Known Steam is condensed at steady state by interacting with a separate liquid water stream.
Find Determine the ratio of the mass flow rate of the cooling water to the mass flow rate of the steam and the energy transfer from the steam to the cooling water per kg of steam passing through the condenser.
Schematic and Given Data:
Engineering Model
1. Each of the two control volumes shown on the accompanying sketch is at steady state.
2. There is no significant heat transfer between the overall condenser and its surroundings. \dot{W}_{ cv }=0.
3. Changes in the kinetic and potential energies of the flowing streams from inlet to exit can be ignored.
4. At states 2, 3, and 4, h ≈ hf(T) (see Eq. 3.14).
h(T, p) \approx h_{ f }(T) (3.14)
Analysis The steam and the cooling water streams do not mix. Thus, the mass rate balances for each of the two streams reduce at steady state to give
\dot{m}_{1}=\dot{m}_{2} \quad \text { and } \quad \dot{m}_{3}=\dot{m}_{4}
a. The ratio of the mass flow rate of the cooling water to the mass flow rate of the condensing steam, \dot{m}_{3} / \dot{m}_{1} can be found from the steady-state form of the energy rate balance, Eq. 4.18, applied to the overall condenser as follows:
0=\dot{Q}_{ cv }-\dot{W}_{ cv }+\sum_{i} \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g z_{i}\right)-\sum_{e} \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g z_{e}\right) (4.18)
0=\underline{\dot{Q}_{ cv }}-\underline{\dot{W}_{ c }}+\dot{m}_{1}\left(h_{1}+\underline{\frac{ V _{1}^{2}}{2}+g z_{1}}\right)+\dot{m}_{3}\left(h_{3}+\underline{\frac{ V _{3}^{2}}{2}+g z_{3}}\right)
The underlined terms drop out by assumptions 2 and 3. With these simplifications, together with the above mass flow rate relations, the energy rate balance becomes simply
0=\dot{m}_{1}\left(h_{1}-h_{2}\right)+\dot{m}_{3}\left(h_{3}-h_{4}\right)
Solving, we get
\frac{\dot{m}_{3}}{\dot{m}_{1}}=\frac{h_{1}-h_{2}}{h_{4}-h_{3}}
The specific enthalpy h1 can be determined using the given quality and data from Table A-3. From Table A-3 at 0.1 bar, h_{ f 1}=191.83 kJ / kg \text { and } h_{ g 1}=2584.7 kJ / kg, so
h_{1}=191.83+0.95(2584.7-191.83)=2465.1 kJ / kg
1 Using assumption 4, the specific enthalpy at 2 is given by h_{2} \approx h_{ f }\left(T_{2}\right)=188.45 kJ/kg. Similarly, h_{3} \approx h_{ f }\left(T_{3}\right) \text { and } h_{4} \approx h_{ f }\left(T_{4}\right), \text { giving } h_{4}-h_{3}=62.7 kJ/kg. Thus,
\frac{\dot{m}_{3}}{\dot{m}_{1}}=\frac{2465.1-188.45}{62.7}=36.3
b. For a control volume enclosing the steam side of the condenser only, begin with the steady-state form of energy rate balance, Eq. 4.20a.
0=\dot{Q}_{ cv }-\dot{W}_{ cv }+\dot{m}\left[\left(h_{1}-h_{2}\right)+\frac{\left( V _{1}^{2}- V _{2}^{2}\right)}{2}+g\left(z_{1}-z_{2}\right)\right] (4.20a)
2 0=\dot{Q}_{ cv }-\underline{W_{ cv }}+\dot{m}_{1}\left[\left(h_{1}-h_{2}\right)+\underline{\frac{\left( V _{1}^{2}- V _{2}^{2}\right)}{2}}+\underline{g\left(z_{1}-z_{2}\right)}\right]
The underlined terms drop out by assumptions 2 and 3. The following expression for the rate of energy transfer between the condensing steam and the cooling water results:
\dot{Q}_{ cv }=\dot{m}_{1}\left(h_{2}-h_{1}\right)
Dividing by the mass flow rate of the steam, \dot{m}_{1}, and inserting values
\frac{\dot{Q}_{ cv }}{\dot{m}_{1}}=h_{2}-h_{1}=188.45-2465.1=-2276.7 kJ / kg
where the minus sign signifies that energy is transferred from the condensing steam to the cooling water.
1 Alternatively, \left(h_{4}-h_{3}\right) can be evaluated using the incompressible
liquid model via Eq. 3.20b.
h_{2}-h_{1}=c\left(T_{2}-T_{1}\right)+\underline{v\left(p_{2}-p_{1}\right)} (3.20b)
2 Depending on where the boundary of the control volume is located, two different formulations of the energy rate balance are obtained. In part (a), both streams are included in the control volume. Energy transfer between them occurs internally and not across the boundary of the control volume, so the term \dot{Q}_{ cv } drops out of the energy rate balance. With the control volume of part (b), however, the term \dot{Q}_{ cv } must be included.
Skills Developed
Ability to…
• apply the steady-state mass and energy rate balances to a control volume.
• develop an engineering model.
• retrieve property data for water.
Quick Quiz
If the mass flow rate of the condensing steam is 125 kg/s, determine the mass flow rate of the cooling water, in kg/s. Ans. 4538 kg/s.
