In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini
tially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg K and 4.4 kJ/kg K, respectively, determine the final equilibrium temperature after quenching, in K.
1 answer:
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Answer:d
Explanation:
Given
Temperature
Also
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.
(a)
Mach no.
Mach no.=0.63
(b)
Mach no.
Mach no.=0.31
(c)
Mach no.
Mach no.=1.27
(d)
Mach no.
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds