A gasoline engine on a ship has a power of 50 kW. The combustion chamber receives 15 kg/h of fuel and 215 kg/h of air. The water that cools the engine absorbs heat at a rate of 42 kJ/s, and there is heat loss due to radiation. We need to find the heat loss due to radiation.

1. Problem Description

A gasoline engine on a ship has a power of 50 kW. The combustion chamber receives 15 kg/h of fuel and 215 kg/h of air. The water that cools the engine absorbs heat at a rate of 42 kJ/s, and there is heat loss due to radiation. We need to find the heat loss due to radiation.

2. Solution Steps

First, convert the power of the engine from kW to kJ/s:

$Power = 50 \text{ kW} = 50 \text{ kJ/s}$

The total heat input to the engine is the heat released from the fuel during combustion. The amount of fuel consumed per second is:

$\text{Fuel flow rate} = 15 \frac{\text{kg}}{\text{h}} = 15 \frac{\text{kg}}{3600 \text{ s}} = \frac{15}{3600} \frac{\text{kg}}{\text{s}}$

The energy balance equation is:

$Q_{in} = Q_{out}$

$Q_{in} = \text{Heat released by fuel combustion per second}$.

The heat output consists of the engine power, the heat absorbed by the cooling water, and the heat loss due to radiation. Let $Q_{rad}$ be the heat loss due to radiation.

Then, the energy balance can be written as:

$Q_{fuel} = Power + Q_{water} + Q_{rad}$

$Q_{rad} = Q_{fuel} - Power - Q_{water}$

From the textbook (Fig. 4.12) given, we assume the calorific value of fuel to be $42700 kJ/kg$. So,

$Q_{fuel} = \text{fuel flow rate} \times \text{calorific value}$

$Q_{fuel} = \frac{15}{3600} \frac{\text{kg}}{\text{s}} \times 42700 \frac{\text{kJ}}{\text{kg}} = \frac{15 \times 42700}{3600} \text{ kJ/s} = \frac{640500}{3600} \text{ kJ/s} = 177.916666... \text{ kJ/s}$

$Q_{water} = 42 \text{ kJ/s}$

$Power = 50 \text{ kJ/s}$

Therefore, the heat loss due to radiation is:

$Q_{rad} = 177.916666... - 50 - 42 = 177.916666... - 92 = 85.916666... \text{ kJ/s}$

3. Final Answer

The heat loss due to radiation is approximately 85.92 kJ/s.