Piessunzed water (pin-10 bar, Tin= 1 10°C) enters the hottom of an L 12-m-long vertical tube of diameter D110 mm at a mass flow rate of 1.5kg/s. The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at o 7 bar, T 600°C. Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and Row work,
(b) the mechanical energy, and
(c) the total energy of the water. Also, (d) determine the heat trans- fer rate. q. Hint: Relevant properties may be obtained from a thermodynamics text.
Consider the tube and inlet conditions above Heat transfer at a rate of 3.89 MW is delivered to the tube. For an exit pressure of p 8 bar, determine
(a) the temperature of the water at the outlet as well as the change in
(b) combined thermal and flow work,
(c) mechanical energy, and (d) total energy of the water from the inlet to the outlet of the tube. Hint: As a first estimate, neglect the change in mechanical energy in solving part (a). Relevant properties may be obtained from a thermodynamics text

Now, specific enthalpy of the water at the inlet [tex]=[/tex] (specific enthalpy of the water at the saturation temperature) [tex]-[/tex] (Latent heat capacity).

[tex]\Rightarrow h_i=h_{sw}-h_l[/tex]

[tex]\Rightarrow h_i=762.50-292.587=469.912kJ/kg[/tex]

The specific volume of the water at intel is the same as the specific volume at the saturation temperature as volume remains unchanged on the addition of latent heat.

So, [tex]v_i=v_{ws}=0.00127 m^3/kg[/tex].

The outlet temperature, [tex]T_o=600^{\circ} C[/tex]and pressure, [tex]P_o=7[/tex] bar. From the superheated water table, we have

The specific volume of water, [tex]v_o=0.5738 m^3/kg[/tex]

The specific enthalpy of water, [tex]h_{wo}=3700.9 kJ/kg[/tex]

The given mass flow rate,[tex]\dot{m} =1.5 kg/s[/tex].

The inlet radius and outlet diameter are the same, i.e

[tex]d_i=d_o=110 mm=0.11m[/tex].

So, Inlet and outlet areas, [tex]A_i=A_f=9.5033\times 10^{-3} m^2[/tex].

Let the inlet and outlet velocities be [tex]V_i[/tex] and [tex]V_o[/tex] respectively.

For the given specific volume, [tex]v[/tex], and mass flow rate, [tex]\dot{m}[/tex], the velocity, [tex]V[/tex], at any cross-section having an area [tex]A[/tex] is

[tex]V=\frac{v\dot{m}}{A}[/tex].

So, the inlet velocity,

[tex]V_i=\frac{v_i \dot{m}}{A_i}[/tex]

[tex]\Rightarrow V_i=\frac{0.00127\times 1.5}{9.5033\times 10^{-3}}[/tex]

[tex]\Rightarrow V_i=0.20 m/s[/tex].

Similarly, the outlet velocity,

[tex]V_o=\frac{v_o \dot{m}}{A_o}[/tex]

[tex]\Rightarrow V_o=\frac{0.5738\times 1.5}{9.5033\times 10^{-3}}[/tex]

[tex]\Rightarrow V_0=90.57 m/s[/tex].

(1) The change in combined thermal energy and work flow [tex]=[/tex] Change in the thermal energy [tex]+[/tex] Change in the flow work

[tex]\Delta E_{tw}= \dot{m}(u_f-u_i)+\dot{m} (P_fv_f-P_iv_i)[/tex]

[tex]\Rightarrow \Delta E_{tw}=\dot{m}[(u_f+P_fv_f) - (u_i+P_iv_i)[/tex]

[tex]\Rightarrow \Delta E_{tw}=\dot{m} (h_f-h_i)[/tex]

[tex]\Rightarrow \Delta E_{tw}=1.5(3700,20-469.912)=4845.43 kW[/tex]

(2)The change in mechanical energy

[tex]\Delta E_m=[/tex] Change in kinetic energy + change in potential energy

[tex]\Rightarrow \Delta E_m=\left(\frac 1 2 \dot{m} V_f^2-\frac 1 2 \dot{m} V_i^2\right)+(\dot{m} g z_f-\dot{m} g z_i)[/tex]

[tex]\Rightarrow \Delta E_m=\frac 1 2 \dot{m}(V_f^2-V_i^2)+\dot{m} g (z_f-z_i)[/tex]

[tex]\Rightarrow \Delta E_m=\frac 1 2\times 1.5((90.57)^2-(0.2)^2)+1.5 \times 9.81 \times12[/tex]

[tex]\Rightarrow \Delta E_m=6328.74 J/s=6.329 kW[/tex]

(3) The change in the total energy of water,

[tex]\Delta E_t=[/tex]chnge in the thernal energy + change in the flow work + change in the mechanical energy

[tex]\Rightarrow \Delta E_t=4845.43+6.329= 4851.759 kW[/tex] [from part (1) and (2)]

(4) Now, as there is no work done by the water, so, the heat input only caused the change in the total energy.

Hence, the rate of heat transfer, [tex]q= 4851.759 kW[/tex] [from part (3).