Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Answer:
not sure if this helps but
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:
Through Goodman's equations the combined effort by fatigue and compression is expressed as:
Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then
Re-arrenge for 
We know that the stress is representing as,
Then,
Where 
=Max Moment
I= Intertia
The inertia for this object is
Then replacing and re-arrenge for
Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
Answer:
0.008
Explanation:
From the question, the parameters given are:
Velocity V = 5 m/s
Pressure = 10 pa
But pressure = F/A
10 = F/A
F = 10A
Substitute all the parameters into the formula below
Coefficient of viscosity (η) = F × r /[AV]
Where
F = tangential force,
r = distance between layers,
A = Area, and
V = velocity
(η) = 10A × 0.004 /[A × 5]
The A will cancel out
(η) = 10 × 0.004 /[5]
(η) = 0.04 /5
(η) = 0.008
Therefore, the coefficient of viscosity of the fluid is 0.008
