Your second customer wants to determine the dissipated electrical power by a personal computer, which has a fan that draws 5 L/s of air at 101.325 kPa and 21°C through the box containing the CPU and other components. After that air leaves at 101.325 kPa and 30 °C.
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In a steady flow process, the condition of a fluid flow is within a control volume such that it does not vary with time,
i.e. the mass flow rate, pressure, volume, work and rate of heat transfer are not the function of time.
i.e., for a steady flow
(dm/dt)entrance = (dm/dt)exit ; i.e, dm/dt = constant
dP/dt = dV/dt = dρ/dt = dEchemical = 0
Assumptions:
The following conditions must be there in a steady flow process.
(a) The mass flow rate through the system remains constant.
(b) The rate of heat transfer is constant.
(c) The rate of work transfer is constant.
(d) The state of working substance at any point within the system is same at all times.
(e) There is no change in the chemical composition of the system. If any one condition is not satisfied, the process is called unsteady process.
Let; A1, A2 = Cross sectional Area at inlet and outlet
ρ1,ρ2 = Density of fluid at inlet and outlet
m1,m2 = Mass flow rate at inlet and outlet
u1,u2 = I.E. of fluid at inlet and outlet
P1,P2 = Pressure of mass at inlet and outlet
ν1, ν2 = Specific volume of fluid at inlet and outlet
V1,V2 = Velocity of fluid at inlet and outlet
Z1, Z2 = Height at which the mass enter and leave Q = Heat transfer rate W = Work transfer rate Consider open system; we have to consider mass balanced as well as energy balance.
In the absence of any mass getting stored the system we can write; Mass flow rate at inlet = Mass flow rate at outlet i.e., mf1 = mf2
since mf = density X volume flow rate = density X Area X velocity
= ρ.A.V ρ1.A1.V1 = ρ2.A2.V2
or, A1.V1/v1 = A2.V2/v2; v1, v2
= specific volume Now total energy of a flow system consist of P.E, K.E., I.E., and flow work
Hence, E = PE + KE + IE + FW = h + V2/2 + gz
Now; Total Energy rate cross boundary as heat and work = Total energy rate leaving at (2) - Total energy rate
leaving at (1)![Q-W= m2[h, + V22/2 + gz,l- m, lh, + V,/2 + gz] For steady flow process m = m = m) Q-W = m/h, - h,) + ( V22 -V,2) + g(2, -2)]](https://media.cheggcdn.com/coop/7d2/7d249d27-9420-40d8-a098-3f88bf35e8c9/1608044629616_1.PNG)
Assumptions:
The following assumptionsare made in the system analysis:
(i) The mass flow through the system remains constant.
(ii) Fluid is uniform in composition.
(iii) The only interaction between the system and surroundings are work and heat.
(iv) The state of fluid at any point remains constant with time.
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