Question

The closed feedwater heater of a regenerative Rankine cycle is to heat 7000 kPa feedwater from 2608C to a saturated liquid. The turbine supplies bleed steam at 6000 kPa and 3258C to this unit. This steam is condensed to a saturated liquid before entering the pump. Calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.

Answer 1

the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s

Step-by-step 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:

`h_1 = h_f = 1134.8 \ kJ/kg`

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;

`(325-300)/(350-300)=\frac{h_(325^0)-{h_(300^0)}}{{h_(350^0)}- {h_(300^0)}}`

`(325-300)/(350-300)=(h_(325^0)-2885.6)/(3043.9-2885.6 )}`

`(25)/(50)=(h_(325^0)-2885.6)/(3043.9-2885.6 )}`

`h_(325^0) = 2885.6 + (25)/(50)({3043.9-2885.6 )`

`h_(325^0) = 2885.6 + 0.5({3043.9-2885.6 )`

`h_(325^0) =2964.75 \ kJ/kg`

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.

`h_6 = h_f = 1267.5 \ kJ/kg`

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.

`v_4 = v_f = 0.001319\ m^3 /kg`

However; the specific work pump can be determined by using the formula;

`W_p = v_4 (P_5-P_4)`

where;

`P_4` = pressure at state 4

`P_5` = pressure at state 5

`W_p = 0.001319 (7000-6000)`

`W_p = 0.001319 (1000)`

`W_p =1.319 \ kJ/kg`

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:

`E_(in) = E_(out) \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6`

where;

`m_1 = 1 \ kg`

Replacing our other value as derived above into the energy balance equation ; we have:

`1 * 1134.8 +m_3 * 2964.75 + m_3 * 1.319 = (1+m_3)* 1267.5`

`1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3`

Collect like terms

`2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8`

`1698.569 \ m_3 =132.7`

`\ m_3 = (132.7)/(1698.569)`

`\mathbf{ m_3 = 0.078 \ kg/s}`

Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s