# Temperatures, pressures, heat quantities and flow rates for the lithium bromide-water cycle

If it is assumed that the temperature of the liquid refrigerant (water) leaving the condenser

Is tc and that the temperature of evaporation is fe, then it is easy to calculate the mass flow

Rate of refrigerant, mr, which has to be circulated per kW of refrigeration. This calculation depends on the further assumption

Fcve = 2463 + 1.89fe kJ kg"1 (14.1)

Where hvc is the enthalpy of the refrigerant vapour leaving the evaporator.

This is approximately true for the pressures and temperatures in common use in air conditioning.

Similarly, for the enthalpy of the liquid leaving the condenser, /ilc, we can write (in the case of water),

Hic — tcci

= 4.19fc kJ kg-1 (14.2)

Mr(hve — hlc ) = kW of refrigeration

And

Mr = 1/(2463 + 1.89fe — 4.19rc) (14.3)

EXAMPLE 14.1

If a vapour absorption system using water as a refrigerant evaporates at 1°C and condenses

At 32°C, determine the mass flow rate of refrigerant circulated per kW of refrigeration.

By equation (14.3)

MT = 1/(2463 + 1.89 x 1 — 4.19 x 32)

= 0.000 429 0 kg s’1 kW“1

From psychrometric tables the condensing and evaporating pressures are 4.754 kPa and 0.657 kPa, respectively. Saturation temperature °C Fig. 14.2 Saturation vapour pressure of water vapour.

Considering now the absorber-generator part of the system, it is assumed that the temperature in the absorber is /a and that the generator is operating at a temperature tg. Knowing these temperatures and pressures pc and pe, the following data are obtained from tables or from a chart such as Figure 14.3, which gives the properties of lithium bromide-water solutions:

Solution leaving absorber concentration Ca%

(at pe and fa) enthalpy h, d kJ kg“1

Solution leaving generator concentration Cg%

(at pc and /g) enthalpy hg kJ kg-1

EXAMPLE 14.2

Determine the concentrations and enthalpies of the solution in the absorber and the generator, for example 14.1.

Referring to Figure 14.3, we read off the following:

At 0.657 kPa and 32°C, Ca = 57.5% and ha = -165 kJ kg"1

At 4.754 kPa and 74°C, Cg = 60% and h& = -84 kJ kg’1

It is now possible to calculate the mass flow rate of solution that must be circulated to meet the needs of the cycle. If msa is the mass flow rate of solution leaving the absorber Concentration, percent of lithium bromide, by weight Fig. 14.3 Use of a chart giving properties of lithium bromide-water.

And msg the flow rate leaving the generator, both in kg s_1 kW-1, then for a mass balance the following must hold—

Qm sa = Cgwsg (14.4)

"*sa = + mr (14.5)

Substituting for msa from equation (14.5) in equation (14.4), we obtain Camr

 M„„ = (14.6)

EXAMPLE 14.3

Using the data of the previous examples, determine the mass flow rate of solution that must be circulated in the absorber and in the generator.

From equation (14.6),

Msg = = 0.009 87 kg s“1 kW“1

And using equation (14.5),

Msa = 0.009 87 + 0.000 429 = 0.010 30 kg s“1 kW’1

An equation similar to (14.1) can be used to determine hvg, the enthalpy of the superheated water vapour leaving the generator,

 (14.7) Hy g = 2463 + 1.89fg

EXAMPLE 14.4

Using the earlier data and equation (14.7), determine the enthalpy of the water vapour leaving the generator.

Hvg = 2463 + 1.89 x 74 = 2603 kJ kg“1

With the information now available it is possible to calculate the heat balance for the whole cycle.

 (14.8) In the absorber

 Heat of Entering Water Heat of Heat of ‘ Heat to be + Entering — Leaving = Removed at Solution Solution Absorber Vapour J
 Mrhve + msghg — msa/za = H3 kW per kW of refrigeration

In the generator

 Heat of leaving water vapour Heat of Heat of Heat to be + Leaving Solution Entering Solution — Supplied to generator

Mrhvg + msghg — msaha = Hg kW per kW of refrigeration In the condenser

 Heat of entering water vapour

 Heat of leaving liquid

 Heat to be removed at condenser

MThvg — mrhic = Hc kW per kW of refrigeration EXAMPLE 14.5

Using the data of preceding examples, calculate Ha, Hg and Hc. Check the heat balance. Answer

From equations (14.8) and (14.1) and referring to Figure 14.1,

//a = 0.000 429(2463 + 1.89 x 1) + 0.009 87 x (-84) — 0.010 30 x (-165)

= 1.0574-0.8291 + 1.6995 = 1.9278 kW per kW of refrigeration

From equation (14.9),

Hg = 0.000 429 x 2603 + 0.009 87 x (-84) — 0.010 30 x (-165)

= 1.1167-0.8291 + 1.6995 = 1.987 kW per kW of refrigeration

From equation (14.10) and (14.2),

Hc = 0.000 429(2603 — 4.19 x 32)

= 1.0592 kW per kW of refrigeration Summarising these results,

(14.10)

 1.928 1.059 2.987 KW/kW 1.987 1.000 2.987 kW/kW Heat removed From absorber From condenser

Total

Heat added To generator To evaporator

Posted in Engineering Fifth Edition