Question

4. (20 points) A geothermal heat pump carries steam at 150°C underwater, where the surrounding temperature and pressure are measured as 15°C and 200 kPa, respectively. Currents in the water maintain a consistent velocity of 1 cm/s. The pipe outer diameter is 20 cm, and the system can be assumed to be at steady-state. Water, T = 15°C, U = 1 cm/s Touter = 85°C Steam, T = 150°C Douter = 20 cm (a) (10 points) Determine the convection coefficient between the pipe surface and the surrounding water. (b) (10 points) The conductivity of the pipe is 15 W/(m*K), and it has a thickness of 2.5 cm. Determine the temperature at the inner surface of the pipe. HINT 1: You will need to set up a thermal circuit here to evaluate. HINT 2: You may assume a pipe length to help you set up/solve the problem. Use EES to solve and submit formatted equations and solution. Screenshots for either will be fine as long as the text is sufficiently legible.

          4. (20 points) A geothermal heat pump carries steam at 150°C underwater, where the
surrounding temperature and pressure are measured as 15°C and 200 kPa, respectively.
Currents in the water maintain a consistent velocity of 1 cm/s. The pipe outer diameter is
20 cm, and the system can be assumed to be at steady-state.
Water, T = 15°C,
U = 1 cm/s
Touter = 85°C
Steam, T =
150°C
Douter = 20 cm
(a) (10 points) Determine the convection coefficient between the pipe surface and the
surrounding water.
(b) (10 points) The conductivity of the pipe is 15 W/(m*K), and it has a thickness of 2.5 cm.
Determine the temperature at the inner surface of the pipe.
HINT 1: You will need to set up a thermal circuit here to evaluate.
HINT 2: You may assume a pipe length to help you set up/solve the problem.
Use EES to solve and submit formatted equations and solution. Screenshots for either will
be fine as long as the text is sufficiently legible.
        
Show more…
4. (20 points) A geothermal heat pump carries steam at 150°C underwater, where the
surrounding temperature and pressure are measured as 15°C and 200 kPa, respectively.
Currents in the water maintain a consistent velocity of 1 cm/s. The pipe outer diameter is
20 cm, and the system can be assumed to be at steady-state.
Water, T = 15°C,
U = 1 cm/s
Touter = 85°C
Steam, T =
150°C
Douter = 20 cm
(a) (10 points) Determine the convection coefficient between the pipe surface and the
surrounding water.
(b) (10 points) The conductivity of the pipe is 15 W/(m*K), and it has a thickness of 2.5 cm.
Determine the temperature at the inner surface of the pipe.
HINT 1: You will need to set up a thermal circuit here to evaluate.
HINT 2: You may assume a pipe length to help you set up/solve the problem.
Use EES to solve and submit formatted equations and solution. Screenshots for either will
be fine as long as the text is sufficiently legible.

Added by Beth K.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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4. 20 points A geothermal heat pump carries steam at 150°C underwater, where the surrounding temperature and pressure are measured as 15°C and 200 kPa, respectively. Currents in the water maintain a consistent velocity of 1 cm/s. The pipe outer diameter is 20 cm, and the system can be assumed to be at steady-state. Water, T = 15°C, U = 1 cm/s Outer diameter = 20 cm Steam, T = 150°C Inner diameter = 20 cm (a) 10 points: Determine the convection coefficient between the pipe surface and the surrounding water. (b) 10 points: The conductivity of the pipe is 15 W/(m*K), and it has a thickness of 2.5 cm. Determine the temperature at the inner surface of the pipe. HINT 1: You will need to set up a thermal circuit here to evaluate. HINT 2: You may assume a pipe length to help you set up/solve the problem. Use EES to solve and submit formatted equations and solution. Screenshots for either will be fine as long as the text is sufficiently legible.
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Transcript

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00:01 So in this question, total heat loss will be, first we draw the fpd of this question.
00:06 So total heat loss is equal to teta is equal to t s minus t0 upon ln, r1 plus t upon r1 upon 2 pi kl plus 1 upon h0 2 pi kl plus 1 upon h0 2 pi r1 plus t into l so theta will be theta is equal to 110 minus 20 into 2 pi 5 upon lm 55 upon 50 upon 2 pi kl is 20 plus 1 upon 100 into 0 .5 plus 0 .5 plus 0 .0 .0...
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