Question Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 1.8 kg/s. If the mixture leaves the mixing chamber at 60°C, determine the mass flow rate of the superheated steam required.

FNV00Z The Asker · Mechanical Engineering

Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 1.8 kg/s. If the mixture leaves the mixing chamber at 60°C, determine the mass flow rate of the superheated steam required.

Transcribed Image Text: Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 1.8 kg/s. If the mixture leaves the mixing chamber at 60°C, determine the mass flow rate of the superheated steam required.

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A)-Assumptions:- 1. Steady ffow 2 KERP.E changes neglect 3. There is no work interactions u - device is adiabatc properfics:- Nute that T<T_{\text {sat } Q \text { Jook-1a }}=133.52^{\circ} \mathrm{C}, the cold water stean and the mixture exists as a campressed liqvid, which can be aproximated as satuvated (igid at the given temperature.so\begin{array}{l}h_{t}=h_{f \odot 2_{a}^{\circ}}=83.91 \mathrm{~kJ} / \mathrm{kg} \\h_{3}=h_{f @ 60^{\circ}}=251.18 \mathrm{~kJ} / \mathrm{kg}\end{array}and\left.\begin{array}{l}P_{2}=300 \mathrm{kra} \\T_{2}=300^{\circ} \mathrm{c}\end{array}\right\} h_{2}=3069.6 \mathrm{k \sigma} k gMass batance:-\begin{array}{l}\Rightarrow \dot{m}_{\text {in }}=\dot{m}_{0}+\vec{\Rightarrow} \Rightarrow \dot{m}_{1}+\dot{m}_{2}=\dot{m}_{3} . \pi_{2}=30 \% \\\end{array}energy bolance:-\hat{E}_{\text {in }}-E_{\text {out }}=\Delta E_{\text {syst }}=0 \text { (steads state) }\begin{array}{l}\dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\\dot{m}_{1} h_{1}+\dot{m}_{2} h_{2}=\dot{m}_{3} h_{3} \quad(\sin C e Q=\omega=\Delta k e=\Delta P E=\end{array}combining (1) > (2), we get in h_{1}+\dot{m}_{2} h_{2}=\left(\dot{m}_{1}+m_{2}\right) h_{3}\begin{aligned}\dot{m}_{2} & =\frac{h_{1}-h_{3}}{h_{3}-h_{2}} \dot{m}_{1} \\& =\frac{83.91-251.18}{251.18-3009.6} \times 1.8=0.107 \mathrm{~kg} / \mathrm{s} \\\therefore \dot{m}_{2} & =0.107 \mathrm{~kg} / \mathrm{s}\end{aligned}pleoge rate themibs up, if you likett. ...