ME 200 – Thermodynamics I – Spring 2020

Homework 23: More Reversible Cycles

Part (i): Residential ground source heat pump

Part (ii): Industrial drying process

Part (i): Residential heat pump

Given:

A heat pump provides heat transfer to a home maintained at 23°C and receives heat transfer from a ground well held at 5°C. The heat pump provides heat transfer at a rate of 120,000kJ/h. Over a period of 7 days an electric meter records 700kWh of electricity provided to the heat pump.

In [2]:
# Given Inputs
T_H = 23 + 273.15     # Temperature of the house (high temperature sink) [K]
T_C = 5 + 273.15      # Temperature of the well water (low temperature source) [K]
Q_dot_H = 120000      # Steady state heat transfer rate to the house from the heat pump [kJ/h]
W_net_in = 700*3600   # Electrical work provided to the heat pump over 7 days [kJ]

Find:

a) Determine the total heat transfer from the well to the heat pump that occurred over the 7-day period, in kJ.

b) Determine the overall coefficient of performance of the heat pump for this 7-day period.

c) Determine the cost of operating the heat pump over the 7-day period $(\$)$ for a cost of electricity of 0.15 $\$$/kWh.

d) Calculate the ratio of the coefficient of performance for this heat pump to the coefficient of performance for a reversible heat pump cycle operating between 5°C and 23°C.

System Sketch:

The heat pump is the system. It receives heat from the well water and rejects heat to a the room.

Assumptions:

1) closed system, 2) heat pump operates at steady state over 7 days, 3) sink and source temperatures are constant

Basic Equations:

$$COP_{HP}=\dfrac{Q_H}{W_{net,in}}$$$$COP_{HP,rev}=\dfrac{T_H}{T_H-T_C}$$$$\dfrac{dE}{dt}= \dot Q - \dot W$$

Solution:

Part a)

The overall heat pump is a closed system and is assumed to operate continuously at steady state over the 7-day period. At any point in time, an energy balance on the heat pump under steady state conditions ($dE/dt=0$) is

$$0 = \dot Q - \dot W$$

The above energy balance follows the conventional sign convention for application of the first law. In terms of the cycle energy transfer inputs and outputs shown in the system sketch, $\dot Q = \dot Q_C - \dot Q_H$ and $\dot W = -\dot W_{net,in}$. Substituting these expressions into the energy balance, the net power input to the cycle can be expressed as the difference between the heat transfer rate from the heat pump to the house and the heat transfer from the ground well to the heat pump.

$$\dot W_{net,in} = \dot Q_H - \dot Q_C$$

Both sides of this equation can be integrated over the 7-day period and solved for the total heat transfer from the ground well ($Q_C$) to give

$$Q_C = Q_H - W_{net,in}$$

where $W_{net,in}$ is the specified electrical energy input to the heat pump over the 7 days, $Q_H = \dot Q_H \cdot \Delta t$, and $\Delta t$ is the number of hours over 7 days.

In [3]:
Q_H = Q_dot_H*24*7    # Total heat transfer to house over 7-day period [kJ]
Q_C = Q_H - W_net_in  # Total heat transfer from ground well to heat pump [kJ]

print('Q_C = ',round(Q_C,2),'kJ')

Q_C =  17640000 kJ

Part b)

The coefficient of performance of a heat pump is the heat transfer delivered to the high temperature sink (house in this case) divided by the net work input to cycle or

$$COP_{HP}=\dfrac{Q_H}{W_{net,in}}$$
In [4]:
COP_HP = Q_H/W_net_in

print('COP_HP = ',round(COP_HP,2))

COP_HP =  8.0

This is a pretty good COP for a heat pump, but ground-source heat pumps are quite efficient becuase of a relatively high source temperature associated with the ground.

Part c)

The overall cost of operating the heat pump over the 7-day period is the product of the total electrical usage in kWh and the cost of electricity in $/kWh.

