Solution Naval Numerical 11
Numerical
11: A ship of 14900 tonne displacement has a shaft power of 4460 Kw at
14.55 knots. The shaft power is reduced to 4120 kW and the fuel
consumption at the same displacement is 541 kg/h. Calculate the fuel
coefficient for the ship.
Solution:(A)
(B). Given that
Δ = 14900 tonne
sp1 = 4460 kW
V1 = 14.55 knots.
sp2 =4120 kW
C = 541 kg/hr
Fuel coeff.=?
We require to find V2.
Using the relation,
^{3} "\small \frac{sp1}{sp2}=\left (\frac{V1}{V2} \right )^{3}")
^{3} "\small \frac{4460}{4120}=\left (\frac{14.55}{V2} \right )^{3}")
V2=14.17knots
Now at 14.17knots,
fuel consumption =541kg/hr
We know that fuel consumption/day
the consumption in tonne per day = 541 x 24/1000
=12.98tonne/day
thus, the fuel coefficient is
Fuel coefficient = 132700.
Solution:(A)
(B). Given that
Δ = 14900 tonne
sp1 = 4460 kW
V1 = 14.55 knots.
sp2 =4120 kW
C = 541 kg/hr
Fuel coeff.=?
We require to find V2.
Using the relation,
V2=14.17knots
Now at 14.17knots,
fuel consumption =541kg/hr
We know that fuel consumption/day
the consumption in tonne per day = 541 x 24/1000
=12.98tonne/day
thus, the fuel coefficient is
Fuel coefficient = 132700.
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