Heat Transfer Miscellaneous
- Cold water flowing at 0.1 kg/s is heated from 20°C to 70°C in a counter flow type heat exchange by a hot water stream flowing at 0.1 kg/s and entering at 90°C. The specific heat of water is 4200 J/(kgK) and density is 1000 kg/m3 . If the overall heat transfer coefficient U for the heat exchange is 2000 W/(m2 K), the required heat exchange area (in m2 ) is
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Ch = Cc = 0.1 × 4200 = 420 J/s
Cc (Tc1 – Tc2) = Ch (Th1 – Th2) (70 – 20) = 90° Th2
LMTD = θ1 = θ2 (∵ θ1 = θ2)
Cc (Tc2 – Tc1) = UA(LMTD)
or A = 0.525 m2.Correct Option: B
Ch = Cc = 0.1 × 4200 = 420 J/s
Cc (Tc1 – Tc2) = Ch (Th1 – Th2) (70 – 20) = 90° Th2
LMTD = θ1 = θ2 (∵ θ1 = θ2)
Cc (Tc2 – Tc1) = UA(LMTD)
or A = 0.525 m2.
- An industrial gas (c = 1 kJ/kgK) enters a parallel flow heat exchange at 250°C with a flow rate of 2 kg/s to heat a water stream. The water stream (c = 4 kJ/kgK) enters the heat exchange at 50°C with a flow rate of 1 kg/s. The heat exchange has an effectiveness of 0.75. The gas stream exit temperature will be
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ε = Ch(Th1 - Th2) Cmin(Thi - Tci) 0.75 = 2 ×(250 - Th2) 2 ×(250 - 50)
Th2 = 100° CCorrect Option: B
ε = Ch(Th1 - Th2) Cmin(Thi - Tci) 0.75 = 2 ×(250 - Th2) 2 ×(250 - 50)
Th2 = 100° C
- In a condenser, water enters at 30°C and flows at the rate 1500 kg/hr. The condensing steam is at a temperature of 120°C and cooling water leaves the condenser at 80°C. Specific heat of water is 4.187 kJ/kgK. If the overall heat transfer coefficient is 2000 W/m2 K, the heat transfer area is
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Here, θ1 = 120 – 30 = 90ºC
θ2 = 120 – 80 = 40ºC
Log mean temperature difference,
Q = mcpw ∆T1500 × 1 × 4.187 × (50) × 103 3600
= 87.22 × 103 Joule/s
Again
Q = UAθm
or 87.22 × 103 = 2000 × A × 61.65
∴ A = 0.707 m2Correct Option: A
Here, θ1 = 120 – 30 = 90ºC
θ2 = 120 – 80 = 40ºC
Log mean temperature difference,
Q = mcpw ∆T1500 × 1 × 4.187 × (50) × 103 3600
= 87.22 × 103 Joule/s
Again
Q = UAθm
or 87.22 × 103 = 2000 × A × 61.65
∴ A = 0.707 m2
- In a counter flow heat exchange, for the hot fluid the heat capacity = 2 kJ/kgK, mass flow rate = 5 kg/s, inlet temperature = 150°C, outlet temperature = 100°C. For the cold fluid, heat capacity = 4 kJ/kg K, mass flow rate = 10 kg/s, inlet temperature = 20°C. Neglecting heat transfer to the surroundings, the outlet temperature of the cold fluid in °C is
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Heat lost by fluid = Heat gained by cold fluid
5 × 2000 × 150 – 100) = 10 × 4000 (Tco – Tci)
∴ Tco = 32.5°cCorrect Option: B
Heat lost by fluid = Heat gained by cold fluid
5 × 2000 × 150 – 100) = 10 × 4000 (Tco – Tci)
∴ Tco = 32.5°c
- A hemispherical furnace of 1 m radius has the inner surface (emissivity, ε = 1) of its roof maintained at 800 K, while its floor (ε = 0.5) is kept at 600 K. Stefan-Boltzmann constant is 5.668 × 110-8W/m2K4. The net radiative heat transfer (in kW) from the roof to the floor is _______.
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Let the base be (1) and hemispherical furnace be (2)
∴ F11 + F12 = 1 ...(1)
F21 + F22 = 1 ...(2)
∵ F11 = 0
∴ F12 = 2
A1 F12 = A2 F21∴ F21 = A1 F12 = πR2 F12 = 0.5F12 A2 2πR2
∴ F21 = 0.5 × 2.0.5
∴ F22 = 0.5
So, F11 = 0, F12 = 1, F21 = 0.5 < F22 = 0.5
Now Radiative heat transfer,
⇒ Q = A1 F12ε2 × 6(8004 – 6004) watt
∴ Q = π × 12 × 1 × 0.5 × 5.668 × 10– 8(8004 – 6004) watt
or Q = 24.9 kW.Correct Option: C
Let the base be (1) and hemispherical furnace be (2)
∴ F11 + F12 = 1 ...(1)
F21 + F22 = 1 ...(2)
∵ F11 = 0
∴ F12 = 2
A1 F12 = A2 F21∴ F21 = A1 F12 = πR2 F12 = 0.5F12 A2 2πR2
∴ F21 = 0.5 × 2.0.5
∴ F22 = 0.5
So, F11 = 0, F12 = 1, F21 = 0.5 < F22 = 0.5
Now Radiative heat transfer,
⇒ Q = A1 F12ε2 × 6(8004 – 6004) watt
∴ Q = π × 12 × 1 × 0.5 × 5.668 × 10– 8(8004 – 6004) watt
or Q = 24.9 kW.