Known A pump system operates at steady state with known inlet and exit conditions. The rate of heat transfer from the pump is specified as a percentage of the power input.
Find Determine the velocities of the water at the inlet and exit of the pump system and the power required.
Schematic and Given Data:
Engineering Model
1. A control volume encloses the pump, inlet pipe, and delivery hose.
2. The control volume is at steady state.
3. The magnitude of the heat transfer from the control volume is 5% of the power input.
4. There is no significant change in temperature or pressure.
5. For liquid water, v \approx v_{f}(T) (Eq. 3.11) and Eq. 3.13 is used to evaluate specific enthalpy.
v(T, p) \approx v_{ f }(T) (3.11)
h(T, p) \approx h_{ f }(T)+\underline{v_{ f }(T)\left[p-p_{ sat }(T)\right]} (3.13)
6. g=32.2 ft / s ^{2}.
Analysis
a. A mass rate balance reduces at steady state to read \dot{m}_{2}=\dot{m}_{1}. The common mass flow rate at the inlet and exit, \dot{m}, can be evaluated using Eq. 4.4b together with v \approx v_{f}\left(70^{\circ} F \right)=0.01605 ft ^{3} / lb from Table A-2E. That is,
\dot{m}=\frac{ AV }{v} (one-dimensional flow) (4.4b)
\dot{m}=\frac{ AV }{v}=\left(\frac{220 gal / min }{0.01605 ft ^{3} / lb }\right)\left|\frac{1 min }{60 s } \right| \left|\frac{0.13368 ft ^{3}}{1 gal }\right|
= 30.54 lb/s
Thus, the inlet and exit velocities are, respectively,
1 V _{1}=\frac{\dot{m} v}{ A _{1}}=\frac{(30.54 lb / s )\left(0.01605 ft ^{3} / lb \right)}{\pi(5 in .)^{2} / 4}\left|\frac{144 in \cdot{ }^{2}}{1 ft ^{2}}\right|=3.59 ft / s
V _{2}=\frac{\dot{m} v}{ A _{2}}=\frac{(30.54 lb / s )\left(0.01605 ft ^{3} / lb \right)}{\pi(1 in .)^{2} / 4}\left|\frac{144 in .^{2}}{1 ft ^{2}}\right|=89.87 ft / s
b. To calculate the power input, begin with the one-inlet, oneexit form of the energy rate balance for a control volume at steady state, Eq. 4.20a. That is,
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)
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]
2 Introducing \dot{Q}_{ cv }=(0.05) \dot{W}_{ cv }, and solving for \dot{W}_{ cv }
\dot{W}_{ cv }=\frac{\dot{m}}{0.95}\left[\left(h_{1}-h_{2}\right)+\left(\frac{ V _{1}^{2}- V _{2}^{2}}{2}\right)+g\left(z_{1}-z_{2}\right)\right] (a)
Using Eq. 3.13, the enthalpy term is expressed as
h_{1}-h_{2}=\left[h_{ f }\left(T_{1}\right)+v_{ f }\left(T_{1}\right)\left[p_{1}-p_{ sat }\left(T_{1}\right)\right]\right]
-\left[h_{ f }\left(T_{2}\right)+v_{ f }\left(T_{2}\right)\left[p_{2}-p_{ sat }\left(T_{2}\right)\right]\right] (b)
Since there is no significant change in temperature, Eq. (b) reduces to
h_{1}-h_{2}=v_{ f }(T)\left(p_{1}-p_{2}\right)
As there is also no significant change in pressure, the enthalpy term drops out of the present analysis. Next, evaluating the kinetic energy term
\frac{ V _{1}^{2}- V _{2}^{2}}{2}=\frac{\left[(3.59)^{2}-(89.87)^{2}\right]\left(\frac{ ft }{ s }\right)^{2}}{2}\left|\frac{1 Btu }{778 ft \cdot lbf }\right|
\times\left|\frac{1 lbf }{32.174 lb \cdot ft / s ^{2}}\right|=-0.1614 Btu / lb
Finally, the potential energy term is
g\left(z_{1}-z_{2}\right)=\left(32.2 ft / s ^{2}\right)(0-35) ft \left|\frac{1 Btu }{778 ft \cdot lbf }\right|
\times\left|\frac{1 lbf }{32.174 lb \cdot ft / s ^{2}}\right|=-0.0450 Btu / lb
Inserting values into Eq. (a)
\dot{W}_{ cv }=\left(\frac{30.54 lb / s }{0.95}\right)[0-0.1614-0.0450]\left(\frac{ B tu }{ lb }\right)
= -6.64 Btu/s
Converting to horsepower
\dot{W}_{ cv }=\left(-6.64 \frac{ B tu }{ s }\right)\left|\frac{1 hp }{2545 \frac{ Btu }{ h }}\right|\left|\frac{3600 s }{1 h }\right|=-9.4 hp
where the minus sign indicates that power is provided to the pump.
1 Alternatively, V _{1} can be evaluated from the volumetric flow rate at 1. This is left as an exercise.
2 Since power is required to operate the pump, \dot{W}_{ cv } is negative in accord with our sign convention. The energy transfer by heat is from the control volume to the surroundings, and thus \dot{Q}_{ cv } is negative as well. Using the value of \dot{W}_{ cv } found in part (b), \dot{Q}_{ cv }=(0.05) \dot{W}_{ cv }=-0.332 Btu / s (-0.47 hp ).
Skills Developed
Ability to…
• apply the steady-state energy rate balance to a control volume.
• apply the mass flow rate expression, Eq. 4.4b.
• develop an engineering model.
• retrieve properties of liquid water.
Quick Quiz
If the nozzle were removed and water exited directly from the hose, whose diameter is 2 in., determine the velocity at the exit, in ft/s, and the power required, in hp, keeping all other data unchanged. Ans. 22.47 ft/s, −2.5 hp.
