Answer:
a) 3.607 m
b) 1.5963 m
Explanation:
See that attached pictures for explanation.
Answer:
COMMON ENGINEERING DOCUMENTS
Inspection or trip reports.
Research, laboratory, and field reports.
Specifications.
Proposals.
Progress reports.
ect...
Explanation:
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT


ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system

For ideal gas
Δh = CpΔT
by putting the values


Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
Answer:

Explanation:
Given
Airline flying at 34,000 ft.
Cabin pressurized to an altitude 8,000 ft.
We know that at standard condition ,density of air

We know that pressure difference
ΔP=ρ g ΔZ
Here ΔZ=34,000-8,000 ft
ΔZ=26,000 ft

ΔP=0.074 x 32.2 x 26,000

So pressure difference will be
.
Answer:
Answer D
Explanation:
They need to be checked monthly