Answer:
the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
Explanation:
Given that:
Pressure of the feed water = 7000 kPa
Temperature of the closed feedwater heater = 260 ° C
Pressure of of the turbine = 6000 kPa
Temperature of the turbine = 325 ° C
The objective is to calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
From the table A-4 of saturated water temperature table at temperature 260° C at state 1 ;
Enthalpies:
[tex]h_1 = h_f = 1134.8 \ kJ/kg[/tex]
From table A-6 superheated water at state 3 ; the value of the enthalpy relating to the pressure of the turbine at 6000 kPa and temperature of 325° C is obtained by the interpolating the temperature between 300 ° C and 350 ° C
At 300° C; enthalpy = 2885.6 kJ/kg
At 325° C. enthalpy = 3043.9 kJ/kg
Thus;
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-{h_{300^0}}}{{h_{350^0}}- {h_{300^0}}}[/tex]
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]\dfrac{25}{50}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]h_{325^0} = 2885.6 + \dfrac{25}{50}({3043.9-2885.6 )[/tex]
[tex]h_{325^0} = 2885.6 + 0.5({3043.9-2885.6 )[/tex]
[tex]h_{325^0} =2964.75 \ kJ/kg[/tex]
At pressure of 7000 kPa at state 6; we obtain the enthalpies corresponding to the pressure at table A-5 of the saturated water pressure tables.
[tex]h_6 = h_f = 1267.5 \ kJ/kg[/tex]
From state 4 ;we obtain the specific volume corresponding to the pressure of 6000 kPa at table A-5 of the saturated water pressure tables.
[tex]v_4 = v_f = 0.001319\ m^3 /kg[/tex]
However; the specific work pump can be determined by using the formula;
[tex]W_p = v_4 (P_5-P_4)[/tex]
where;
[tex]P_4[/tex] = pressure at state 4
[tex]P_5[/tex] = pressure at state 5
[tex]W_p = 0.001319 (7000-6000)[/tex]
[tex]W_p = 0.001319 (1000)[/tex]
[tex]W_p =1.319 \ kJ/kg[/tex]
Using the energy balance equation of the closed feedwater heater to calculate the amount of bleed steam required to heat 1 kg of feed water ; we have:
[tex]E_{in} = E_{out} \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6[/tex]
where;
[tex]m_1 = 1 \ kg[/tex]
Replacing our other value as derived above into the energy balance equation ; we have:
[tex]1 \times 1134.8 +m_3 \times 2964.75 + m_3 \times 1.319 = (1+m_3)\times 1267.5[/tex]
[tex]1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3[/tex]
Collect like terms
[tex]2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8[/tex]
[tex]1698.569 \ m_3 =132.7[/tex]
[tex]\ m_3 = \dfrac{132.7}{1698.569}[/tex]
[tex]\mathbf{ m_3 = 0.078 \ kg/s}[/tex]
Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
