Answer:
∆S1 = 0.5166kJ/K
∆S2 = 0.51826kJ/K
Explanation:
Check attachment for solution
Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 ×
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85
)²× 9
× sin18 × cos18
Qd = 94.305 ×
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 ×
× π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 ×
m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 ×
- 85.2 ×
Qx = 9.10
m³/s
Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.
Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:
Without any extra adjustments or corrections, either 125% of the continuous load, OR
When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).
This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.
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Answer: the half-angle "alpha" of the Mach cone = 30⁰
Explanation:
To calculate the half-angle "alpha" of the Mach cone.
we say ;
Sin∝ = 1 / Ma
given that Ma = 2
now we substitute
Sin∝ = 1 / 2
Sin∝ = 0.5
∝ = Sin⁻¹ 0.5
∝ = 30⁰
Therefore, the half-angle "alpha" of the Mach cone is 30⁰