Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
Answer:
Explanation:
You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.
Answer:
The strength coefficient is K = 591.87 MPa
Explanation:
We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by
where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.
Solving for strength coefficient
From the strain hardening equation we can solve for K
And we can replace values
Thus we get that the strength coefficient is K = 591.87 MPa
Answer:
The value of critical length = 3.46 mm
The value of volume of fraction of fibers = 0.43
Explanation:
Given data
= 800 M pa
D = 0.017 mm
L = 2.3 mm
= 5500 M pa
= 18 M pa
= 13.5 M pa
(a) Critical fiber length is given by
Put all the values in above equation we get
mm
This is the value of critical length.
(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by
Put all the values in above formula we get
= 0.43
This is the value of volume of fraction of fibers.
