the energy at all, only the size of the machine, because gas trapped in the clearance volume re-expands on the suction stroke and steals induction capacity. That is why a low-pressure-ratio ammonia plant like this one can have a high COP and still need a very large cylinder. The brine loop, finally, is a deliberate thermodynamic penalty accepted for a practical reason: it puts a heat exchanger between the ammonia and the food, at the cost of the extra temperature difference that shows up as a lower evaporating temperature.

Common pitfall. Using $h_f$ at the condensing temperature instead of the stated undercooled liquid enthalpy, which throws away the undercooling the question explicitly supplies. The other classic error is in the swept volume: applying the volumetric efficiency to the mass flow rather than to the volume, or forgetting that a twin-cylinder machine completes two induction strokes per revolution, either of which puts the cylinder size out by a factor of two.

Source. Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., Ch. 10 (vapour-compression cycle) and Ch. 11 (reciprocating compressors, volumetric efficiency); ASHRAE Handbook — Refrigeration, Ch. 2.

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