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## Q. 2.14

Water at 90°C is pumped from a storage tank at a rate of $3 L·s^{−1}$. The motor for the pump supplies work at a rate of $1.5 kJ·s^{−1}$. The water goes through a heat exchanger, giving up heat at a rate of $670 kJ·s^{−1}$, and is delivered to a second storage tank at an elevation 15 m above the first tank. What is the temperature of the water delivered to the second tank?

## Verified Solution

This is a steady-state, steady-flow process for which Eq. (2.31) applies. The initial and final velocities of water in the storage tanks are negligible, and the term $Δu^{2}/2$

can be omitted. All remaining terms are expressed in units of $kJ·kg^{−1}$. At 90°C the density of water is $0.965 kg·L^{−1}$ and the mass flow rate is:

$\Delta H + \frac{\Delta u^2}{2} + g \Delta z=Q+W_s$                 (2.31)

$\dot{m}=(3)(0.965)= 2.895 kg⋅s^{-1}$

For the heat exchanger,

$Q = − 670 / 2.895 = − 231.4 kJ⋅ kg^{−1}$

For the shaft work of the pump,

$W_{s}=1.5 / 2.895 = 0.52 kJ⋅ kg ^{−1}$

If g is taken as the standard value of $9.8 m·s^{−2}$ , the potential-energy term is:

$g\Delta z= (9.8) (15)= 147 m^{2} .s^{-2}= 147 J⋅ kg^{−1} = 0.147 kJ⋅ kg^{−1}$

The steam-table value for the enthalpy of liquid water at 90°C is:

$H_{1} = 376.9 kJ⋅ kg ^{−1}$

Thus,

$ΔH = H_{2} − ​H_{1} = H_{2} − 376.9 = − 231.0$

and

$H_{2} = 376.9 − 231.0 = 145.9 kJ⋅ kg ^{−1}$

The temperature of water having this enthalpy is found from the steam tables:

t = 34.83° C

In this example, $W_{s}$ and gΔz are small compared with Q, and for practical purposes could have been neglected.