## JeanetteBaker 2 years ago 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.

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1. electrokid

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

2. abb0t

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

3. electrokid

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

4. Dalbertjam

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