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I am reading thermodynamics from Cengel and Boles. I am currently studying flow work. The following image contains text of the book: enter image description here

The work done is calculated by considering the pressure of upstream fluid but the downstream fluid also applies pressure on it,but that is not considered why? According to me,the work done should be (P2-P1)*V where P2 is pressure of fluid upstream and P1 is pressure downstream at the boundary of control volume.

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  • $\begingroup$ Are you sure Cengel and Boles don't get around to taking the downstream pressure into account if you read on for another page or two? Either way, remember that a given parcel of fluid is not guaranteed to have the same volume at exit that it had at entrance, so it's $P_2V_2-P_1V_1$, not $\left(P_2-P_1\right)V$. $\endgroup$ May 4 at 16:07
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PA is force . F*L=PAL=PV is work. if you consider per Kg, then W= Pv When you write the energy balance equation at point 1 and 2, you write P1*v left and P2*v right along with other terms (internal, kinetic, potential). See, you get your desired term by reorganizing: 0=(P2-P1)*v + (ke2-ke1) + (pe2-pe1) + (u2-u1). Considering no heat flow. heat=work+change of internal/potential/kinetic energy. Well, that's the first law. Think from Thermodynamic System perspective, the surrounding (upstream) does work on the system and the system does work on the surrounding (downstream). You can get the net work from difference

It is interesting that unlike other work quantities, flow work is expressed in terms of properties. In fact, it is the product of two properties of the fluid.

Now if you imagine this as combination property , it becomes easy to write energy balance equation. or otherwise you separately add the net work. (P2-P1)*v

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In your final sentence you say the work should be (P2-P1)*A. The units of your statement are force, which does not carry the same units as work. If you multiplied your statement by a distance, you would have the correct units. It just so happens to be that the product of pressure and specific volume at a control surface carries the same units as specific work. Since pressure is a force per unit area and specific volume is volume per unit mass, the unit volume and unit area cancel out to leave a length term.

Flow work is thought of as the work required to move the fluid through the control volume due to its own mass. In a real-life situation, the difference in pressure are influenced by many other forces then just the force exerted by the mass of the fluid. I hope this helps you understand the physical significance of flow work. Shihabus' answer gives the mathematical reason why flow work is a function of only thermodynamic properties.

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