Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, 
= 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X 
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
= 
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m
Answer:
The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.
Explanation:
The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform. In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.
Can I get a picture? Of the question so I can understand it more?
Answer:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Explanation:
Calculation to estimate the upper and lower bounds of the modulus of this composite.
First step is to calculate the maximum modulus for the combined material using this formula
Modulus of Elasticity for mixture
E= EcuVcu+EwVw
Let pug in the formula
E =( 110 x 0.40)+ (407 x 0.60)
E=44+244.2 GPa
E=288.2GPa
Second step is to calculate the combined specific gravity using this formula
p= pcuVcu+pwTw
Let plug in the formula
p = (19.3 x 0.40) + (8.9 x 0.60)
p=7.72+5.34
p=13.06
Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness
UPPER BOUNDS
Using this formula
Upper bounds=E/p
Let plug in the formula
Upper bounds=288.2/13.06
Upper bounds=22.07 GPa
LOWER BOUNDS
Using this formula
Lower bounds=EcuVcu/pcu+EwVw/pw
Let plug in the formula
Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3
Lower bounds=(44/8.9)+(244.2/19.3)
Lower bounds=4.94+12.65
Lower bounds=17.59 GPa
Therefore the Estimated upper and lower bounds of the modulus of this composite will be:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
So we equate the COP of our heater with COP of Carnot heater
Rearrange the equation
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C
