Answer:
Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib
Answer : moment of inertia = 186.7 Ib - in
Explanation:
Given data
weight of the mailbox = 3.2 Ib
weight of the uniform cross member = 10.3 Ib
The origin is of mailbox and cross member is 0
The perpendicular distance from Y axis of centroid of the mailbox
= 4 + (25/2) = 16.5"
The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"
therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)
= 52.8 + 133.9 = 186.7 Ib-in
Answer:
Explanation:
g = Acceleration due to gravity at sea level =
R = Radius of Earth = 6371000 m
h = Altitude of observatory = 4205 m
Change in acceleration due to gravity due to change in altitude is given by
Weight at sea level
Weight at the given height
Change in weight
Her weight reduces by .
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:
It is necessary to get the Reynold's number:
Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:
The overall heat transfer coefficient:
Here
Substituting values: