Respuesta :
Explanation:
Given that,
Speed = 5x m/s
Pressure = 10x² N/m²
(a). We need to calculate the time rate of change of pressure at the fixed location x = 1
Using formula of rate of change of pressure at fixed location
[tex]\dfrac{\delta P}{\delta t}=\dfrac{\delta}{\delta t}(10x^2)[/tex]
[tex]\dfrac{\delta P}{\delta t}=0[/tex]
(b). We need to calculate the the time rate of change of pressure for a fluid particle flowing past x = 1
Using formula of rate of change of pressure at the given instant
[tex]\dfrac{DP}{Dt}=\dfrac{\delta P}{\delta t}+u\dfrac{\delta P}{\delta x}+\dfrac{\delta P}{delta y}+\dfrac{\delta P}{\delta z}[/tex]
[tex]\dfrac{DP}{Dt}=u\dfrac{\delta P}{\delta x}[/tex]
Put the value into the formula
[tex]\dfrac{DP}{Dt}=5x\dfrac{\delta}{\delta x}(10x^2)[/tex]
[tex]\dfrac{DP}{Dt}=5x\times20x[/tex]
At x = 1,
[tex]\dfrac{DP}{Dt}=100\ N/m^2 s[/tex]
(c). The velocity is instantaneous at x = 1 for a part and for b part the velocity is steady flow.
Therefore, the answer is different for a and b part.
Hence, This is the required solution.
