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4. Consider a Vapour Absorption Refrigeration System. Neglecting pump work, show that the maximum COP of the system is the product of the COP of a reverse Carnot refrigeration system and the Carnot efficiency of a heat engine. A heat pipe has a length of 30-cm and uses water at mass flow rate of 0.01 g/s. The permeability of the wick is $3 \times 10^{-10}$ m$^2$. Calculate the cross-sectional area of the wick for a pressure drop of 400 Pa. What is the APR of the heat pipe? Assume the properties of water as $h_{fg} = 2260$ kJ/kg; $\rho_l = 1000$ kg/m$^3$; $\mu_l = 2.75 \times 10^{-4}$ N-s/m$^2$; $\sigma_l = 5.89 \times 10^{-2}$ N/m. 

          4. Consider a Vapour Absorption Refrigeration System. Neglecting pump work, show that the maximum
COP of the system is the product of the COP of a reverse Carnot refrigeration system and the Carnot
efficiency of a heat engine.
A heat pipe has a length of 30-cm and uses water at mass flow rate of 0.01 g/s. The permeability of
the wick is $3 \times 10^{-10}$ m$^2$. Calculate the cross-sectional area of the wick for a pressure drop of 400 Pa.
What is the APR of the heat pipe? Assume the properties of water as $h_{fg} = 2260$ kJ/kg; $\rho_l = 1000$
kg/m$^3$; $\mu_l = 2.75 \times 10^{-4}$ N-s/m$^2$; $\sigma_l = 5.89 \times 10^{-2}$ N/m.
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4. Consider a Vapour Absorption Refrigeration System. Neglecting pump work, show that the maximum COP of the system is the product of the COP of a reverse Carnot refrigeration system and the Carnot efficiency of a heat engine. A heat pipe has a length of 30-cm and uses water at mass flow rate of 0.01 g/s. The permeability of the wick is 3 × 10^-10 m^2. Calculate the cross-sectional area of the wick for a pressure drop of 400 Pa. What is the APR of the heat pipe? Assume the properties of water as hfg = 2260 kJ/kg; = 1000 kg/m^3; = 2.75 × 10^-4 N-s/m^2; = 5.89 × 10^-2 N/m.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A heat pipe has a length of 30 cm and uses water at a mass flow rate of 0.01 g/s. The permeability of the wick is 3 x 10^-10 m. Calculate the cross-sectional area of the wick for a pressure drop of 400 Pa. What is the APR of the heat pipe? Assume the properties of water as hfg = 2260 kJ/kg; ρ = 1000 kg/m^3; μ = 2.75 x 10^-3 N-s/m^2; σ = 5.89 x 10^-2 N/m.
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Transcript

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00:01 In this question it is given that water enters in a heat exchanger and the mass flow rate of the water is 10 kilogram per second.
00:11 In the heat exchanger, its temperature increases from 100 degrees centigrade to 500 degrees centigrade and the pressure remains constant that is 2 ,000 kilo -pascal.
00:26 Air is used to heat this water.
00:30 The inlet temperature of the air is 1 ,400 kelvin and the outlet temperature of the air is 460 kelvin.
00:41 We are required to calculate the second law efficiency.
00:45 So let's see how to solve this question.
00:47 First of all, let's draw a rough diagram of the heat exchanger and the heat exchanger is shown below.
00:54 So here we have the heat exchanger.
00:57 Water is entering from this point let's say this is 1 and it is leaving from this point let's say it is 2 air is entering from this point let's say this is 3 and air is leaving from this point let's say this is 4 now let's apply the energy equation to calculate the mass flow rate of air so we can write m .w multiplied by h2 minus h1 this will be equals to m .d.
01:28 Air, this is the mass flow rate of air, multiplied by h3 minus h4.
01:34 So from here we can write, mass flow rate of air is equals to mass flow rate of water multiplied by h2 minus h1 divided by h3 minus h4.
01:47 Now we will substitute all the values from this steam table.
01:51 So we can write m .d .a is equals to m .w, that means, 10, multiplied by 3 ,467 .55 minus 420 .45 divided by 1 ,515 .27 minus 462 .34.
02:17 So when we further calculate we get mass flow rate of air is equal to 28 .9 .9 kilogram per second...
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