First Law of Thermodynamics - Energy Equation - Introduction to Fluid Mechanic | CE 321, Study notes of Fluid Mechanics

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Lecture 16 Energy Equation First Law of Thermodynamics What is energy? • The ability of an object (or the capacity of a physical system) to do work. • Energy can take a variety of forms • Units: Joules or ergs • Joule: the amount of energy needed to apply a force of 1 newton over a distance of 1 m. 2kg.m kg.m 1J 1N m 1 m 1 2 2 . . s s = = = The First Law of Thermodynamics • States that energy is never lost but is converted from one form to another sys net net DE Q W D    . • There are many forms of energy Vsys t E em e d   • For a system of fluid particles, the first law   2 / sys Ve E m u gz can be written as: • e: total energy 2   Internal Energy u • The internal energy of a system is the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energy of atoms within l lmo ecu es • It includes the energy in all the chemical bonds and the energy of the free conduction, , electrons in metals. The First Law of Thermodynamics Th ti t f i N t ti t f N t ti t f     e me ra e o ncrease e me ra e o energy e me ra e o energy of the total stored energy addition by heat transfer addition by work transfer of the system into the system in                 to the system            in out in out syssys sys DE Q Q W W Dt                      net in net insys sys sys DE Q W Dt E e d          sys Reynolds Transport Theorem for Energy DE e d e A V e A V      out out out out in in in in cv TRANSFER ACROSS BOUNDARIESSYSTEM CONTROL VOLUME Dt t    Time rate of increase the time rate of i of the total stored energ       ncrease of the total stored energ       y of the system      y of the contents of the control volume the net rate of flow       of the total stored energy out of the control volume        through the control surface  First Law for a Control Volume  out out out out in in in in cv e d e A V e A V t         net in net in cvQ W   Adiabatic Processes • Many engineering processes are adiabatic iQ n • The work transfer rate is called power Work Transfer … • Work transfer also occurs at a CS when force associated with fluid pressure acts over a distance (p A V): l t i l t i i i i iW F V p A V norma s ress n norma s ress n n n n n normal stressout normal stressout out out out outW F V p A V   normal stress in in in out out outW p A V p A V   Work Transfer … • Can also occur due to tangential stress forces. • But this is generally zero at inlets and outlets (as flow is normal to the surfaces) First Law for a CV i i i ie d e A V e A V       out out out out n n n n cv t i h ft t i i i i t t t t Q W p A V p A V       ne n s a ne n n n n ou ou ou Now let us expand the “e” terms … Enthalpy ph u   1D Energy Equation for Steady Flows in in in out out outA V A V m    2 2 ( )out inout in out in V Vm h h g z z           2 Q W     net in shaft net in  Steady Flow with Zero Shaft Power 2 2    ( ) 2 out in out in out in out in V Vp pu u g z z                   net in net in Q q m     Heat transfer per unit mass 2 2 ( ) 2 2 out out in in out in out in net in p V p Vgz gz u u q             loss 5.53 A water siphon having a constant inside diameter of 3 in. is arranged as shown in Fig. PS.53. If the friction loss be- tween A and B is 0.6V7/2, where V is the velocity of flow in the siphon, determine the flowrate involved. 7o determine the flowrate, Q, we use @<=<AV= tl > Vv (1) To obtain Vo we apply the energy eguation (EG. 5.56) befuter Ponts A and CM fhe sketch above, Thus, 5 fit ler 93, fiifein, 16 /oss Ct in or ; V z 49%, = Gz, - 06 ue 7aus = (z, ~ Zz. Ff V a fas *e) G eae) = (794 ft Ss and with Eg. / a Q- i 17.94 ft) = 059) #8 144 on. *) S fp Problem 5.54 5.54 Water flows through a valve (see Fig. P5.54) with a weight flowrate, mg, of 1000 lb/s. The pressure just upstream of the valve is 90 psi, and the pressure drop across the valve is 5 psi. The inside diameters of the valve inlet and exit pipes are 12 and 24 in. If the flow through the valve occurs in a horizontal plane, determine the loss in available energy across the valve. t cv este, contre! yolvne k— 12 in. * a FIGURE P5.54 Problem 5.55 5.55 Water flows steadily from one location to another in the inclined pipe shown in Fig. P5S.55. At one section, the static pressure is 8 psi. At the other section, the static pressure is 5 psi. Which way is the water flowing? Explain. P= &Spsi 10 100 ft a FIGURE P5.55 To determme. the durecton of wakr flow we apply the energy eguchon (4. $.56) for flow ton, Sechons (A) (8) and How from sechins(3) WA) the loss obtained wilh &. S56 13 positive for fhe Coneeh Flow direchon but negative for the incovvect flaw direction . For flow tom sections (A) (8), &]. 5.56 |€ads fp Joss = Fa~B, vity, By 9C%- 1+ ue F. z * mane toss = (Bpsi-Spsi) (44 tn. f# Mb 32.2 tt i) G22 2) ewnyie 7 # Hea) d OM oss = = sea a Stee For tlow fam sections (B) to (A), 2g. $56 lads r i] or loss = af - oe a) *ug 3 tig. Ti a4 a) 4? + 99.3 tt4 Shes Tne watey flow is ftom Section 6) to Sechim (A). ov and loss = Problem 5.57 5.57 What is the maximum possible power output of the hydroelectric turbine shown in Fig. P5.57?