For a Pelton wheel with a given water jet velocity, the maximum output power from the Pelton wheel is obtained when the ratio of the bucket speed to the water jet speed is _______ (correct to two decimal places).

The correct answer is 0.50.

Step-by-step Derivation and Explanation:

A Pelton wheel is an impulse turbine where the pressure energy of water is converted into kinetic energy in a nozzle before striking the buckets of the runner. Let us define the variables involved:
Let $V$ (or $V_1$) be the velocity of the water jet striking the bucket.
Let $u$ be the peripheral velocity (bucket speed) of the Pelton wheel.

The relative velocity of the jet entering the bucket is given by:
$V_{r1}=V-u$

Assuming smooth, frictionless buckets and neglecting any losses, the relative velocity at the exit remains the same:
$V_{r2}=V_{r1}=V-u$

The blade outlet angle is denoted by $\beta$, and the deflection angle of the jet is $\theta =180°-\beta$. The whirl velocity component at the inlet is:
$V_{w1}=V$

The whirl velocity component at the outlet is:
$V_{w2}=V_{r2}\cos \beta -u=(V-u)\cos \beta -u$

The work done per second (power output $P$) by the jet on the runner is given by:
$P=\rho Q(V_{w1}+V_{w2})u$
where $\rho$ is the density of water and $Q$ is the volumetric flow rate.

Substituting the whirl velocities:
$P=\rho Q[V+(V-u)\cos \beta -u]u$
$P=\rho Q(V-u)(1+\cos \beta )u$

To find the bucket speed $u$ that maximizes the output power, we differentiate $P$ with respect to $u$ and set it to zero:
$\frac{dP}{du}=0$

Since $\rho$, $Q$, and $(1+\cos \beta )$ are constants, we only need to differentiate the product terms involving $u$:
$\frac{d}{du}\left[\right(V-u\left)u\right]=0$
$\frac{d}{du}(Vu-u^2)=0$
$V-2u=0$
$u=\frac{V}{2}$

Therefore, the ratio of bucket speed ($u$) to water jet speed ($V$) is:
$\frac{u}{V}=0.5$

Thus, the ratio of the bucket speed to the water jet speed for maximum output power is 0.50 (correct to two decimal places).