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Refrigerant Compressor Efficiency Calculations

This document contains two examples involving refrigeration compressors. The first example provides operating data for a six-cylinder refrigerant compressor and asks to calculate the clearance volumetric efficiency, actual volumetric efficiency, and compression efficiency. The second example asks to calculate the speed of the impeller tip needed to compress two different refrigerants from 10°C saturated vapor to a condensing temperature of 30°C.

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2K views3 pages

Refrigerant Compressor Efficiency Calculations

This document contains two examples involving refrigeration compressors. The first example provides operating data for a six-cylinder refrigerant compressor and asks to calculate the clearance volumetric efficiency, actual volumetric efficiency, and compression efficiency. The second example asks to calculate the speed of the impeller tip needed to compress two different refrigerants from 10°C saturated vapor to a condensing temperature of 30°C.

Uploaded by

allovid
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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1

4. Compressors



Example No. 1
Catalog data for a six-cylinder refrigerant 22 compressor operating at 29 r/s indicate a
refrigerating capacity of 96.4 kW and a power requirement of 28.9 kW at an operating
temperature of 5 C and condensing temperature of 50 C. The performance data are based
on 3 C liquid subcooling and 8 C superheating of the suction gas entering the compressor.
The cylinder bore is 67 mm and the piston stroke is 57 mm. Compute (a) the clearance
volumetric efficiency if the clearance volume is 4.8 percent, (b) the actual volumetric
efficiency, and (c) the compression efficiency.

Given:
Six-cylinder refrigerant 22 compressor,
29 r/s, 96.4 kW refrigerating capacity,
28.9 kW power requirements,
Operating temperature of 5 C,
Condensing temperature of 30 C,
3 C liquid subcooling, 8 C superheating
Cylinder bore = 67 mm
Piston stroke = 57 mm
Clearance volume = 4.8%
Required:
Solution:


At state 1:
5 C saturation temperature (584 kPa), t
1
= 5 C + 8 C = 13 C.
h
1
= 413.1 kJ/kg; v
1
= 43.2 L/kg; and s
1
= 1.7656 kJ/kg-K.
At state 2:
50 C saturation temperature (1942 kPa), s2 = s1.
h
2
= 444.5 kJ/kg and v
2
= 14.13 L/kg.
At state 3, and state 4, at 47 C sat. (t3 = 50 C 3 C = 47 C).
2

h
3
= h
4
= 259.1 kJ/kg.
(a) The clearance volumetric efficiency is
|
|

\
|
=
|
|

\
|
= 1 100 1 100
2
1
v
v
c
v
v
c
dis
suc
vc

% 1 . 90 1
13 . 14
2 . 43
8 . 4 100 = |

\
|
=
vc

(b) The compressor displacement rate is
( )( )
( )
( ) s L s m r cyl m s r cyl V
D
97 34 03497 0 057 0
4
067 0
29 6
3 3
2
. . .
.
= =
(

=


The actual rate of refrigerant flow is
( )kW h h m q
4 1
=
s kg
kg kJ
kW
m 6260 0
1 259 1 413
4 96
.
. .
.
=

=
The actual volumetric flow rate of the refrigerant measured at the compressor suction is
( )( ) s L kg L s kg V 04 27 2 43 6260 0
1
. . . = =
The actual volumetric efficiency is, then,
100
3
3
1
= =
s m compressor of rate nt displaceme
s m compressor entering rate flow volume
V
V
D
va
,
,

% 3 . 77 100
97 . 34
04 . 27
= =
s L
s L
va

(c) The compression efficiency is the isentropic work of compression divided by the actual
work of compression. The latter is

kW W n compressio of work actual 9 28. = =
so that
100 =
n compressio of work actual
n compressio of work isentropic
c


( )
% 100
1 2

=
W
h h m
c


( ) ( )( )
%
.
. . .
68 100
9 28
1 413 5 444 6260 0
1 2
=

=
kW
kg kJ s kg
W
h h m
c




Example No. 2
Calculate the speed of the impeller tip in order to compress the following refrigerants from
saturated vapour at 10 C to pressure corresponding to a condensing temperature of 30 C
when the refrigerant is (a) refrigerant 11 and (b) ammonia.

Given:
Evaporating temperature of 10 C
Condensing temperature of 30 C
Required:
Speed of the impeller tip for (a) refrigerant 11 and (b) ammonia.
3

Solution:
(a) In the isentropic compression of refrigerant 11 from saturated vapour at 10 C to a
saturated condensing temperature of 30 C
kg kJ h
i
8 . 12 9 . 393 7 . 406 = =
The tip speed is
( ) s m V
t
1 . 113 8 . 12 1000
2
= =
(b) For ammonia
kg kJ h
i
88 1472 1560 = =
The tip speed is
( ) s m V
t
297 88 1000
2
= =

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