DM}}\)\(\Rightarrow \frac{{d{W_P}}}{{dm}} = {h_4} - {h_3} = dh\) __________(i)and REVERSIBLE adiabatic compression as = dh – vdp = 0⇒ dh = cdpnow from EQUATION (1)\(\begin{array}{l} \frac{{d{W_P}}}{{dm}} = dh = vdp = \frac{1}{\rho }\left( {{P_1} - {P_2}} \right)\\ \frac{{d{W_P}}}{{dm}} = \frac{{3000 - 70}}{{1000}} = 2.93\;kJ/kg \END{array}\)

"> DM}}\)\(\Rightarrow \frac{{d{W_P}}}{{dm}} = {h_4} - {h_3} = dh\) __________(i)and REVERSIBLE adiabatic compression as = dh – vdp = 0⇒ dh = cdpnow from EQUATION (1)\(\begin{array}{l} \frac{{d{W_P}}}{{dm}} = dh = vdp = \frac{1}{\rho }\left( {{P_1} - {P_2}} \right)\\ \frac{{d{W_P}}}{{dm}} = \frac{{3000 - 70}}{{1000}} = 2.93\;kJ/kg \END{array}\)

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Assume the above turbine to be part of a simple Rankine cycle. The density of water at the inlet to the pump is 1000 kg/m3. Ignoring kinetic and potential energy effects, the specific work (in kJ/kg) supplied to the pump is

Fluid Mechanics First Law Thermodynamics in Fluid Mechanics . 7 months ago

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S.F.E.E for pump \(\Rightarrow {h_4} = {h_3} + \frac{{d{W_P}}}{{DM}}\)\(\Rightarrow \frac{{d{W_P}}}{{dm}} = {h_4} - {h_3} = dh\) __________(i)and REVERSIBLE adiabatic compression as = dh – vdp = 0⇒ dh = cdpnow from EQUATION (1)\(\begin{array}{l} \frac{{d{W_P}}}{{dm}} = dh = vdp = \frac{1}{\rho }\left( {{P_1} - {P_2}} \right)\\ \frac{{d{W_P}}}{{dm}} = \frac{{3000 - 70}}{{1000}} = 2.93\;kJ/kg \END{array}\)

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Posted on 22 Nov 2024, this text provides information on Fluid Mechanics related to First Law Thermodynamics in Fluid Mechanics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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