Answer to Question #275810 in Mechanical Engineering for julie

Question

A steam main of 150mm outside diameter containing wet steam at 28bar is insulted with an inner layer of diatomaceous earth, 40mm thick and an outer layer of 85% magnesium 25mm thick. The inside surface of the pipe is at the steam temperature and heat transfer coefficient for the outside surface of the lagging is 17w/m^2k. The thermal conductivities of diatomaceous earth and 85% magnesium are 0.09 and 0.06 w/mk respectively.

Neglecting radiation and thermal resistance of the pipped wall calculate : the rate of heat loss per unit length of the pipe and temperature of the outside surface of the lagging when room temperature is 20 degrees celcius

Answer 1

Momentum after release of steam P.F._f 

P_f = mv + m'v'       (1)

Where mass of empty launch steam m=7500 kg

m' is mass of satellite , m'=250 kg

v is final velocity of launch steam after releasing satellite, v=900 m/s

v' is final velocity of the satellite after release

Putting all the values in equation (1), we get

P_f =(7500 kg)(900 m/s) + (250 kg)v'

P_f = 6.750×10⁶ kg•m/s + (250 kg)v'

Now applying conservation of momentum

Momentum before release of satellite = momentum after release of satellite

P_i = P_f

7.750×10₆ kg•m/s = 6.750×10⁶ kg•m/s + (250 kg)v'

(7.750×10₆ - 6.750×10⁶ )kg•m/s= (250 kg)v'

10⁶ m/s = 250v'

v' = (10⁶/250) m/s

v' = 4000 m/s