Answer:
(1) ![\Delta E = 4845.43 kW](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%204845.43%20kW)
(2) ![\Delta E_{m} = 5.7319 kW](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bm%7D%20%3D%205.7319%20kW)
(3) ![\Delta E_{t} = 4839.69 kW](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bt%7D%20%3D%204839.69%20kW)
(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, ![h_{ws} = h_{f} = 762.5 kJ/kg](https://tex.z-dn.net/?f=h_%7Bws%7D%20%3D%20h_%7Bf%7D%20%3D%20762.5%20kJ%2Fkg)
At inlet:
Saturated temperature of water, ![T_{sw} = 179.88^{\circ}C](https://tex.z-dn.net/?f=T_%7Bsw%7D%20%3D%20179.88%5E%7B%5Ccirc%7DC)
Specific volume of water,
Using super heated water table corresponding to a temperature of
and at 7 bar:
At outlet:
Specific volume of water, ![V_{wso} = 0.5738 m^{3}/kg](https://tex.z-dn.net/?f=V_%7Bwso%7D%20%3D%200.5738%20m%5E%7B3%7D%2Fkg)
Specific enthalpy of water, ![h_{wo} = 3700.2 kJ/kg](https://tex.z-dn.net/?f=h_%7Bwo%7D%20%3D%203700.2%20kJ%2Fkg)
Now, at inlet, water's specific enthalpy is given by:
![h_{i} = C_{p}(T - T_{sw}) + h_{ws}](https://tex.z-dn.net/?f=h_%7Bi%7D%20%3D%20C_%7Bp%7D%28T%20-%20T_%7Bsw%7D%29%20%2B%20h_%7Bws%7D)
![h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5](https://tex.z-dn.net/?f=h_%7Bi%7D%20%3D%204.187%28110%5E%7B%5Ccirc%7D%20-%20179.88%5E%7B%5Ccirc%7D%29%20%2B%20762.5)
![h_{i} = -292.587 + 762.5= 469.912 kJ/kg](https://tex.z-dn.net/?f=h_%7Bi%7D%20%3D%20-292.587%20%2B%20762.5%3D%20469.912%20kJ%2Fkg)
(1) Now, the change in combined thermal energy and work flow is given by:
![\Delta E = E_{o} - E_{i}](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20E_%7Bo%7D%20-%20E_%7Bi%7D)
![\Delta E = m(h_{wo} - h_{i})](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20m%28h_%7Bwo%7D%20-%20h_%7Bi%7D%29)
![\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%201.5%283700.2%20-%20469.912%29%20%3D%204845.43%20kW)
(2) The mechanical energy can be calculated as:
velocity at inlet, ![v_{i} = \rho A V_{wi}](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%20%5Crho%20A%20V_%7Bwi%7D)
![v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%20%5Cfrac%7BmV_%7Bwi%7D%7D%7Bfrac%7B%5Cpi%20d%5E%7B2%7D%7D%7B4%7D%7D)
![v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%20%5Cfrac%7BmV_%7Bwi%7D%7D%7Bfrac%7B%5Cpi%20d%5E%7B2%7D%7D%7B4%7D%7D)
![v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%20%5Cfrac%7B1.5%5Ctimes%200.00127%7D%7Bfrac%7B%5Cpi%20%2863%5Ctimes%2010%5E%7B-%203%7D%29%5E%7B2%7D%7D%7B4%7D%7D)
![v_{i} = 0.542 m/s](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%200.542%20m%2Fs)
Similarly,, the velocity at the outlet,
![v_{o} = \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%20%20%5Cfrac%7B1.5%5Ctimes%200.57378%7D%7Bfrac%7B%5Cpi%20%2863%5Ctimes%2010%5E%7B-%203%7D%29%5E%7B2%7D%7D%7B4%7D%7D)
![v_{o} = 276.099 m/s](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%20%20276.099%20m%2Fs)
Now, change in mechanical energy:
![\Delta E_{m} = E_{mo} - E_{mi}](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bm%7D%20%3D%20E_%7Bmo%7D%20-%20E_%7Bmi%7D)
![\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})]](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bm%7D%20%3D%20m%5B%28%5Cfrac%7Bv_%7Bo%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20gz_%7Bo%7D%29%20-%20%28%5Cfrac%7Bv_%7Bi%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20gz_%7Bi%7D%29%5D)
![\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}]](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bm%7D%20%3D%201.5%5B%28%5Cfrac%7B276.099%5E%7B2%7D%7D%7B2%7D%20%2B%209.8%28z_%7Bo%7D%20-%20z_%7Bi%7D%29%20-%20%28%5Cfrac%7B0.542%5E%7B2%7D%7D%7B2%7D%5D)
![\Delta E_{m} = 57319 J = 5.7319 kW](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bm%7D%20%3D%2057319%20J%20%3D%205.7319%20kW)
(3) The total energy of water is given by:
![\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bt%7D%20%3D%20E%20-%20E_%7Bm%7D%20%3D%204845.43%20-%205.7319%20%3D%204839.69%20kW)
(4) The rate of heat transfer:
q = ![\Delta E_{t} = 4839.69 kW](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bt%7D%20%3D%204839.69%20kW)