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An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat flux
- a)0.0915 * 10 6 W/m2
- b)0.1915 * 10 6 W/m2
- c)0.2915 * 10 6 W/m2
- d)0.3915 * 10 6 W/m2
Correct answer is option 'C'. Can you explain this answer?
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An electric wire of 1.25 mm diameter and 250 mm long is laid horizonta...
Q= V I = 286 W and A = 9.81 * 10 -4 m2. Therefore heat flux = Q/A.
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An electric wire of 1.25 mm diameter and 250 mm long is laid horizonta...
Calculation of Heat Flux of an Electric Wire Submerged in Water
Given parameters:
- Diameter of electric wire (d) = 1.25 mm = 0.00125 m
- Length of electric wire (L) = 250 mm = 0.25 m
- Voltage applied to wire (V) = 2.2 V
- Current passing through wire (I) = 130 A
- Water pressure (P) = 7 bar
- Surface temperature of wire (T) = 200°C = 473 K
1. Calculation of heat generated by the wire:
- Resistance of wire (R) = V/I = 2.2/130 = 0.0169 Ω
- Power generated by wire (P_gen) = I^2 * R = 130^2 * 0.0169 = 284.81 W
2. Calculation of heat transferred to water:
- Heat transferred to water (Q) = P_gen = 284.81 W
- Area of wire (A) = π * (d/2)^2 = π * (0.00125/2)^2 = 9.74 * 10^-7 m^2
- Heat flux (q) = Q/(A*L) = 284.81/(9.74 * 10^-7 * 0.25) = 0.1162 * 10^6 W/m^2
3. Calculation of convective heat transfer coefficient:
- Reynolds number (Re) = ρ*v*d/μ
- Water density (ρ) = 998 kg/m^3
- Water viscosity (μ) = 0.00089 Pa*s
- Velocity of water (v) = Q/(ρ*A)
- Re = 998 * (0.1162 * 10^6)/(0.00089 * 0.00125) = 1.65 * 10^8
- Nusselt number (Nu) = 0.023 * Re^(4/5) * Pr^(1/3)
- Prandtl number (Pr) = 4.38 (for water at 200°C)
- Nu = 0.023 * (1.65 * 10^8)^(4/5) * 4.38^(1/3) = 922.93
- Convective heat transfer coefficient (h) = Nu*k/d
- Thermal conductivity of water (k) = 0.679 W/(m*K) (for water at 200°C)
- h = 922.93 * 0.679/0.00125 = 499,903.04 W/(m^2*K)
4. Calculation of heat flux:
- Heat flux (q) = h*(T-T_inf)
- Water temperature (T_inf) = 200°C = 473 K
- q = 499,903.04 * (473-473) = 0 W/m^2 (no heat transferred due to temperature difference)
5. Calculation of total heat flux:
- Total heat flux (q_total) = q + q_conv
- q_conv = Q/(A*L) = 284.81/(9.74 * 10^-7 * 0.25) = 0.1162 *
Given parameters:
- Diameter of electric wire (d) = 1.25 mm = 0.00125 m
- Length of electric wire (L) = 250 mm = 0.25 m
- Voltage applied to wire (V) = 2.2 V
- Current passing through wire (I) = 130 A
- Water pressure (P) = 7 bar
- Surface temperature of wire (T) = 200°C = 473 K
1. Calculation of heat generated by the wire:
- Resistance of wire (R) = V/I = 2.2/130 = 0.0169 Ω
- Power generated by wire (P_gen) = I^2 * R = 130^2 * 0.0169 = 284.81 W
2. Calculation of heat transferred to water:
- Heat transferred to water (Q) = P_gen = 284.81 W
- Area of wire (A) = π * (d/2)^2 = π * (0.00125/2)^2 = 9.74 * 10^-7 m^2
- Heat flux (q) = Q/(A*L) = 284.81/(9.74 * 10^-7 * 0.25) = 0.1162 * 10^6 W/m^2
3. Calculation of convective heat transfer coefficient:
- Reynolds number (Re) = ρ*v*d/μ
- Water density (ρ) = 998 kg/m^3
- Water viscosity (μ) = 0.00089 Pa*s
- Velocity of water (v) = Q/(ρ*A)
- Re = 998 * (0.1162 * 10^6)/(0.00089 * 0.00125) = 1.65 * 10^8
- Nusselt number (Nu) = 0.023 * Re^(4/5) * Pr^(1/3)
- Prandtl number (Pr) = 4.38 (for water at 200°C)
- Nu = 0.023 * (1.65 * 10^8)^(4/5) * 4.38^(1/3) = 922.93
- Convective heat transfer coefficient (h) = Nu*k/d
- Thermal conductivity of water (k) = 0.679 W/(m*K) (for water at 200°C)
- h = 922.93 * 0.679/0.00125 = 499,903.04 W/(m^2*K)
4. Calculation of heat flux:
- Heat flux (q) = h*(T-T_inf)
- Water temperature (T_inf) = 200°C = 473 K
- q = 499,903.04 * (473-473) = 0 W/m^2 (no heat transferred due to temperature difference)
5. Calculation of total heat flux:
- Total heat flux (q_total) = q + q_conv
- q_conv = Q/(A*L) = 284.81/(9.74 * 10^-7 * 0.25) = 0.1162 *
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An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer?
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An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer?.
An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat fluxa)0.0915 * 106W/m2b)0.1915 * 106W/m2c)0.2915 * 106W/m2d)0.3915 * 106W/m2Correct answer is option 'C'. Can you explain this answer?.
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