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A condenser is to be designed to condense 225.0 kg of steam per hour at a pressure of 0.15 bar. A square array of 400 tubes, each of 6 mm in diameter, is available for the task. If the tube surface temperature is to be maintained at 26 degree Celsius, make calculations for the length of the tube
- a)4.353 m
- b)3.353 m
- c)2.353 m
- d)1.353 m
Correct answer is option 'D'. Can you explain this answer?
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A condenser is to be designed to condense 225.0 kg of steam per hour a...
t sat = 54 degree Celsius and h f g = 2373000 J/kg, t f = 40 degree Celsius and h v = 0.725 [k 3 g p 2 h f g/l δ (t sat – t s)] 0.25 = 5202.34 W/m2 K. Total S.A of 400 tubes = 7.536l and heat flow rate = m h f g.
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A condenser is to be designed to condense 225.0 kg of steam per hour a...
Given data:
Mass flow rate of steam, m = 225 kg/hr
Pressure of steam, P = 0.15 bar
Number of tubes, n = 400
Diameter of each tube, d = 6 mm
Tube surface temperature, Ts = 26°C
We need to find out the length of tubes required for the given condenser.
Calculation:
The heat transfer rate in the condenser can be calculated by using the following formula:
Q = m × hfg
where,
hfg = latent heat of vaporization of steam at given pressure
From steam tables, hfg = 2075.8 kJ/kg
Q = 225 × 2075.8 = 467557.5 kJ/hr
The heat transfer coefficient (h) can be calculated by using the Dittus-Boelter equation:
Nu = 0.023 × Re^0.8 × Pr^0.4
where,
Nu = Nusselt number
Re = Reynolds number
Pr = Prandtl number
Re = (4 × m) / (π × d × μ)
where,
μ = viscosity of the fluid at condensing temperature
From steam tables, at P = 0.15 bar, the saturation temperature of steam is 51.76°C. Assuming the tube surface temperature to be constant at 26°C, the average temperature difference between steam and tube surface can be calculated as:
ΔT = (51.76 - 26) / ln(51.76 / 26) = 17.95°C
The properties of steam at saturation temperature can be calculated from steam tables as:
Density of steam, ρg = 11.17 kg/m³
Dynamic viscosity of steam, μg = 0.017 Ns/m²
Re = (4 × 225) / (π × 0.006 × 0.017) = 2.5 × 10^6
Pr = μg × Cp,g / kg
where,
Cp,g = specific heat capacity of steam at constant pressure
kg = thermal conductivity of steam
From steam tables, Cp,g = 1.86 kJ/kg K and kg = 0.036 W/m K
Pr = (0.017 × 1.86) / 0.036 = 0.88
Nu = 0.023 × (2.5 × 10^6)^0.8 × 0.88^0.4 = 319.7
The heat transfer coefficient can be calculated as:
h = Nu × kg / d = 319.7 × 0.036 / 0.006 = 1918.2 W/m² K
The required heat transfer area can be calculated as:
A = Q / (h × ΔT) = 467557.5 / (1918.2 × 17.95) = 14.45 m²
The total length of tubes required can be calculated as:
L = A / (n × π × d² / 4) = 14.45 / (400 × π × 0.006² / 4) = 1.353 m
Therefore, the length of tubes required for the given condenser is 1.353 m.
Mass flow rate of steam, m = 225 kg/hr
Pressure of steam, P = 0.15 bar
Number of tubes, n = 400
Diameter of each tube, d = 6 mm
Tube surface temperature, Ts = 26°C
We need to find out the length of tubes required for the given condenser.
Calculation:
The heat transfer rate in the condenser can be calculated by using the following formula:
Q = m × hfg
where,
hfg = latent heat of vaporization of steam at given pressure
From steam tables, hfg = 2075.8 kJ/kg
Q = 225 × 2075.8 = 467557.5 kJ/hr
The heat transfer coefficient (h) can be calculated by using the Dittus-Boelter equation:
Nu = 0.023 × Re^0.8 × Pr^0.4
where,
Nu = Nusselt number
Re = Reynolds number
Pr = Prandtl number
Re = (4 × m) / (π × d × μ)
where,
μ = viscosity of the fluid at condensing temperature
From steam tables, at P = 0.15 bar, the saturation temperature of steam is 51.76°C. Assuming the tube surface temperature to be constant at 26°C, the average temperature difference between steam and tube surface can be calculated as:
ΔT = (51.76 - 26) / ln(51.76 / 26) = 17.95°C
The properties of steam at saturation temperature can be calculated from steam tables as:
Density of steam, ρg = 11.17 kg/m³
Dynamic viscosity of steam, μg = 0.017 Ns/m²
Re = (4 × 225) / (π × 0.006 × 0.017) = 2.5 × 10^6
Pr = μg × Cp,g / kg
where,
Cp,g = specific heat capacity of steam at constant pressure
kg = thermal conductivity of steam
From steam tables, Cp,g = 1.86 kJ/kg K and kg = 0.036 W/m K
Pr = (0.017 × 1.86) / 0.036 = 0.88
Nu = 0.023 × (2.5 × 10^6)^0.8 × 0.88^0.4 = 319.7
The heat transfer coefficient can be calculated as:
h = Nu × kg / d = 319.7 × 0.036 / 0.006 = 1918.2 W/m² K
The required heat transfer area can be calculated as:
A = Q / (h × ΔT) = 467557.5 / (1918.2 × 17.95) = 14.45 m²
The total length of tubes required can be calculated as:
L = A / (n × π × d² / 4) = 14.45 / (400 × π × 0.006² / 4) = 1.353 m
Therefore, the length of tubes required for the given condenser is 1.353 m.
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A condenser is to be designed to condense 225.0 kg of steam per hour at a pressure of 0.15 bar. A square array of 400 tubes, each of 6 mm in diameter, is available for the task. If the tube surface temperature is to be maintained at 26 degree Celsius, make calculations for the length of the tubea)4.353 mb)3.353 mc)2.353 md)1.353 mCorrect answer is option 'D'. Can you explain this answer?
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