Explanation:
a) Term of entering steam will be Tsaturated of corresponding pressure
b)
A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 mg/L as CaCO3 of Mg2 . All of the
hardness is carbonate hardness. Using the stoichiometry of the lime soda ash softening equations, what is the daily sludge production (in dry weight, kg/day) if the plant treats water at a rate of 2.935 m^3/s. Assume that the effluent water contains no carbon dioxide, 30.0 mg. L^-1 as CaCO3 of Ca2+ and 10.0 mg.L^1 as CaCO3 of Mg2+. Be sure to calculate the mass of CaC03 and Mg(OH)2 sludge produced each day.
2 answers:
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Answer:
a. T₂ = 837.2 K
b. ηt = 91.4%
Explanation:
Assuming k = 1.4
KNOWN: Air expands adiabatically through a turbine operating at steady state. Operating data are known.
FIND: Determine the exit temperature and the isentropic turbine efficiency.
ENGINEERING MODEL (See the pics):
(1) The control volume is at state.
(2) For the control volume, Qcv = 0 and kinetic and potential energy effects can be neglected.
(3) The air is modeled as an ideal gas with constant specific heats: k = 1.4.
ANALYSIS:
(a) Mass and energy rate balances reduce to give:
0 = - Wcv + m*(h₁ – h₂).
With h₁ – h₂ = cp*(T₁ – T₂)
then
cp = k*R/(k-1) = 1.4*(8.314/28.97)/(1.4 – 1) = 1.004 kJ/kg∙K and
T₂ = T₁ - (Wcv/(m*cp))
⇒ T₂ = 1040 K – (1120 kW)/[(5.5 kg/s)(1.004 kJ/kg∙K)(1 kJ/s/ 1 kW)
⇒ T₂ = 837.2 K
b) The isentropic efficiency is
ηt = (h₁ – h₂) / (h₁ – h₂s) = cp*(T₁ – T₂) / (cp*(T₁ – T₂s)).
To get T₂s we note that for an isentropic process of an ideal gas with constant specific heats
(T₂s / T₁) = (p₂ / p₁)∧((k-1)/k) ⇒ T₂s = T₁*(p₂ / p₁)∧((k-1)/k)
⇒ T₂s = (1040 K)*(120 / 278)∧((1.4-1)/1.4) = 818.1 K
Thus, the isentropic efficiency is
ηt = (1040 – 837.2)/(1040 – 818.1) = 0.914
ηt = 91.4%
