Answer:
Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb
Vg = specific volume of saturated steam = 0.430129 ft^3/lb
Explanation:
Given data:
water temperature is given as 287-degree Celcius
we have to find Vf and Vg
Vf = specific volume of saturated liquid
Vg = specific volume of saturated steam
we know that from the saturated steam table we can find these value
therefore for temperature 287-degree Celcius
Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb
Vg = specific volume of saturated steam = 0.430129 ft^3/lb
Answer: The answer is given below
Explanation:
Ductile to brittle transition temperature (DBTT) is the temperature when there is a pronounced reduction in the ability of a material to absorb force without being fractured.
It shuld be noted that when this happens, the material transitions from ductile to brittle which is known as ductile to brittle transition temperature(DBTT).
DBTT
Answer:
Given that
LHS of above given equation have dimension ![[M^oL^{1}T^{-1}]](https://tex.z-dn.net/?f=%5BM%5EoL%5E%7B1%7DT%5E%7B-1%7D%5D)
.
Now find the dimension of RHS
Dimension of P = ![[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
.
Dimension of d= ![[M^{0}L^{1}T^{0}]](https://tex.z-dn.net/?f=%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D)
.
Dimension of μ = ![[ML^{-1}T^{-1}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D)
.
Dimension of L= .
So
![\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5Cdfrac%7B%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%5E2%7D%7B%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%7D)
![\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5BM%5E0L%5E%7B1%7DT%5E%7B-1%7D%5D)
It means that both sides have same dimensions.
Answer:
False......................
