Pleaae find $$EGL$.$ It is not $+$ or $-0\mathrm{.}42.$

A diffuser in a pipe flow is basically a slow expansion of the pipe diameter, which slows down the fluid velocity and increases the pressure $($the Bernoulli effect$).$ Water at room temperature flows with a volume flow rate of $0\mathrm{.}0250\frac{m^3}{s}$ through a horizontal diffuser in which the pipe diameter increases gradually from $D_1=6\mathrm{.}00$ to $D_2=11\mathrm{.}00cm.$ The irreversible head loss through the diffuser is estimated to be $0\mathrm{.}42m.$ The flow is turbulent, and the kinetic energy correction factors at both the inlet and the outlet of the diffuser are assumed to be $1\mathrm{.}05.$

It may be surprising that the answer to part $($a$)$ is positive, i$.$e$.,$ the pressure rises downstream. How is this possible? Explain by calculating the change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}\text{E}\text{G}\text{L}}$ and the change in hydraulic grade line $\frac{?}{{?}_{\frac{?}{?}}HGL}$ from the upstream to the downstream location. In particular, does EGL go up or down, and does HGL go up or down?

The change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}\text{E}\text{G}\text{L}}$ is calculated to be m$.$

The change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}HGL}$ is calculated to be m$.$

A diffuser in a pipe flow is basically a slow expansion of the pipe diameter, which slows down the fluid velocity and increases the pressure $($the Bernoulli effect$).$ Water at room temperature flows with a volume flow rate of $0\mathrm{.}0250\frac{m^3}{s}$ through a horizontal diffuser in which the pipe diameter increases gradually from $D_1=6\mathrm{.}00$ to $D_2=11\mathrm{.}00cm.$ The irreversible head loss through the diffuser is estimated to be $0\mathrm{.}42m.$ The flow is turbulent, and the kinetic energy correction factors at both the inlet and the outlet of the diffuser are assumed to be $1\mathrm{.}05.$

It may be surprising that the answer to part $($a$)$ is positive, i$.$e$.,$ the pressure rises downstream. How is this possible? Explain by calculating the change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}\text{E}\text{G}\text{L}}$ and the change in hydraulic grade line $\frac{?}{{?}_{\frac{?}{?}}HGL}$ from the upstream to the downstream location. In particular, does EGL go up or down, and does HGL go up or down?

The change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}\text{E}\text{G}\text{L}}$ is calculated to be $-0\mathrm{.}42m.$

The change in energy grade line $\frac{?}{{?}_{\frac{?}{?}}HGL}$ is calculated to be

$m.$