A steam stream of 400 m3 /s, standard (standard cubic meter per second) is cooled by adding 7 m3 /s cold water. The fluid leaving the system is liquid water. The crosssectional area of the outlet port is 0.5 m2 . What is the velocity of the leaving water stream? (Hint: steam is a gas. Its specific gravity is about 18/29.)

510 m/s

Explanation: specific gravity of steam is 18/29 = 0.620

It is the ratio of the density of steam over density of water

400m3/s of steam =

400m3ms * 0.620 of water

= 248m3/s of water

Total flow rate Q = 248 + 7 = 255m3/s

Using Q = AV

Where A is area and V is velocity

V = Q/A

V = 255/0.5 = 510m/s

this means that the temperature of both substances will be equal

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.