Steam produced in a boiler is frequently \"wet\" -- it consists of a mist compos

Question

Steam produced in a boiler is frequently "wet" -- it consists of a mist composed of saturated vapor and entrained liquid droplets. The quality of the wet steam is defined as the fraction of the mixture by mass that is vapor.

A wet steam at a pressure of 14.0 bar with a quality of 0.78 is isothermally "dried" by evaporating the entrained liquid. The flow rate of the dried steam is 85.1 m^3 /h.

a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry streams, and the total mass flow rate of the process stream.

b) Calculate the heat input (kW) required for the evaporation process

Explanation / Answer

a)

From steam tables, looking at P = 14 bar, we get

T = 195.07 degrees celcius

Initial conditions(Wet steam):

P = 14 bar

T = 195.07 degrees celcius

Quality , x = 0.78

h(i) = hs + x(hg) = 2358.87 kJ/Kg

Final conditions: (Dry steam)

P = 14 bar

T = 195.07 degrees celcius

Quality = 1

h(f) = 2790 kJ/Kg

Specifric volume = 0.14084 m^3/Kg

Given Volume flow rate of dried steam = 85.1 m3/h

Mass flow rate = (Volume flowrate /Specific volume) = 85.1/0.14084 = 604.23 Kg/h = 0.168 Kg/s

b)

Applying first law of the thermodynamics,

Q + mh(i) = W + mh(f)

Here work done, W = 0

So,

Q = m[h(f)-h(i)] =0.168(2790-2358.87) =72.43 kJ/s

Heat input required, Q = 72.43 kW

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