Question

A lead bullet is shot down from the altitude of 100 m above the ground. What must be the
minimal initial velocity of the bullet so that it completely melts after a perfectly inelastic collision
with the ground? The initial temperature of the bullet is 500 K, melting temperature of lead is 600
K. Speci c heat capacity of lead is c = 130 J/(kgK), speci c heat of fusion of lead is r = 24 kJ/kg.
The heat released after the collision is distributed equally between the bullet and the ground.

Answer 1

We will have to do energy balance for this case

General energy balance on per unit mass basis is given by

Now we apply this general energy balance to our case

The system given above is a steady state process also there is no Heat transfer or work done

Hence we are left with initial and final flow work

Hin + (1/2)*Vin² + g*Zin - Hout - (1/2)*Vout² - g*Zout = 0

Now we will take reference temperature as initial temperature = 500 K

Hence Hin = 0

and we will take ground as datum, hence, Zout= 0 and finally after collision it will come to rest, hence Vout = 0

hence the equation reduces to

(1/2)*Vin² + g*Zin - Hout = 0

Now Hout = Hsensible + Hfusion = c*(Tout-Tin) + Hfusion = 130*(600-500) + 24*10³ = 37000 J/Kg

(1/2)*Vin²+ 9.81*100 = Hout = 37000

Vin = 268.39 m/s