Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
![E(t)=E_o[1-a\frac{T}{T_m}]](https://tex.z-dn.net/?f=E%28t%29%3DE_o%5B1-a%5Cfrac%7BT%7D%7BT_m%7D%5D)
where,
E(t) is the modulus of elasticity at any temperature 'T'
is the modulus of elasticity at absolute zero.
is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.
Answer:
1) the final temperature is T2 = 876.76°C
2) the final volume is V2 = 24.14 cm³
Explanation:
We can model the gas behaviour as an ideal gas, then
P*V=n*R*T
since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:
P*V^k = constant = C, k= adiabatic coefficient for air = 1.4
then the work will be
W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)
W = (P1*V1/T1)*(T2-T1)/(1-k)
T2 = (1-k)W* T1/(P1*V1) +T1
replacing values (W=-450 J since it is the work done by the gas to the piston)
T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C
the final volume is
TV^(k-1)= constant
therefore
T2/T1= (V2/V1)^(1-k)
V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³
Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".
Obtain the following properties at 10kPa from the table "saturated water"
Calculate the enthalpy at exit of the turbine using the energy balance equation.
Since, the process is isentropic process 
Use the isentropic relations:
Calculate the enthalpy at isentropic state 2s.
a.)
Calculate the isentropic turbine efficiency.
b.)
Find the quality of the water at state 2
since 
at 10KPa <
<
at 10KPa
Therefore, state 2 is in two-phase region.
Calculate the entropy at state 2.
Calculate the rate of entropy production.
since, Q = 0
Answer:
Complete question is:
write the following decorators and apply them to a single function (applying multiple decorators to a single function):
1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”
2. The decorator will return the wrapper per usual.
3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.
4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.
5. Apply both decorators (by @ operator to greetings()).
6. Invoke the greetings() function and capture the result.
Code :
def strong_decorator(func):
def func_wrapper(name):
return "<strong>{0}</strong>".format(func(name))
return func_wrapper
def em_decorator(func):
def func_wrapper(name):
return "<em>{0}</em>".format(func(name))
return func_wrapper
@strong_decorator
@em_decorator
def Greetings(name):
return "{0}".format(name)
print(Greetings("Hello"))
Explanation:
