(a) A shell and tube type condenser is employed in a steam power plant to handle 35000 kg/h of dry and saturated steam at 50 °C. The cooling water enters the condenser at 15 °C and leaves at 25 °C. The tubes are of 22·5 mm inside diameter and 25 mm o

Question

(a) A shell and tube type condenser is employed in a steam power plant to handle 35000 kg/h of dry and saturated steam at 50 °C. The cooling water enters the condenser at 15 °C and leaves at 25 °C. The tubes are of 22·5 mm inside diameter and 25 mm outside diameter. The water flows through the tubes at an average velocity of 2 m/s. The heat transfer coefficient on steam side is 5001 W/m²-K.

Calculate the following :

(i) The mass flow rate of water (in kg/s)

(ii) The heat transfer surface area (based on U₀)

(iii) The number of tubes required for water flow

(iv) The number of tube passes in condenser if the length of each tube pass should not be more than 2·5 m

Assume that condensate coming out from the condenser is saturated water and resistance of tube wall is negligible. For waterside heat transfer coefficient, use the correlation Nu = 0·023 Re⁰·⁸ Pr⁰·⁴. Take latent heat of steam as 2374 kJ/kg. The properties of water at mean bulk temperature of 20 °C are as follows :

Pr = 6·98
ρ = 998·9 kg/m³
Cₚ = 4·180 kJ/kg-K
ν = 1·0006 × 10⁻⁶ m²/s
kƒ = 0·59859 W/m-K

Assume fully developed flow through tubes and a single shell is used. (20 marks)

(b) A stationary gas turbine plant operates on a Brayton cycle and delivers 20 MW to an electric generator. The maximum temperature is 1200 K and the minimum temperature is 290 K. The minimum pressure is 95 kPa and the maximum pressure is 380 kPa. If the isentropic efficiencies of the turbine and compressor are 0·85 and 0·8, respectively, find—

(i) the mass flow rate of air to the compressor;

(ii) the volume flow rate of air to the compressor;

(iii) the fraction of turbine work output needed to drive the compressor;

(iv) the cycle efficiency.

If a regenerator of 75% effectiveness is added to the plant, what would be the changes in the cycle efficiency and net work output?

Assume Cp and Cv for air as 1·005 kJ/kg-K and 0·718 kJ/kg-K, respectively. (20 marks)

(c)

The heat transfer rate due to free convection from a vertical surface, 1 m high and 0·6 m wide, to quiescent air that is 20 K colder than the surface is known. What is the ratio of the heat transfer rate for that situation to the rate corresponding to a vertical surface, 0·6 m high and 1 m wide, when the quiescent air is 20 K warmer than the surface? Neglect heat transfer by radiation and any influence of temperature on the relevant thermophysical properties of air.

The correlation between Nusselt number and Rayleigh number is given as NūL = 0·10 RaL^0·25. 10 marks

The correlation between Nusselt number and Rayleigh number is given as
NūL = 0·10 RaL^0·25. (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i) By steady-flow energy balance on the steam side:
Q = (35000/3600) kg/s × 2374 kJ/kg = 23080.56 kW.
Water flow rate:
mw = Q/(Cp ΔT) = 23080.56/(4.180 × 10) = 552.17 kg/s.

(a)(ii) Using the Dittus–Boelter correlation for the water side:
Re = V Di/ν = 2 × 0.0225/(1.0006 × 10⁻⁶) = 4.497 × 10⁴.
Nu = 0.023 Re⁰·⁸ Pr⁰·⁴ = 0.023 × (4.497 × 10⁴)⁰·⁸ × 6.98⁰·⁴ = 264.0.
hi = Nu kf/Di = 264.0 × 0.59859/0.0225 = 7024 W/m²K.

Using the LMTD method, overall coefficient based on outside area, wall resistance negligible:
1/U₀ = 1/5001 + (25/22.5)/7024 = 3.581 × 10⁻⁴.
U₀ = 2792 W/m²K.

LMTD = (35 − 25)/ln(35/25) = 29.72 K.
Area: A₀ = Q/(U₀ LMTD) = 23080556/(2792 × 29.72) = 278.1 m².

(a)(iii) Volume flow of water:
V̇ = mw/ρ = 552.17/998.9 = 0.5528 m³/s.
Area per tube:
At = π Di²/4 = π(0.0225)²/4 = 3.976 × 10⁻⁴ m².
Number of tubes:
N = V̇/(V At) = 0.5528/(2 × 3.976 × 10⁻⁴) = 695.1.
Use integer next higher: N = 696 tubes.

(a)(iv) Total tube length required:
Ltot = A₀/(π D₀) = 278.1/(π × 0.025) = 3541 m.
For 696 tubes per pass, length per pass with p passes is L = 3541/(696p).
For p = 2, L = 2.544 m > 2.5 m; hence p = 3 gives L = 1.696 m.
Number of tube passes = 3.

(b)(i) Using air-standard Brayton cycle analysis:
Pressure ratio: rp = 380/95 = 4.
R = Cp − Cv = 0.287 kJ/kg-K; exponent = R/Cp = 0.287/1.005 = 0.28557.
T2s = 290 × 4^0.28557 = 430.85 K.
ηc = (T2s − T1)/(T2 − T1), so T2 = 290 + (430.85 − 290)/0.8 = 466.07 K.
wc = Cp(T2 − T1) = 1.005(466.07 − 290) = 176.95 kJ/kg.

T4s = 1200/4^0.28557 = 807.70 K.
ηt = (T3 − T4)/(T3 − T4s), so T4 = 1200 − 0.85(1200 − 807.70) = 866.55 K.
wt = Cp(T3 − T4) = 1.005(1200 − 866.55) = 335.12 kJ/kg.
wnet = wt − wc = 158.18 kJ/kg.
Mass flow: m = 20000/158.18 = 126.44 kg/s.

(b)(ii) At compressor inlet, P1 = 95 kPa, T1 = 290 K:
V̇ = m R T1/P1 = 126.44 × 0.287 × 290/95 = 110.78 m³/s.

(b)(iii) Back-work ratio = wc/wt = 176.95/335.12 = 0.528 = 52.8%.

(b)(iv) Heat added: qin = Cp(T3 − T2) = 1.005(1200 − 466.07) = 737.60 kJ/kg.
η = wnet/qin = 158.18/737.60 = 0.2145 = 21.45%.

Regenerator effectiveness ε = 0.75:
T2r = T2 + ε(T4 − T2) = 466.07 + 0.75(866.55 − 466.07) = 766.43 K.
qin,r = 1.005(1200 − 766.43) = 435.74 kJ/kg.
ηr = 158.18/435.74 = 0.3630 = 36.30%.
Change in efficiency = +14.85 percentage points.
Net work per kg is unchanged, so net work output remains 20 MW for the same pressure ratio and mass flow; change in net work output = 0.

(c) Using the free-convection correlation NūL = 0.10 RaL^0.25:
Q = h A ΔT = (Nu k/L)A ΔT.
Nu = 0.10 Ra^0.25, with Ra ∝ L³ |ΔT|.
Thus Q ∝ L⁻¹/⁴ A |ΔT|^1.25.
Here A and |ΔT| = 20 K are the same for both cases.
Q₁/Q₂ = (L₁/L₂)⁻¹/⁴ = (1/0.6)⁻¹/⁴ = 0.6¹/⁴ = (3/5)¹/⁴.
Q₁/Q₂ = 0.880 approximately, taking heat-transfer rates as magnitudes. If signed surface-to-air heat transfer is used, the second rate has opposite sign.

(a) A shell and tube type condenser is employed in a steam power plant to handle 35000 kg/h of dry and saturated steam at 50 °C…

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