Question

thermo A large container ship is propelled by a heat engine that uses fuel oil as a heat source and sea water as a heat sink. The engine produces 75000 W while consuming 13600 Litres per hour of fuel oil. The fuel oil releases 43000 kJ per litre burned. Calculate the rate of heat transfer to the ocean and the thermal efficiency of the ships heat engine. (5 points)

Answer 1

Answer:

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$, $\dot Q_{out} = 162369.444\,kW$

Explanation:

The definition of thermal efficiency follows to this expression:

$\eta = \frac{\dot W}{\dot Q_{in}}$

$\eta = \frac{75\,kW}{\left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}$

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$

The rate of heat transfer to the ocean is:

$\dot Q_{out} = \dot Q_{in}-\dot W$

$\dot Q_{out} = \left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)-75\,kW$

$\dot Q_{out} = 162369.444\,kW$