The hot water needs of a household are met by an electric 60-L hot water tank equipped with a 1.6-kW heating element. The tank is initially filled with hot water at 80°C, and the cold water temperature is 20°C. Someone takes a shower by mixing constant flow rates of hot and cold waters. After a

Question

The hot water needs of a household are met by an electric 60-L hot water tank equipped with a 1.6-kW heating element. The tank is initially filled with hot water at 80°C, and the cold water temperature is 20°C. Someone takes a shower by mixing constant flow rates of hot and cold waters. After a showering period of 8 minutes, the average water temperature in the tank is measured to be 60°C. The heater is kept on during the shower and hot water is replaced by cold water. If the cold water is mixed with the hot water stream at a rate of 0.06 kg/s, determine the flow rate of hot water and the average temperature of mixed water used during the shower.

Accepted Answer

Somebody takes a shower using a mixture of hot and cold water. The mass flow rate of hot water and the average temperature of mixed water are to be determined.

Assumptions The hot water temperature changes from 80°C at the beginning of shower to 60°C at the end of shower. We use an average value of 70°C for the temperature of hot water exiting the tank.

Properties The properties of liquid water are [latex] C_{p}=4.18 kJ / kg \cdot{ }^{\circ} C ext { and } ho=977.6 kg / m ^{3}[/latex] (Table A-2).

Analysis We take the water tank as the system. The energy balance for this system can be expressed as

[latex] E_{ ext {in }}-E_{ ext {out }}=\Delta E_{ ext {sys }} [/latex]

[latex] \left[\dot{W}_{ e , in }+\dot{m}_{ hot } C\left(T_{ in }-T_{ ext {out }} ight) ight] \Delta t=m_{ tank } C\left(T_{2}-T_{1} ight) [/latex]

where [latex]T_{ ext {out }}[/latex] is the average temperature of hot water leaving the tank: (80+70)/2=70°C and

[latex] m_{ ext {tank }}= ho V=\left(977.6 kg / m ^{3} ight)\left(0.06 m ^{3} ight) = 58.656 kg [/latex]

Substituting,

[latex] \left[1.6 kJ / s +\dot{m}_{ hot }\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(20-70){ }^{\circ} C ight](8 imes 60 s )[/latex]

[latex] = (58.656 kg )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(60-80)^{\circ} C [/latex]

[latex] \dot{m}_{ ext {hot }} = 0.0565 kg / s [/latex]

To determine the average temperature of the mixture, an energy balance on the mixing section can be expressed as

[latex] \dot{E}_{ in } = \dot{E}_{ out } [/latex]

[latex] \dot{m}_{ ext {hot }} C T_{ ext {hot }}+\dot{m}_{ ext {cold }} C T_{ ext {cold }}=\left(\dot{m}_{ ext {hot }}+\dot{m}_{ ext {cold }} ight) C T_{ ext {mixture }} [/latex]

[latex] (0.0565 kg / s )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)\left(70^{\circ} C ight)+(0.06 kg / s )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)\left(20^{\circ} C ight)[/latex]

[latex] = (0.0565 + 0.06 kg / s )\left(4.18 kJ / kg .{ }^{\circ} C ight) T_{ ext {mixture }}[/latex]

[latex] T_{ ext {mixture }}= 4 4 . 2 ^{\circ} C [/latex]

Question 1.136: The hot water needs of a household are met by an electric 60...

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