Respuesta :
Answer:
(1) [tex]\Delta E = 4845.43 kW[/tex]
(2) [tex]\Delta E_{m} = 5.7319 kW[/tex]
(3) [tex]\Delta E_{t} = 4839.69 kW[/tex]
(4) q = 4839.69 kW[/tex]
Solution:
Using Saturated water-pressure table corresponding to pressure, P = 10 bar:
At saturated temperature, Specific enthalpy of water, [tex]h_{ws} = h_{f} = 762.5 kJ/kg[/tex]
At inlet:
Saturated temperature of water, [tex]T_{sw} = 179.88^{\circ}C[/tex]
Specific volume of water, [tex]V_{wi} = V_{f} = 0.00127 m^{3}/kg[/tex]
Using super heated water table corresponding to a temperature of [tex]600^{\circ}C[/tex] and at 7 bar:
At outlet:
Specific volume of water, [tex]V_{wso} = 0.5738 m^{3}/kg[/tex]
Specific enthalpy of water, [tex]h_{wo} = 3700.2 kJ/kg[/tex]
Now, at inlet, water's specific enthalpy is given by:
[tex]h_{i} = C_{p}(T - T_{sw}) + h_{ws}[/tex]
[tex]h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5[/tex]
[tex]h_{i} = -292.587 + 762.5= 469.912 kJ/kg[/tex]
(1) Now, the change in combined thermal energy and work flow is given by:
[tex]\Delta E = E_{o} - E_{i}[/tex]
[tex]\Delta E = m(h_{wo} - h_{i})[/tex]
[tex]\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW[/tex]
(2) The mechanical energy can be calculated as:
velocity at inlet, [tex]v_{i} = \rho A V_{wi}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{i} = 0.542 m/s[/tex]
Similarly,, the velocity at the outlet,
[tex]v_{o} = \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{o} = 276.099 m/s[/tex]
Now, change in mechanical energy:
[tex]\Delta E_{m} = E_{mo} - E_{mi}[/tex]
[tex]\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})][/tex]
[tex]\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}][/tex]
[tex]\Delta E_{m} = 57319 J = 5.7319 kW[/tex]
(3) The total energy of water is given by:
[tex]\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW[/tex]
(4) The rate of heat transfer:
q = [tex]\Delta E_{t} = 4839.69 kW[/tex]
