Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is
, and the steam properties as, A-4E:

Using the energy balance for the system:

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R
Answer:
3540.5N
Explanation:
Step one:
given data
mass m= 0.196kg
speed v= 31m/s
distance r= 5.32cm = 0.0532m
Step two
The expression relating force, mass, velocity and distance is
F= mv^2/r
substitute we have
F=0.196*31^2/0.0532
F=0.196*961/0.0532
F=188.356/0.0532
F=3540.5N