Question

Pressurized water ( 10 bar, 110°C) enters the bottom of an 10-m-long vertical tube of diameter 63 mm at a mass flow rate of 1.5 kg/s. The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at 7 bar, 600°C. Determine the change in the rate at which the following quantities enter and exit the tube: (1) the combined thermal and flow work, (2) the mechanical energy, and (3) the total energy of the water. Also, (4) determine the heat transfer rate, . Hint: Relevant properties may be obtained from a thermodynamics text.

Answer 1

(1)
`\Delta E = 4845.43 kW`

(2)
`\Delta E_(m) = 5.7319 kW`

(3)
`\Delta E_(t) = 4839.69 kW`

(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`

At inlet:

Saturated temperature of water,
`T_(sw) = 179.88^(\circ)C`

Specific volume of water,
`V_(wi) = V_(f) = 0.00127 m^(3)/kg`

Using super heated water table corresponding to a temperature of
`600^(\circ)C` and at 7 bar:

At outlet:

Specific volume of water,
`V_(wso) = 0.5738 m^(3)/kg`

Specific enthalpy of water,
`h_(wo) = 3700.2 kJ/kg`

Now, at inlet, water's specific enthalpy is given by:

`h_(i) = C_(p)(T - T_(sw)) + h_(ws)`

`h_(i) = 4.187(110^(\circ) - 179.88^(\circ)) + 762.5`

`h_(i) = -292.587 + 762.5= 469.912 kJ/kg`

(1) Now, the change in combined thermal energy and work flow is given by:

`\Delta E = E_(o) - E_(i)`

`\Delta E = m(h_(wo) - h_(i))`

`\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW`

(2) The mechanical energy can be calculated as:

velocity at inlet,
`v_(i) = \rho A V_(wi)`

`v_(i) = \frac{mV_(wi)}{frac{\pi d^(2)}{4}}`

`v_(i) = \frac{mV_(wi)}{frac{\pi d^(2)}{4}}`

`v_(i) = \frac{1.5* 0.00127}{frac{\pi (63* 10^(- 3))^(2)}{4}}`

`v_(i) = 0.542 m/s`

Similarly,, the velocity at the outlet,

`v_(o) =  \frac{1.5* 0.57378}{frac{\pi (63* 10^(- 3))^(2)}{4}}`

`v_(o) =  276.099 m/s`

Now, change in mechanical energy:

`\Delta E_(m) = E_(mo) - E_(mi)`

`\Delta E_(m) = m[((v_(o)^(2))/(2) + gz_(o)) - ((v_(i)^(2))/(2) + gz_(i))]`

`\Delta E_(m) = 1.5[((276.099^(2))/(2) + 9.8(z_(o) - z_(i)) - ((0.542^(2))/(2)]`

`\Delta E_(m) = 57319 J = 5.7319 kW`

(3) The total energy of water is given by:

`\Delta E_(t) = E - E_(m) = 4845.43 - 5.7319 = 4839.69 kW`

(4) The rate of heat transfer:

q =
`\Delta E_(t) = 4839.69 kW`