Mixing of Hot and Cold Waters in a Shower Consider an ordinary shower where hot water at 140°F is mixed with cold water at 50°F. If it is desired that a steady stream of warm water at 110°F be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from

Question

Mixing of Hot and Cold Waters in a Shower

Consider an ordinary shower where hot water at 140°F is mixed with cold water at 50°F. If it is desired that a steady stream of warm water at 110°F be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 20 psia.

Accepted Answer

In a shower, cold water is mixed with hot water at a specified temperature. For a specified mixture temperature, the ratio of the mass flow rates of the hot to cold water is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus [latex]\Delta m_{\mathrm{CV}}=0[/latex] and [latex]\Delta E_{\mathrm{CV}}=0[/latex]. 2 The kinetic and potential energies are negligible, [latex]k e \cong p e \cong 0[/latex]. 3 Heat losses from the system are negligible and thus [latex]\dot Q \cong 0[/latex]. 4 There is no work interaction involved.

Analysis We take the mixing chamber as the system (Fig. 5-33). This is a control volume since mass crosses the system boundary during the process. We observe that there are two inlets and one exit.

Under the stated assumptions and observations, the mass and energy balances for this steady-flow system can be expressed in the rate form as follows:
Mass balance:

[latex]\dot{m}_{ ext {in }}-\dot{m}_{ ext {out }} =d m_{ ext {system }} / d t earrow^{0 ext { (steady) }} =0[/latex]

[latex]\dot{m}_{ ext {in }}=\dot{m}_{ ext {out }} ightarrow \dot{m}_{1}+\dot{m}_{2}=\dot{m}_{3}[/latex]

Energy balance:

[latex]\underbrace{E_{ ext {in }}-E_{ ext {out}}}_{\begin{array}{c} ext { Rate of net energy transfer} \ ext { by heat, work, and mass }\end{array}} =\underbrace{{dE_{ ext {system }}}/dt earrow ^{0( ext{steady})}}_{\begin{array}{c} ext {Rate of Change in internal, kinetic, } \ ext { potential, etc., energies }\end{array}}=0[/latex]

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

[latex]\dot{m}_{1} h_{1}+\dot{m}_{2} h_{2} =\dot{m}_{3} h_{3}( ext { since } \dot{Q} \cong 0, \dot{W}=0, \mathrm{ke} \cong \mathrm{pe} \cong 0)[/latex]

Combining the mass and energy balances,

[latex]\dot{m}_{1} h_{1}+\dot{m}_{2} h_{2}=\left(\dot{m}_{1}+\dot{m}_{2} ight) h_{3}[/latex]

Dividing this equation by [latex]\dot{m}_{2}[/latex] yields

[latex]y h_{1}+h_{2}=(y+1) h_{3}[/latex]

where [latex]y=\dot{m}_{1} / \dot{m}_{2}[/latex] is the desired mass flow rate ratio.
The saturation temperature of water at 20 psia is 227.92°F. Since the temperatures of all three streams are below this value [latex]\left(T

[latex]h_{1} \cong h_{f @ 140^{\circ} \mathrm{~F}}=107.99 \mathrm{Btu} / \mathrm{lbm} [/latex]

[latex]h_{2} \cong h_{f @ 50^{\circ} \mathrm{F}}=18.07 \mathrm{Btu} / \mathrm{lbm} [/latex]

[latex]h_{3} \cong h_{f @ 110^{\circ} \mathrm{F}}=78.02 \mathrm{Btu} / \mathrm{lbm}[/latex]

Solving for y and substituting yields

[latex]y=\frac{h_{3}-h_{2}}{h_{1}-h_{3}}=\frac{78.02-18.07}{107.99-78.02}=2.0[/latex]

Discussion Note that the mass flow rate of the hot water must be twice the mass flow rate of the cold water for the mixture to leave at 110°F.

Question 5.9: Mixing of Hot and Cold Waters in a Shower Consider an ordina...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions