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A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air are
k =3.64 * 10 -2 W/m K
v =30.09 * 10 -6 m2/s
P r =0.682
Β =1/160 + 273 = 2.31 * 10 -3 per K
k =3.64 * 10 -2 W/m K
v =30.09 * 10 -6 m2/s
P r =0.682
Β =1/160 + 273 = 2.31 * 10 -3 per K
- a)1156.5 W
- b)1146.5 W
- c)1136.5 W
- d)1126.5 W
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A steam pipe 50 mm diameter and 2.5 m long has been placed horizontall...
Explanation: Q = h A d t, where h = 14.60 k/D.
Most Upvoted Answer
A steam pipe 50 mm diameter and 2.5 m long has been placed horizontall...
We can use the heat transfer equation for a horizontal cylinder exposed to still air:
Q = (h)(A)(ΔT)
where Q is the rate of heat loss, h is the convective heat transfer coefficient, A is the surface area of the cylinder, and ΔT is the mean temperature difference between the cylinder and the ambient air.
First, we need to find the surface area of the cylinder:
A = πDL + πD^2/4
= π(0.05 m)(2.5 m) + π(0.05 m)^2/4
= 0.425 m^2
Next, we need to find the convective heat transfer coefficient:
Nu = 0.60 + 0.387(Re^0.62)(Pr^0.33)(1 + (0.559/Pr)^0.25)^0.27
Re = ρVD/μ
Pr = μCp/k
where Nu is the Nusselt number, Re is the Reynolds number, Pr is the Prandtl number, ρ is the density of air, V is the velocity of air, D is the diameter of the cylinder, μ is the dynamic viscosity of air, Cp is the specific heat of air at constant pressure, and k is the thermal conductivity of air.
Assuming that the velocity of air is negligible (i.e. still air), we can simplify the Reynolds number to Re = 0. We can also use the given thermo-physical properties of air to find the Prandtl number:
Pr = (μCp)/k
= (30.09*10^-6 m^2/s)(1005 J/kg*K)/(3.64*10^-2 W/m*K)
= 0.831
Plugging in these values, we get:
Nu = 0.60 + 0.387(0)^0.62(0.831)^0.33(1 + (0.559/0.831)^0.25)^0.27
= 0.60 + 0
= 0.60
h = Nu*k/D
= (0.60)(3.64*10^-2 W/m*K)/(0.05 m)
= 4.32 W/m^2*K
Finally, we can calculate the rate of heat loss:
Q = (h)(A)(ΔT)
= (4.32 W/m^2*K)(0.425 m^2)(160 K)
= 292.61 W
Therefore, the rate of heat loss from the steam pipe is 292.61 W.
Q = (h)(A)(ΔT)
where Q is the rate of heat loss, h is the convective heat transfer coefficient, A is the surface area of the cylinder, and ΔT is the mean temperature difference between the cylinder and the ambient air.
First, we need to find the surface area of the cylinder:
A = πDL + πD^2/4
= π(0.05 m)(2.5 m) + π(0.05 m)^2/4
= 0.425 m^2
Next, we need to find the convective heat transfer coefficient:
Nu = 0.60 + 0.387(Re^0.62)(Pr^0.33)(1 + (0.559/Pr)^0.25)^0.27
Re = ρVD/μ
Pr = μCp/k
where Nu is the Nusselt number, Re is the Reynolds number, Pr is the Prandtl number, ρ is the density of air, V is the velocity of air, D is the diameter of the cylinder, μ is the dynamic viscosity of air, Cp is the specific heat of air at constant pressure, and k is the thermal conductivity of air.
Assuming that the velocity of air is negligible (i.e. still air), we can simplify the Reynolds number to Re = 0. We can also use the given thermo-physical properties of air to find the Prandtl number:
Pr = (μCp)/k
= (30.09*10^-6 m^2/s)(1005 J/kg*K)/(3.64*10^-2 W/m*K)
= 0.831
Plugging in these values, we get:
Nu = 0.60 + 0.387(0)^0.62(0.831)^0.33(1 + (0.559/0.831)^0.25)^0.27
= 0.60 + 0
= 0.60
h = Nu*k/D
= (0.60)(3.64*10^-2 W/m*K)/(0.05 m)
= 4.32 W/m^2*K
Finally, we can calculate the rate of heat loss:
Q = (h)(A)(ΔT)
= (4.32 W/m^2*K)(0.425 m^2)(160 K)
= 292.61 W
Therefore, the rate of heat loss from the steam pipe is 292.61 W.
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A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer?
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A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. 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 A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. 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 A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer?.
A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. 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 A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. 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 A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
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A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A steam pipe 50 mm diameter and 2.5 m long has been placed horizontally and exposed to still air at 25 degree Celsius. If the pipe wall temperature is 295 degree Celsius, determine the rate of heat loss. At the mean temperature difference of 160 degree Celsius, the thermo-physical properties of air arek =3.64 * 10-2W/m Kv =30.09 * 10-6m2/sPr=0.682Β =1/160 + 273 = 2.31 * 10-3per Ka)1156.5 Wb)1146.5 Wc)1136.5 Wd)1126.5 WCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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