Question

An experimental spacecraft consumes a special fuel at a rate of 353 L/min . The density of the fuel is 0.700 g/mL and the standard enthalpy of combustion of the fuel is − 57.9 kJ/g . Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1 kW = 1 kJ/s

Answer 1

Explanation:

The given data is as follows.

        Space craft fuel rate = 353 L/min

As 1 liter equals 1000 ml and 1 min equals 60 seconds.

So,     $353 \times \frac{1000 ml}{60 sec}$

           = 5883.33 ml/sec

It is also given that density of the fuel is 0.7 g/ml and standard enthalpy of combustion of fuel is -57.9 kJ/g.

Fuel rate per second is 5883.33 ml.

             $5883.33 ml \times 0.7 g/ml$

               = 4118.33 g

Hence, calculate the maximum power as follows.

          Power = Fuel consumption rate × (-enthalpy of combustion)

                      = 4118.33 g/s \times 57.9 kJ/g

                      = 238451.36 kJ/s

or,                  = 238451.36 kW

Thus, we can conclude that maximum power produced by given spacecraft is 238451.36 kW.