Saturated liquid water at 20 bar is fed to a heater at 1 kg/s. It is heated isob
ID: 482213 • Letter: S
Question
Saturated liquid water at 20 bar is fed to a heater at 1 kg/s. It is heated isobarically to 300 degree C. a) Starting from the general energy balance equation (Eq. 3.34), cross out the terms that are negligible under this condition. d/dt {M (U + upsilon^2/2 + gh)} = sigma_j = 16j = J {m_j, in (H_j + upsilon_j^2/2 + gh_j)} - sigma + k = 1^k = K {m-k, out (H_k + upsilon_k62/2 + gh_k)} + W_s + W_EC + Q b) How much power (kW) is needed? A stream of superheated steam at 500 degree C and 200 bar enters an adiabatic turbine and a saturated vapor exits the turbine at 1 bar. a) Express the energy balance equation appropriate for this process. b) If the shaft work is -1000 kW, what is the mass flow rate (kg/s)?Explanation / Answer
Hence m*(Hout-Hin)= Q
At 20 bar, inltet enthalpy = 910 Kj/Kg ( from steam table)
Exit enthalpy ( 20 bar correspond to a saturation temperature of 213 deg.c, since during heating pressure remained constant and heated to 300 deg.c, it is super heated at the exit)
Hout= 3138.6 Kj/Kg ( from super heated steam table)
Q= 1kg/s*(3138.6-910) KJ/Kg=2229 Kj/sec
d/dt [M*(U+v2/2+gh)] = m*( H+V2/2+gh)in –m*(H+V2/2+gh)out +WS+WEc+Q
Work is produced during the process. During the expansion, there is change in enthalpy. The simplified energy balance ( since process is adiabatic, Q=0, there is no change in potential energy or kinetic energy, there is no accumulation( steady state process)
Hence m*(Hin-Hout)+WS= 0
2. Hin ( from super heated steam table)=3241.17 Kj/kg
Hout = 2675.3 Kj/kg
Hence m*(3214.17-2675.3)-1000 =0
Hence m*= 1.85 kg/s
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