Question 7.15: A water pump runs steadily at a flowrate of Q = 80 gal/min wi...

This problem is concerned with the power input to run a pump under ideal and actual conditions. Figure 7.27 shows an appropriate mixed CV containing the pump and the fluid within it and the cut across the shaft connecting the pump to the pump motor. We will assume turbulent flow at the inlet and outlet ports with uniform properties, and that the inlet and outlet elevations are the same. A mass balance shows that \dot M = ρ_1A_1\bar V_1 = ρ_2A_2 \bar V_2. Since the density is constant and the areas are the same, the average velocity at each port is also the same, i.e., \bar V_1 = \bar V_2. For this steady constant density flow with uniform properties Eq. 7.36 gives the energy balance. We will drop the gravitational potential energy terms, and since there is no additional energy input, and no fluid power input (all moving parts of the pump in contact with the fluid are inside the mixed control volume), the corresponding terms in Eq. 7.36 will also be dropped leaving us with

\dot M\left[\left(u_2+\frac{p_2}{\rho } +\frac{1}{2}\bar V^2_2\right)-\left(u_1+\frac{p_1}{\rho } +\frac{1}{2}\bar V^2_1\right)\right]=\dot W_{shaft}+\dot Q_C

Since we know \bar V_1 = \bar V_2, the applicable form of the energy balance is

\dot M\left[\left(u_2+\frac{p_2}{\rho } \right)-\left(u_1+\frac{p_1}{\rho }\right)\right]=\dot W_{shaft}+\dot Q_C                            (A)

We will now apply this equation to analyze the two relevant cases.

1. Minimum power input: We are told the minimum power input corresponds to an ideal pumping process in which heat transfer and internal energy changes are absent. Thus the energy balance (A) becomes

\dot M\left[\left(\frac{p_2}{\rho } \right)-\left(\frac{p_1}{\rho }\right)\right]=\dot W_{shaft}

Since the shaft work in this case is the ideal shaft work required to run the pump we will write this as \dot M[(p_2/ρ)−(p_1/ρ)]=\dot W_{ideal} ideal. Solving for the ideal shaft work, and noting that \dot M= ρQ, where Q is the volume flowrate, we find

\dot W_{ideal}=Q(p_2-p_1)                                (B)

Equation (B) is often used to provide an estimate of the power required to pump a constant density fluid. It corresponds to a perfectly efficient pump. Substituting the given data into (B) we obtain

\dot W_{ideal}=80\ gal/min\left(\frac{1\ min}{60\ s} \right)\left(\frac{1\ \mathsf{ft}^3}{7.48\ gal} \right)(40\ lb_{\mathsf{f}}/n.^2)\left(\frac{12\ in.}{1\ \mathsf{ft}} \right)^2

=[1027(\mathsf{ft}-lb_{\mathsf{f}})/s]\left[\frac{1\ hp}{550(\mathsf{ft}-lb_{\mathsf{f}})/s}\right]=1.87\ hp

Thus under ideal conditions we need \dot W_{ideal}= 1.87 hp to run the pump.

2. Pump efficiency: To consider an actual pumping process, we start with the energy balance (A), and use the volume flowrate to write

\dot W_{shaft}=\dot M(u_2 −u_1)+ Q(p_2 − p_1)−\dot Q_C                       (C)

Since we defined \dot W_{ideal}=Q(p_2-p_1), the energy balance for this process canbe written as

\dot W_{shaft}=\dot W_{ideal}+\dot M(u_2 −u_1)−\dot Q_C                         (D)

Equation (D) shows that the power required in the actual pumping process is increased over that in an ideal process because of the increase in the internal energy of the fluid caused by viscous effects (friction) and any heat transfer that leaves the CV. The efficiency of the pump can be defined as the ratio of the ideal to actual power requirement or

\eta =\frac{\dot W_{ideal}}{\dot W_{shaft}}                                 (E)

For the pump in this example, the efficiency can be calculated from the data as \eta =\dot W_{ideal}/\dot W_{shaft} = 1.87/2 = 0.935, or approximately 94%.

We are told that the heat transfer is negligible in the operation of the pump. To calculate the temperature increase in the water passing through the pump, we drop the heat transfer term in (D) and solve for the internal energy change. Noting that for a liquid u₂ − u₁ = c(T₂ − T₁), we find

T_2-T_1=\frac{\dot W_{shaft}-\dot W_{ideal}}{\dot Mc}                             (F)

We can use the data to calculate the mass flowrate as

\dot M = ρQ =1.94\ slugs/\mathsf{ft}^3(80\ gal/min)\left(\frac{1\ min}{60\ s} \right)\left(\frac{1\ \mathsf{ft}^3}{7.48\ gal}\right)

= 3.46 × 10⁻¹ slug/s

We also know that

\dot W_{shaft}-\dot W_{ideal}=(2-1.87)hp\left[\frac{550(\mathsf{ft}-lb_{\mathsf{f}})/s}{1\ hp} \right]=71.5\ (\mathsf{ft}-lb_{\mathsf{f}})/s

Converting the specific heat for water from Table 2.3 to BG, we have

c =4186\ J/(kg-K)\left(\frac{0.7376\ \mathsf{ft}-lb_{\mathsf{f}}}{1\ J} \right)\left(\frac{14.59\ kg}{1\ slug} \right)\left(\frac{1\ K}{1.8^\circ F} \right)

= 2.5 × 10⁴ (ft-lb_f)/(slugs-°F)

which is the same as the standard value for water of c = 1 Btu/(lb_m-°F) in EE. The temperature increase is now calculated as

We see that the temperature increase is imperceptible because of the large heat capacity of the water. The temperature increase in the fluid due to viscous dissipation of energy is responsible for the heat transfer out of the CV. Since the temperature increase is tiny, the heat transfer is negligible.

3. Useful power input: To determine the useful power input to the water passing through the pump, note that the total energy in the water per unit mass is given by (u + p/ρ + \frac{1}{2}\bar V^2 + gz). For a constant density fluid we can consider the value of the mechanical energy per unit mass (p/ρ + \frac{1}{2}\bar V^2 + gz) to represent useful or available energy, since this energy content can be extracted by devices to produce shaft work. In this case the water pressure has increased in traveling through the pump, and thus the pressure potential energy per unit mass has increased. There is no change in the kinetic or potential energy of the water, so the useful power input to the water is \dot M[(p_2/ρ)−(p_1/ρ)]= Q(p_2 − p_1). We see that this is the same as the ideal shaft work calculated earlier or \dot W_{ideal} =1.87\ hp. This calculation shows that of the 2 hp delivered to the pump, 1.87 hp increases the useful energy of the water passing through the pump. Losses in the pumping process absorb 0.13 hp, and this portion of the total shaft work input causes the internal energy, and hence temperature of the water to rise. We cannot make use of this energy to produce useful work.

TABLE 2.3 Thermal Conductivity and Specific Heat Values for Various Substances at 1 atm and 25°C