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anonymous
 3 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.
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0applying 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
 3 years ago
Best ResponseYou've already chosen the best response.0Whoa. 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
 3 years ago
Best ResponseYou've already chosen the best response.0I did not use a differential eq. what you've got is the same as mine but with divided by the density

anonymous
 one year ago
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