 2021-11-10

Steam is lo be condensed on the shell side of a heat exchanger at ${75}^{\circ }F$. Cooling water enters the tubes at ${50}^{\circ }F$ at a rate of 43 lbm/s and leaves at ${65}^{\circ }F$. Assuming the heat exhanger to be well-insulated, determine the rate of heat transfer in the heat exchanger and the rate of condensation of the steam. Jozlyn

The rate form of the energy balance for this system with steady-state flow is
${E}_{\in }-{E}_{out}=\mathrm{△}{E}_{system}$
we have
${E}_{\in }={E}_{out}$
therefore
${Q}_{\in }+m{h}_{1}=m{h}_{2}$
we have
${Q}_{\in }=m{c}_{p}\left({T}_{2}-{T}_{1}\right)$
When this happens, the heat exchanger's rate of heat transfer to the cold water becomes
$Q={\left[m{c}_{p}\left({T}_{out}-{T}_{\in }\right)\right]}_{{H}_{2}O}$
we replace the values in the equation
$Q=\left(451\frac{lbm}{s}\right)\left(1.0\frac{Btu}{lbm{\cdot }^{\circ }F}\right)\left({65}^{\circ }C-{50}^{\circ }C\right)\to Q=675\frac{Btu}{s}$
the rate of condensation of the steam in the heat exchanger is determined from solve for ${m}_{steam}$
$Q={\left(m{h}_{fg}\right)}_{steam}$
we replace the values in the equation
${m}_{steam}+\frac{675\frac{Btu}{s}}{1050.9\frac{Btu}{lbm}}\to {m}_{steam}=0.642\frac{lbm}{s}$

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