Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
Answer and Explanation:
The crack formation growth that takes place in an environment corrosive.
Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.
Some of the characteristic features of stress corrosion cracks are:
- These occur at high temperatures.
- Occurrence of failures in metals mechanically.
- Occurrence of sudden and unexpected failures under tensile stress.
- The rate of work hardening of the metal alloy is high.
- Time
- An environment that is specific for stress corrosion cracking.
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore
and radial component of given velocity is zero
we have
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
so
solvingt for 
therefore eccentrcity of orbit is 0.22
Answer:
Check the explanation
Explanation:
The loop invariant has to satisfy some amount of requirements to be of good use. Another complex factor as to why a loop is the question of loop termination. A loop that doesn’t terminate can’t invariably be correct, and in fact the computation in whatever form amounts to nothing. The total axiomatic description of a while construct will have to involve all of the following to be true, in which I is the loop invariant:
P => I
{I and B} S {I}
(I and (not B)) => Q
Then the loop terminates
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