Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at 20 bar and 400o C. The steam is then reheated at constant pressure to 500o C before entering the second-stage turbine. Steam leaves the second stage as saturated vapor at 0.6 bar. For operation at steady state, and neglecting stray heat transfer and kinetic and potential energy effects, determine (a) The mass flow rate of the steam (in kg/h). (b) The total power produced by the two stages of the turbine (in kW). (c) The rate of heat transfer to the steam flowing through the reheater (in kW)

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

answer:

ya

explanation:

sure ill do it