Answer to Question #235642 in Molecular Physics | Thermodynamics for Rehema

Question

10 kg of fluid per minute goes through a reversible steady flow process. The

properties of fluid at the inlet are : p1 = 1.5 bar, ρ1 = 26 kg/m3, C1 = 110 m/s and u1 = 910 kJ/kg and

at the exit are p2 = 5.5 bar, ρ2 = 5.5 kg/m3, C2 = 190 m/s and u2 = 710 kJ/kg. During the passage,

the fluid rejects 55 kJ/s and rises through 55 metres. Determine :

(i) The change in enthalpy (∆ h) ;

(ii) Work done during the process (W).

Answer 1

Flow of fluid = 10 kg/min

Properties of fluid at the inlet :

Pressure $p_1= 1.5 \;bar = 1.5 \times 10^5 \;N/m^2$

Density $ρ_1 = 26 \;kg/m^3$

Velosity $C_1= 110 \;m/s$

Internal energy $u_1=910 \;kJ/kg$

Properties of the fluid at the exit :

Pressure $p_2= 5.5 \;bar = 5.5 \times 10^5 \;N/m^2$

Density $ρ_2 = 5.5 \;kg/m^3$

Velocity $C_2= 190 \;m/s$

Internal energy $u_2=710 \;kJ/kg$

Heat rejected by the fluid, $Q= 55 \;kJ/s$

Rise is elevation of fluid = 55 m

(i) The change in enthalpy,

$Δh=Δu +Δ(pv) \;\;\;\;(i)\\ Δ(pv) = \frac{p_2v_2 -p_1v_1}{1} \\ = \frac{p_2}{ρ_2} -\frac{p_1}{ρ_1} \\ = 1 \times 10^5 -0.0577 \times 10^5 \\ = 10^5 \times 0.9423 \;Nm \\ = 94.23 \;kJ \\ Δu =u_2 -u_1 = 710 -910 = -200 \;kJ/kg$

Substituting the value in eqn. (i), we get

$Δh = -200 + 94.23 = -105.77 \;kJ/kg$

(ii) The steady flow equation for unit mass flow can be written as

$Q=ΔKE + ΔPE +Δh +W$

where Q is the heat transfer per kg of fluid

$Q= 55 \;kJ/s = \frac{55 \;kJ/s}{\frac{10}{60} \;kg/s} = 55 \times 6 = 330 \;kJ/kg \\ ΔKE = \frac{C^2_2 -C^2_1}{2}= \frac{190^2 -110^2}{2} \;Nm = 12000 \;J = 12 \;kJ \\ ΔPE=(Z_2 -Z_1) \;g = (55- 0) \times 9.81 \;Nm = 539.5 \;J = 0.54 \;kJ$

Substituting the value in steady flow equation,

$-330 = 12 +0.54 -105.77 +W \\ W= -236.77 \;kJ/kg$

Work done per second $= -236.77 \times \frac{10}{60} = -39.46 \;kJ/s = -39.46 \;kW$