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Answer:
temperature T2 = 826.9°C
= -1142.7 kJ/kg
Explanation:
given data
initial state temperature = 120°C
final state pressure p1 = 1 bar
pressure p2 = 100 bar
solution
we use here super heated water table A6 that is
specific internal energy u1 = 2537.3 kJ/kg
specific entropy s1 = 7.4668 kJ/kg.K
and
here specific entropy stage 1 = stage 2
so for specific entropy and pressure 100 bar
specific internal energy u2 = 3680 kJ/kg
and temperature T2 = 826.9°C
so here now we get specific work of steam is
ΔU = -W ...........1
m ( u1 - u2) = W
= u1 - u2
= 2537.3 - 3680
= -1142.7 kJ/kg
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
