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mid_term11-20-06-solution

mid_term11-20-06-solution - Mid Term 105A Question 1 A...

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Mid Term 105A 11/20/06 1/4 Question 1 A solar panel receives energy from the Sun and converts it into heat. The panel is 5 meter high and 2 meter wide. It is inclined 40 ° with the horizontal. This panel receives 100 W/m 2 of energy, which warms liquid water from 20 ° C to 50 ° C. We will suppose that water is incompressible and its density equal to 996 kg/m 3 . The area of the pipes is considered constant throughout the system. First we ignore both the kinetic and potential energy of water. a- What is the mass flow rate of water at steady state in kg/s? b- How many liters of water at 50 ° C one collector can provide in one hour? c- How many collectors will be needed to produce the same power as a regular power plant delivering 500MW? What will be the total surface covered by the collectors, supposing there is no free space around them? We now take into account the potential energy of water. d- Supposing the flow of water goes from the top of the panel to the bottom, what is the new mass flow rate of water at steady state in kg/s? We also take into account the fact that the specific volume of water changes as it heats up inside the collector. The water velocity at the outlet is 0.8 m/s. e- Still supposing the flow of water goes from the top of the panel to the bottom, what is the new mass flow rate of water at steady state in kg/s? f- What is the area of the pipe? g- What is the water velocity at the inlet? Using the mass conservation equation to our control volume at steady state we get cv dm dt = ± ± 0 i e m m . Hence = = ± ± ± i e m m m (1). Now using the energy conservation equation at steady state we have cv dE dt = + + + + + ± ± ± ± 0 2 2 2 2 e i cv cv i i i e e e V V Q W m h gz m h gz . Using the equation
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