Working Principle and leakage heat transfer in the nozzle do not
let all the enthalpy drop to be converted to
Steam jet refrigeration is a thermally driven kinetic energy. State 2’ is for isentropic
refrigeration system that operates on the expansion and state 2 is the actual state after
principle of utilizing the expansion of steam to expansion. The nozzle efficiency is
create a cooling effect. Saturation pressure of
water at 5oC is 0.0087 bar, which means that if ηnozzle = (h2′−h1)/(h1−h2)
the pressure in a vessel containing water is The high velocity jet entrains the flash vapours
reduced to 0.0087 bar by a vacuum pump, the from 6. The pressure p2 is equal to p6 but
water will boil at 5oC absorbing its latent heat enthalpy h2 is not same as h6. State 3 is that
from the water in the container. The very low of motive steam after entrainment and state 4
pressure or high vacuum on the surface of the is that of mixture of motive steam and the
water can be maintained by throttling the flash vapour after mixing. At state 3, assume
steam through jets or nozzles. both the motive steam and the flash vapours
Consider a flash chamber contains 100 kg of have velocity V3. momentum conservation
water. The latent heat of water at 5oC is would give
2489.8 kJ/kg and its specific heat is 4.18 kJ/kg- ṁ1V3 + ṁ6V3 = (ṁ1 + ṁ6)*V4 ∴ V 3 = V4 (1)
K. If 1 kg of water is removed by boiling,
pressure is reduced due to throttling of steam. The energy equation for mixing is
Approximately 2385 kJ of heat will be approximated as (neglecting kinetic energies).
removed from the water. The fall in ṁ1h3 + ṁ6h6 = (ṁ1 + ṁ6)*h4 (2)
temperature of the remaining water will be,
Energy balance over control volume including
Q = mCpdT states 1, 5 and 6 gives
dT = 2385/(99* 4.187) = 5.7oC ṁ1h1 + ṁ6h6 = (ṁ1 + ṁ6)*h5 (3)
Evaporating one more kg of water reduces the subtracting (2) from (3)
remaining water temperature by 5.7oC further.
ṁ1*(h1 - h3) = (ṁ1 + ṁ6)*(h5 - h4) (4)
This is the governing equation for the steam-
jet refrigeration system. The left-hand side is
the enthalpy drop of motive steam after
entrainment and the right-hand side is the
energy required for compression of the
mixture of motive steam and flash vapour.
diagram of the steam ejector
Equation is stated as: Energy required for
compression is equal to the enthalpy drop of
motive steam.
Considering the flash chamber and the load as
control volume, we get for refrigeration load,
QE = ṁ6*(h6 - h7) – Wp1 (5)
Energy input to the motive steam is given by,
Qsteam = ṁ1h1 - (ṁ1 + ṁ6)*h12
m = ṁ1h1 - (ṁ1 + ṁ6)*(h11 + wp2)
Thermodynamic cycle of the steam-jet ejector system.
Motive steam is assumed to be dry and
saturated. It expands through the nozzle and
the pressure drops from p1 to p2. The friction
QE
∴ COP = =
QSTEAM
ṁ6∗( h6−h7 ) – W p 1 ¿ ¿
ṁ1 h1−(ṁ1+ ṁ6 )∗(h11 + w p 2 )
References
"Refrigeration and Air Conditioning" by
C.P. Arora
“Refrigeration Cycles 6.8 Steam Jet
Refrigeration System” Prof. U.S.P. Shet,
Prof. T. Sundararajan, and Prof. J.M.
Mallikarjuna
