Man B&W S50ME-B9.3-TII, Project Manual

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MAN B&W 6.04
Page 7 of 12
MAN Diesel
198 90 84-5.0 MAN B&W S50ME-B9.3\-TII
Calculation of Freshwater Production for Derated Engine
Example 2:
Freshwater production from a derated 6S50ME-B9.3-TII with 1 high efficiency MAN TCA55-26
turbocharger , high load and fixed pitch propeller.
Based on the engine ratings below, this example will show how to calculate the expected available jacket
cooling water heat removed from the diesel engine, together with the corresponding freshwater production
from a freshwater generator.
The calculation is made for the service rating (S) of the diesel engine being 80% of the specified MCR.
Nominal MCR, (L
1
) P
L1
: 10,680 kW (100.0%) and 117.0 r/min (100.0%)
Specified MCR, (M) P
M
: 9,612 kW (90.0%) and 111.2 r/min (95.0%)
Service rating, (S) P
S
: 7,689 kW and 103.2 r/min, P
S
= 80.0% of P
M
Reference conditions
Air temperature T
air
.............................................................. 20° C
Scavenge air coolant temperature T
CW
............................... 18° C
Barometric pressure p
bar
...................................................... 1,013 mbar
Exhaust gas backpressure at specified MCR
∆
p
M
............ 300 mm WC
The expected available jacket cooling water heat at service rating is found as follows:
Q
jw,L1
= 1,690 kW from List of Capacities
Q
jw%
= 92.2% using 90.0% power and 95.0%
speed for M in Fig. 6.04.02
By means of equation [1], and using factor 0.885
for actual ambient condition the heat dissipation
in the SMCR point (M) is found:
Q
jw,M
= Q
jw,L1
x
Q
jw%
___
100
x 0.885
= 1,690 x
92.2
____
100
x 0.885 = 1,379 kW
By means of equation [2], the heat dissipation in
the service point (S) i.e. for 80.0% of specified
MCR power, is found:
k
p
= 0.852 using 80.0% in Fig. 6.04.04
Q
jw
= Q
jw,M
x k
p
= 1,379 x 0.852 = 1,175 kW
-15%/0%
For the service point the corresponding expected
obtainable freshwater production from a freshwa-
ter generator of the single effect vacuum evapora-
tor type is then found from equation [3]:
M
fw
= 0.03 x Q
jw
= 0.03 x 1,175 = 35.3 t/24h
-15%/0%