A performance test on a single cylinder, four stroke, reciprocating internalcombustion petrol engine produced the following data:The indicated mean effective pressure 119.7kN/m2The swept volume 0.014m3Engine speed 2000rev/minThe mechanical efficiency 72%Fuel consumption 2 x 10-6 m3/sDensity of fuel 740kg/m3Calorific value of fuel 43,000 kJ/kgAir flow rate into the engine 0.025 m3/sAir density 1.2 kg/m3Exhaust temperature 502oCSpecific heat capacity of exhaust gas 1.01kJ/kgKCooling water inlet temperature 12oCCooling water outlet temperature 52oCCooling water flow rate 0.12kg/sSpecific heat capacity of cooling water 4.2kJ/kgKAir temperature at engine inlet 12oC.(i) Carry out an energy balance on the engine and show that the first law ofthermodynamics is valid.

To carry out an energy balance on the engine and show that the first law of thermodynamics is valid, we need to calculate the energy input, the useful energy output, and the energy losses.

1. Energy Input: The energy input is the energy provided by the fuel. This can be calculated by multiplying the fuel consumption rate by the density of the fuel and the calorific value of the fuel.

   Energy Input = Fuel Consumption Rate x Density of Fuel x Calorific Value of Fuel = 2 x 10^-6 m^3/s x 740 kg/m^3 x 43,000 kJ/kg = 63.64 kJ/s

2. Useful Energy Output: The useful energy output is the work done by the engine. This can be calculated by using the formula for the power of an engine, which is the product of the indicated mean effective pressure, the engine speed, and the swept volume, divided by 2 (since it's a four-stroke engine, it produces work every two revolutions).

   Useful Energy Output = Indicated Mean Effective Pressure x Engine Speed x Swept Volume / 2 = 119.7 kN/m^2 x 2000 rev/min x 0.014 m^3 / 2 = 16.76 kJ/s

3. Energy Losses: The energy losses are due to heat transfer to the cooling water and the exhaust gases. The heat transfer to the cooling water can be calculated by multiplying the cooling water flow rate by the specific heat capacity of the cooling water and the temperature difference between the cooling water outlet and inlet. The heat transfer to the exhaust gases can be calculated by multiplying the air flow rate into the engine by the air density, the specific heat capacity of the exhaust gas, and the temperature difference between the exhaust temperature and the air temperature at the engine inlet.

   Heat Transfer to Cooling Water = Cooling Water Flow Rate x Specific Heat Capacity of Cooling Water x (Cooling Water Outlet Temperature - Cooling Water Inlet Temperature) = 0.12 kg/s x 4.2 kJ/kgK x (52°C - 12°C) = 20.16 kJ/s

   Heat Transfer to Exhaust Gases = Air Flow Rate into Engine x Air Density x Specific Heat Capacity of Exhaust Gas x (Exhaust Temperature - Air Temperature at Engine Inlet) = 0.025 m^3/s x 1.2 kg/m^3 x 1.01 kJ/kgK x (502°C - 12°C) = 14.92 kJ/s

   Total Energy Losses = Heat Transfer to Cooling Water + Heat Transfer to Exhaust Gases = 20.16 kJ/s + 14.92 kJ/s = 35.08 kJ/s

The first law of thermodynamics states that the energy input must be equal to the sum of the useful energy output and the energy losses. Let's check if this is valid:

Energy Input = Useful Energy Output + Energy Losses 63.64 kJ/s = 16.76 kJ/s + 35.08 kJ/s 63.64 kJ/s = 51.84 kJ/s

The energy input is not equal to the sum of the useful energy output and the energy losses. This discrepancy could be due to other energy losses not accounted for in the calculations, such as frictional losses and losses due to incomplete combustion. Therefore, the first law of thermodynamics is not valid in this case.