Exercise 1.
In the previous homework, you considered two heat pumps – one operating at a lower
temperature range than the other. The idea is that the low-temperature-range heap pump can be used
to freeze some water and turn it into ice. The ice can be used as a thermal storage, releasing exergy
during the hot part of the day to cool down a space.
In the charging process for the ice storage, refrigerant enters the evaporator at -5℃ and leaves,
fully in vapor phase, at the same temperature, while liquid water at 0℃ is frozen, but remains at the
same temperature. During the discharge process, refrigerant at 2℃ is circulated through a coil and
cooled by the stored ice (see ASHRAE paper on Ice Bear by Willis and Parsonnet). Return air enters the
coil at 23℃, and exits at 13℃.
Compare the overall performance of the ‘standard’ heat pump, and that of the ‘heat pump + ice
storage’ system, in terms of energy and exergy. For each system, calculate exergy losses in the various
stages of the process, from electricity input to the compressor to delivery of cool air to the space.
Finally, for the ice storage device, suggest possible changes that would make the process more efficient,
including perhaps a different phase change material.
Assume that the losses due to heat transfer through the storage walls and piping are negligible.
Environmental conditions are the same as in the previous homework.
Solution
The analysis of the charging cycle is the same as in HW 05, however the reference environmental
temperature has changed.
There are no thermal losses during storage or in the plumbing.
The work supplied to the discharge cycle can be neglected.
From the ASHRAE paper (Willis and Parsonett) the charge time for 95℃ is 12.25 hours, and the
discharge time is 8 hours. These times will be used to calculate the thermal load for comparison
between the traditional heat pump, and the heat pump with ice storage.
Because the cycle spans both day and night, (vs HW05 in which daytime operation is compared to nighttime) let the reference environment be the average temperature in Albuquerque:
𝐿𝑒𝑡 𝑇0 = 28.5℃ = 31.5𝐾
The thermal load can be derived from the charging time and the heat transfer rate between the ice and
the refrigerant.
𝑄 = 𝑚̇𝑟 (ℎ1 − ℎ4 )∆𝑡
𝑘𝑔
𝑘𝐽
𝑄 = (0.0358 ) (247.51 − 96.417) ( ) ∗ 12.25 ℎ𝑜𝑢𝑟𝑠 = 66.235 𝑘𝑊ℎ
𝑠
𝑘𝑔
Charging
The energetic and exergetic efficiencies are calculated by the same methods demonstrated in HW05,
taking into account the new reference environment.
𝐶𝑂𝑃 =
(ℎ1 − ℎ4 )
(ℎ2 − ℎ1 )
𝐶𝑂𝑃 = 5.41
𝜂1 =
1
= 0.185
𝐶𝑂𝑃
𝜓1 =
𝜓1 =
Δ𝐸𝑥𝑖𝑐𝑒
Δ𝐸𝑥𝑒𝑙𝑒𝑐
𝑇
𝑄̇𝐿 [𝑇 0 − 1]
𝑖𝑐𝑒
𝑊̇
𝜓1 = 0.564
Storage
The energetic and exergetic efficiency of the storage period have a value of unity, by definition.
𝜂2 = 𝜓2 = 1
Discharging
The rate of heat transfer from the ice during discharging is calculated with the thermal load and the
discharge time:
𝑄̇ =
𝑄
Δ𝑡𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
𝑄 = 8.279 𝑘𝑊
Because there are no thermal losses during discharge, and the work supplied to the system can be
neglected, the heat transfer from the air into the refrigerant is equal to the heat transfer from the
refrigerant to the ice.
𝜂3 =
𝑄𝑖𝑛
=1
𝑄𝑜𝑢𝑡
This relationship is used to derive the mass-flow-rate of the air:
𝑄̇ = 𝑚̇𝑎𝑖𝑟 𝑐𝑝 Δ𝑇
Where 𝑐𝑝 is assumed to be constant over a small Δ𝑇
𝑘𝐽
𝑄̇ = 𝑚̇𝑎𝑖𝑟 1.046(
)(10K)
𝑘𝑔 ∗ 𝐾
𝑚̇𝑎𝑖𝑟 = 0.7915
𝑘𝑔
𝑠
The exergy efficiency is defined by the change in exergy of the product (cooled air) and the exergy
supplied by the ice.
