Answer:
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Explanation:
Answer:
106.335 Btu/Ib
Explanation:
Given data :
T1 = 520°R = 288.89K
P1 = 14 Ibf/in^2 = 96526.6 pa
r = 12
k = 1.4
R = 287 J/kg-k
<u>Calculate work done per unit mass of air flowing</u> ( two-stage compressor )
we will apply the equation below
W = 2k / K-1 * ( RT₁ ) *
input values into equation above
W = 247.336 KJ/kg = 106.335 Btu/Ib
Answer:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa
Answer:
Option e)
Explanation:
Dimensional Analysis helps in the analysis of relationship between physical quantities, checking its dimension thus checking the accuracy of a given or to find the dimensions of a quantity or the degree of correctness of an equation.
Dimensional analysis has some of the applications that are listed below:
- To establish a relationship between physical quantities
- To check the accuracy of a given formula
- To derive units of a physical quantity
- To convert system of units in one another.
Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar =
ar =
ar = 22.74 m/s²
so magnitude of total acceleration is
A =
A =
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²