Answer:
u could get hurt or it could not sence in but it easy to work with and u can just relax till u get were ur going.
Explanation:
Answer:
Activation energy for creep in this temperature range is Q = 252.2 kJ/mol
Explanation:
To calculate the creep rate at a particular temperature
creep rate, 
Creep rate at 800⁰C, 
.........................(1)
Creep rate at 700⁰C
.................(2)
Divide equation (1) by equation (2)
![\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%20%2A%2010%5E%7B-4%7D%20%7D%20%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20-%5Cfrac%7B-Q%7D%7B973R%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20%2B%5Cfrac%7BQ%7D%7B973R%7D%20%5D%5C%5CR%20%3D%208.314%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073%2A8.314%7D%20%2B%5Cfrac%7BQ%7D%7B973%2A8.314%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B0.0000115%20Q%5D%5C%5C)
Take the natural log of both sides
Answer:
7.94 ft^3/ s.
Explanation:
So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.
Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.
Therefore; kp/ks = 1/6.
Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.
Reynolds scaling==> Hp/ 700 = (1/6)^2.5.
= 7.94 ft^3/ s
