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The hot-water needs of a household are to be met by heating water at \(55^{\circ} \mathrm{F}\) to \(180^{\circ} \mathrm{F}\) by a parabolic solar collector at a rate of \(4 \mathrm{lbm} / \mathrm{s}\). Water flows through a 1.25 -indiameter thin aluminum tube whose outer surface is blackanodized in order to maximize its solar absorption ability. The centerline of the tube coincides with the focal line of the collector, and a glass sleeve is placed outside the tube to minimize the heat losses. If solar energy is transferred to water at a net rate of \(400 \mathrm{Btu} / \mathrm{h}\) per ft length of the tube, determine the required length of the parabolic collector to meet the hotwater requirements of this house.

Short Answer

Expert verified
Answer: The required length of the parabolic collector is approximately 4500 ft.

Step by step solution

01

Calculate the energy needed to heat the water

First, we need to calculate the energy required to heat 1 lb of water from 55°F to 180°F. To do this, we will use the specific heat capacity of water (c) which is approximately 1 Btu/(lbm °F). Energy to heat 1 lb of water (ΔQ) = m × c × ΔT where: ΔQ = Energy (Btu) m = mass of water (lbm) c = specific heat capacity of water (1 Btu/(lbm °F)) ΔT = Change in temperature (°F) ΔT = 180 - 55 = 125 °F ΔQ = 1 × 1 × 125 = 125 Btu
02

Calculate the energy absorbed per ft length of the collector

We are given the solar energy transferred to water at a net rate of 400 Btu/h per ft length of the tube. To find the energy absorbed per second, we need to convert the units. Energy absorbed per ft length = (400 Btu/h) × (1 h/3600 s) = 400/3600 Btu/s Energy absorbed per ft length ≈ 0.1111 Btu/s
03

Calculate the energy needed to heat the water at 4 lbm/s

The mass flow rate (m_dot) of the water is given as 4 lbm/s. We can calculate the energy needed to heat the water at this mass flow rate using the following formula: Total energy needed (Q_total) = m_dot × ΔQ Q_total = 4 lbm/s × 125 Btu/lbm = 500 Btu/s
04

Determine the required length of the parabolic collector

We will now calculate the required length of the parabolic collector by dividing the total energy needed by the energy absorbed per ft length of the collector. Required length (L) = Q_total / (Energy absorbed per ft length) L = 500 Btu/s / 0.1111 Btu/s L ≈ 4500 ft Hence, the required length of the parabolic collector to meet the hot water requirements of this house is approximately 4500 ft.

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Most popular questions from this chapter

Water flows through a shower head steadily at a rate of \(10 \mathrm{L} / \mathrm{min}\). An electric resistance heater placed in the water pipe heats the water from 16 to \(43^{\circ} \mathrm{C}\). Taking the density of water to be \(1 \mathrm{kg} / \mathrm{L},\) determine the electric power input to the heater, in \(\mathrm{kW}\). In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of \(39^{\circ} \mathrm{C}\) through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case. If the price of the electric energy is 11.5 e \(/ \mathrm{kWh}\), determine how much money is saved during a 10 -min shower as a result of installing this heat exchanger.

Refrigerant 134 a enters a compressor with a mass flow rate of \(5 \mathrm{kg} / \mathrm{s}\) and a negligible velocity. The refrigerant enters the compressor as a saturated vapor at \(10^{\circ} \mathrm{C}\) and leaves the compressor at \(1400 \mathrm{kPa}\) with an enthalpy of \(281.39 \mathrm{kJ} / \mathrm{kg}\) and a velocity of \(50 \mathrm{m} / \mathrm{s}\). The rate of work done on the refrigerant is measured to be \(132.4 \mathrm{kW}\). If the elevation change between the compressor inlet and exit is negligible, determine the rate of heat transfer associated with this process, in \(\mathrm{kW}\).

Air is to be heated steadily by an 8 -kW electric resistance heater as it flows through an insulated duct. If the air enters at \(50^{\circ} \mathrm{C}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\), the exit temperature of air is \((a) 46.0^{\circ} \mathrm{C}\) \((b) 50.0^{\circ} \mathrm{C}\) \((c) 54.0^{\circ} \mathrm{C}\) \((d) 55.4^{\circ} \mathrm{C}\) \((e) 58.0^{\circ} \mathrm{C}\)

A refrigeration system is being designed to cool \(\operatorname{eggs}\left(\rho=67.4 \mathrm{lbm} / \mathrm{ft}^{3} \text { and } c_{p}=0.80 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\right)\) with an average mass of \(0.14 \mathrm{lbm}\) from an initial temperature of \(90^{\circ} \mathrm{F}\) to a final average temperature of \(50^{\circ} \mathrm{F}\) by air at \(34^{\circ} \mathrm{F}\) at a rate of 10,000 eggs per hour. Determine \((a)\) the rate of heat removal from the eggs, in \(\mathrm{Btu} / \mathrm{h}\) and \((b)\) the required volume flow rate of air, in \(\mathrm{ft}^{3} / \mathrm{h}\), if the temperature rise of air is not to exceed \(10^{\circ} \mathrm{F}\).

A \(110-V\) electric hot-water heater warms \(0.1 \mathrm{L} / \mathrm{s}\) of water from 18 to \(30^{\circ} \mathrm{C}\). Calculate the current in amperes that must be supplied to this heater.

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