A liquid water turbine receives 2 kg/s water at 2000 kPa, 20 C with a velocity of 15 m/s. The exit is at 100 kPa, 20 C, and very low velocity. Find the specific work and the power produced.

Question

anonymous · Answer

applying Bernoulli's equation, we have
$$
P_1+{1\over2}\rho v_1^2=P_2+{1\over 2}\rho v_2^2\
P_1=2000\times10^3{\rm Pa}\
P_2=100\times10^3{\rm Pa}\
v_1=15{\rm m/s}\
v_2=?\
\text{then,}\
\text{work done}=\text{change in kinetic energy}\
W={1\over 2}mv_2^2-{1\over 2}mv_2^2\
{\rm then,}\
P={W\over t}
$$

abb0t · Answer

Whoa. Here's another method without using differential equations

The energy equation I found was given by: 
$h_1+\frac{1}{2}V_1^2+gZ_1=h_2\frac{1}{2} V_2^2+gZ_2+w_T$ 
$Z_1 = Z_2$ and $V_2=0$ 

$h_1 = 85.82 \frac{kj}{kg}$ and $h_2 = 83.94$ same units.
which is at 2.3 kPa so u add $\Delta PV$ = $97.7 \times 10^{-2}$
plug in everything, you get 1.99 kj/ kg

and for $W_T$ = $m \times w_T$ = 3.985

specific kinetic energy is not relevent here

anonymous · Answer

I did not use a differential eq.
what you've got is the same as mine but with divided by the density

anonymous · Answer

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