โ๏ธ Absorption Refrigeration
Analyze NH3-H2O and LiBr-H2O absorption refrigeration systems. Compute COP, circulation ratio, and heat duties.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
Thermodynamics
๐ Solver Telemetry
โ ACTIVE
๐๏ธ Views
94
โก Solves
74
๐พ Downloads
378
๐ฆ Fortran Code
11.4 KB
๐
Released
Jun 2026
โฑ๏ธ Latency
< 1 ms
๐ Configuration
Key Equations:
COP = Qevap/(Qgen+Wpump)
COPCarnot = (Te/(TcโTe))ยท(TgโTa)/Tg
f = xs/(xsโxw)
Energy: Qgen+Qevap+Wp = Qcond+Qabs
COP = Qevap/(Qgen+Wpump)
COPCarnot = (Te/(TcโTe))ยท(TgโTa)/Tg
f = xs/(xsโxw)
Energy: Qgen+Qevap+Wp = Qcond+Qabs
๐ Results
Configure inputs and click Analyze to view results.
๐ Methodology
NHโ-HโO System
Ammonia is the refrigerant, water is the absorbent. Can reach sub-zero evaporator temperatures (โ25ยฐC and below). Requires a rectifier to purify ammonia vapor from the generator.
LiBr-HโO System
Water is the refrigerant, LiBr solution is the absorbent. Limited to evaporator temps above 0ยฐC (typically 5โ10ยฐC for air conditioning). Higher COP than NHโ systems at moderate conditions.
Assumptions
- Simplified property correlations (educational).
- No solution heat exchanger (SHX) modeled.
- Steady-state operation.
- Pump work is small relative to Q_gen.
- No rectifier losses for NHโ system.
๐ Calculation Methodology: Single-Effect Absorption Refrigeration (H2O-LiBr / NH3-H2O)
Mathematical Model & Theory
Absorption chillers replace mechanical compression with a thermal compressor loop (generator, absorber, pump, solution heat exchanger) driven by low-grade thermal waste heat:
$$COP_{abs} = \frac{Q_{evap}}{Q_{gen} + W_{pump}} \approx \frac{Q_{evap}}{Q_{gen}}, \quad COP_{Carnot} = \left(\frac{T_{evap}}{T_{cond} - T_{evap}}\right)\left(\frac{T_{gen} - T_{abs}}{T_{gen}}\right)$$
Assumptions
- Steady-state equilibrium in binary solution pair ($H_2O-LiBr$ or $NH_3-H_2O$).
- Negligible solution pump electrical work compared to generator thermal duty.
Academic References
- Herold, K. E., Radermacher, R., & Klein, S. A.: Absorption Chillers and Heat Pumps, CRC Press.
- Moran, M. J. et al.: Fundamentals of Engineering Thermodynamics, Wiley.
Worked Engineering Example
Problem Statement:
An $H_2O-LiBr$ absorption chiller produces $Q_{evap} = 350\text{ kW}$ of cooling at $T_{evap} = 5^\circ\text{C}$ with generator heat input $Q_{gen} = 480\text{ kW}$ at $90^\circ\text{C}$. Calculate actual and Carnot COP ($T_{cond} = T_{abs} = 35^\circ\text{C}$).
Step-by-step Solution:
1. Actual COP: $COP = 350 / 480 = 0.729$.
2. Carnot maximum COP: $COP_{rev} = \frac{278.15}{308.15 - 278.15} \times \frac{363.15 - 308.15}{363.15} = \frac{278.15}{30} \times \frac{55}{363.15} = 9.272 \times 0.1515 \approx 1.405$.
Final Result:
Thermal COP is $\mathbf{0.729}$ (Second-law exergetic efficiency $\mathbf{51.9\%}$).
An $H_2O-LiBr$ absorption chiller produces $Q_{evap} = 350\text{ kW}$ of cooling at $T_{evap} = 5^\circ\text{C}$ with generator heat input $Q_{gen} = 480\text{ kW}$ at $90^\circ\text{C}$. Calculate actual and Carnot COP ($T_{cond} = T_{abs} = 35^\circ\text{C}$).
Step-by-step Solution:
1. Actual COP: $COP = 350 / 480 = 0.729$.
2. Carnot maximum COP: $COP_{rev} = \frac{278.15}{308.15 - 278.15} \times \frac{363.15 - 308.15}{363.15} = \frac{278.15}{30} \times \frac{55}{363.15} = 9.272 \times 0.1515 \approx 1.405$.
Final Result:
Thermal COP is $\mathbf{0.729}$ (Second-law exergetic efficiency $\mathbf{51.9\%}$).