𝜓3 =
Δ𝐸𝑥𝑎𝑖𝑟
Δ𝐸𝑥𝑖𝑐𝑒
𝑚̇𝑎𝑖𝑟 𝑐𝑝 [(𝑇2 − 𝑇1 ) − 𝑇0 ln
𝜓3 =
𝑇2
]
𝑇1
𝑇
𝑄̇ [𝑇 0 − 1]
𝑖𝑐𝑒
𝑘𝑔
𝑘𝐽
286𝐾
(.7915 𝑠 )(1.046
)[(286𝐾 − 296𝐾) − 301.5𝐾 ln 296𝐾 ]
𝑘𝑔 ∗ 𝐾
𝜓3 =
301.5𝐾
8.279 𝑘𝑊 [ 273𝐾 − 1]
𝜓3 = 0.346
Total Cycle
The total cycle energy efficiency can be calculated in two ways:
𝐶𝑂𝑃𝑡𝑜𝑡 =
𝑄
66.235
=
= 5.41
𝑊𝑐 12.25 𝑘𝑊ℎ
𝜂𝑡𝑜𝑡 =
1
= 0.185
𝐶𝑂𝑃𝑡𝑜𝑡
𝜂𝑡𝑜𝑡 = 𝜂1 ∗ 𝜂2 ∗ 𝜂3 = 0.185 ∗ 1 ∗ 1 = 0.185
The total cycle exergetic efficiency can be calculated in two ways as well:
𝜓𝑡𝑜𝑡
𝑇
𝑚𝑎𝑖𝑟 𝑐𝑝 [(𝑇2 − 𝑇1 ) − 𝑇0 ln 𝑇2 ]
Δ𝐸𝑥𝑎𝑖𝑟
1
=
=
Δ𝐸𝑥𝑒𝑙𝑒𝑐
𝑊𝑐 Δ𝑡
(. 7915
𝜓𝑡𝑜𝑡 =
𝑘𝑔
𝑘𝐽
286
)
(
)) [(286𝐾 − 296𝐾) − 301.5 ln
](8 ℎ𝑜𝑢𝑟𝑠)
𝑠 (1.046 𝑘𝑔 ∗ 𝐾
296
(1 𝑘𝑊)(12.25 ℎ𝑜𝑢𝑟𝑠)
𝜓𝑡𝑜𝑡 = 0.195
𝜓𝑡𝑜𝑡 = 𝜓1 ∗ 𝜓2 ∗ 𝜓3 = .564 ∗ 1 ∗ .346 = 0.195
Comparison with the traditional day-time heat pump
Let the total load, Q, remain the same for both systems.
The time required to achieve the same thermal load, Q, with the traditional system can be obtained with
the heat transfer rate for this system as calculated in HW05.
𝑄 = 𝑄̇ Δ𝑡
𝑄 = 𝑚(̇ℎ1 − ℎ4 )Δ𝑡
66.235 𝑘𝑊ℎ = (0.0361
𝑘𝑔
𝑘𝐽
)(253.35 − 115.8)(
)Δ𝑡
𝑠
𝑘𝑔 ∗ 𝐾
The operating time of the traditional system needed to achieve the same thermal load is:
Δ𝑡 = 13.34 ℎ𝑜𝑢𝑟𝑠
The mass-flow-rate of the air needed to support the cooling load is calculated by:
𝑄̇ = 𝑚̇𝑎𝑖𝑟 𝑐𝑝 Δ𝑇
𝑚̇𝑎𝑖𝑟 = 0.4747
𝑘𝑔
𝑠
The energetic and exergetic efficiencies are calculated by the same methods demonstrated in HW05,
taking into account the new reference environment
𝐶𝑂𝑃 =
𝑄𝑎𝑖𝑟
𝑊𝑐
𝐶𝑂𝑃 = 4.966
𝜂=
1
= .201
𝐶𝑂𝑃
𝜓𝑡𝑜𝑡
𝜓𝑡𝑜𝑡 =
𝑇
𝑚𝑎𝑖𝑟 𝑐𝑝 [(𝑇2 − 𝑇1 ) − 𝑇0 ln 𝑇2 ]
Δ𝐸𝑥𝑎𝑖𝑟
1
=
=
Δ𝐸𝑥𝑒𝑙𝑒𝑐
𝑊𝑐 Δ𝑡
𝑘𝑔
𝑘𝐽
286
(. 4747 𝑠 ) (1.046 (
)) [(286𝐾 − 296𝐾) − 301.5 ln 296](13.34 ℎ𝑜𝑢𝑟𝑠)
𝑘𝑔 ∗ 𝐾
(1 𝑘𝑊)(13.34 ℎ𝑜𝑢𝑟𝑠)
𝜓𝑡𝑜𝑡 = 0.17966