Answer:
a) ∝ and β
The phase compositions are :
C
= 5wt% Sn - 95 wt% Pb
C
= 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is
put here value with
turbine is insulated so q = 0
so here
solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
Answer:
The answer will be Rule 61G15-23 F.A.C, relating to Seals.
Explanation:
According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.
Answer:
Test code:
>>u=10;
>>g=9.8;
>>q=100;
>>m0=100;
>>vstar=10;
>>tstar=fzero_rocket_example(u, g, q, m0, vstar)
Explanation:
See attached image
Answer:
So the fraction of proeutectoid cementite is 44.3%
Explanation:
Given that
cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.
We know that
Fraction of proeutectoid cementite phase gievn as
Now by putting the values
So the fraction of proeutectoid cementite is 44.3%
