Answer:
Explanation:
The pressures given are relative
p1 = 2000 psi
P1 = 2014 psi = 13.9 MPa
p2 = 4 psi
P2 = 18.6 psi = 128 kPa
Values are taken from the steam pressure-enthalpy diagram
h2 = 2500 kJ/kg
If the output of the turbine has a quality of 85%:
t2 = 106 C
I consider the expansion in the turbine to adiabatic and reversible, therefore, isentropic
s1 = s2 = 6.4 kJ/(kg K)
h1 = 3500 kJ/kg
t2 = 550 C
The work in the turbine is of
w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg
The thermal efficiency of the cycle depends on the input heat.
η = w/q1
q1 is not a given, so it cannot be calculated.
Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
<em>attached below is a detailed solution</em>
Answer:
(a) Calculate the rod base temperature (°C). = 299.86°C
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m
Explanation:
see attached file below
Answer:
The right choice would be Option b (2.545).
Explanation:
The given values are:
The aggregate blend will be:
= 55%
= 2.631
= 25%
= 2.331
= 20%
= 2.609
Now,
On applying the formula, we get
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
Answer and Explanation:
The explanation is attached below