Question

A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125 lbm/s at 14.7 psia. Make-up water to replenish steam losses from the plant enters at inlet 2 at a rate of 10 lbm/s at 14.7 psia and 60o F. Water exits the tank at 14.7 psia. Neglect kinetic and potential energy effects, determine for the water exiting the the tank (a) the mass flow rate, in lb/s. (b) the specific enthalpy, in Btu/lb (c) the temperature, in F

Answer 1

A

Explanation:I might not might be correct hmu if it is wrong im so sorry

Answer 2

a)
`\dot m_(3) = 135\,(lbm)/(s)`, b)
`h_(3)=168.965\,(BTU)/(lbm)`, c)
`T = 200.829\,^(\textdegree)F`

Step-by-step explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

`\dot m_(1) + \dot m_(2) - \dot m_(3) = 0`

The mass flow rate exiting the tank is:

`\dot m_(3) = \dot m_(1) + \dot m_(2)`

`\dot m_(3) = 125\,(lbm)/(s) + 10\,(lbm)/(s)`

`\dot m_(3) = 135\,(lbm)/(s)`

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

`\dot m_(1)\cdot h_(1) + \dot m_(2) \cdot h_(2) - \dot m_(3)\cdot h_(3) = 0`

`h_(3) = (\dot m_(1)\cdot h_(1)+\dot m_(2)\cdot h_(2))/(\dot m_(3))`

Properties of water are obtained from tables:

`h_(1)=180.16\,(BTU)/(lbm)`

`h_(2)=28.08\,(BTU)/(lbm) + \left(0.01604\,(ft^(3))/(lbm)\right)\cdot (14.7\,psia-0.25638\,psia)`

`h_(2)=29.032\,(BTU)/(lbm)`

The specific enthalpy at outlet is:

`h_(3)=((125\,(lbm)/(s) )\cdot (180.16\,(BTU)/(lbm) )+(10\,(lbm)/(s) )\cdot (29.032\,(BTU)/(lbm) ))/(135\,(lbm)/(s) )`

`h_(3)=168.965\,(BTU)/(lbm)`

c) After a quick interpolation from data availables on water tables, the final temperature is:

`T = 200.829\,^(\textdegree)F`