In [5]:
C_elec = 0.15         # cost of electricity ($/kWh)
Elec = W_net_in/3600  # total electricity usage (kWh)
Cost = C_elec*Elec    # total cost of operating the heat pump ($)

print('Cost = $',round(Cost,2))

Cost = $ 105.0

Part d)

The coefficient of performance of a reversible heat pump cycle is

$$COP_{HP,rev}=\dfrac{T_H}{T_H-T_C}$$

Then, the ratio of the actual COP to the reversible heat pump COP is as follows.

In [6]:
COP_HP_rev = T_H/(T_H-T_C)
COP_ratio = COP_HP/COP_HP_rev

print('COP_ratio = ',round(COP_ratio,3))

COP_ratio =  0.486

A COP that is 40 to 50% of the reversible heat pump COP is a pretty efficient device.

Part (ii): Industrial drying process

Given:

An industrial facility requires heat transfer at a rate ($\dot Q_P$) of 30 MW for a drying process at 200°C and 25 MW of power ($\dot W_P$) for other processes in the plant. Two options are being considered: (A) a reversible heat engine to generate the necessary 25 MW of power operating between a heat source at 1000°C and surroundings at 10°C along with direct heat transfer from the source at 1000°C to the process at 200°C and (B) a reversible heat pump operating between the surroundings at 10°C and the process at 200°C along with a reversible power cycle to generate the total required power for both the heat pump and other purposes. This power cycle also operates between 1000°C and the surroundings at 10°C.

In [7]:
Q_dot_P = 30         # Heat transfer rate required for the drying process [MW]
W_dot_P = 25         # Work extracted from the heat engine in both process A and B [MW]
T_H = 1000 + 273.15  # Temperature of the heat source [K]
T_P = 200 + 273.15   # Temperature of the drying process [K]
T_C = 10 + 273.15    # Temperature of the surroundings [K]

Find:

Determine the total required heat transfer rate (MW) for the high temperature source for both option A and B. Assuming that the same type of fuel would be used for both options, comment on which of the two options would use the least amount of fuel? Using second law arguments, explain why this option is inherently better than the other.

Assumptions:

1) closed heat engine and heat pump, 2) steady state operation, 3) reversible heat engine and heat pump, 4) sink and source temperatures are constant

Basic Equations:

$$\eta_{th} = \frac{\dot W_{net,out}}{\dot Q_{H}}$$$$\eta_{th,rev}= 1 - \frac{T_{C}}{T_{H}}$$$$COP_{HP} = \dfrac{\dot Q_H}{\dot W_{net,in}}$$$$COP_{HP,rev} = \dfrac{T_H}{T_H-T_C}$$

Solution:

First consider option A. The heat engine operates between the high temperature source and low temperature surroundings and is considered to be reversible, such that

$$\eta_{th}= 1 - \frac{T_{C}}{T_{H}}$$
In [8]:
eta_th = 1 - T_C/T_H
print('eta_th = ',round(eta_th,3))

eta_th =  0.778

Based on the general definition of heat engine effficiency, the required heat transfer rate $\dot Q_{H,A}$ to produce the required power output for option A ($\dot W_p$) can be determined as

$$\dot Q_{H,A} = \frac{\dot W_p}{\eta_{th}}$$

Then, the total heat transfer rate from the high temperature source for option A is the sum of the heat engine and drying process heat transfer rates or

$$\dot Q_{total,A} = \dot Q_{H,A} + \dot Q_p$$
In [9]:
Q_dot_H_A = W_dot_P/eta_th
Q_dot_total_A = Q_dot_H_A + Q_dot_P

print('Q_dot_H_A = ',round(Q_dot_H_A,1),'MW')
print('Q_dot_total_A = ',round(Q_dot_total_A,1),'MW')

Q_dot_H_A =  32.2 MW
Q_dot_total_A =  62.2 MW

Next let's consider option B. The heat engine has the same efficiency as option A, but the required work output and heat tranfer input are greater because of the heat pump requirement. The heat pump system is considered to be reversible and operating between T_C and T_P such that

$$COP_{HP} = \dfrac{T_P}{T_P-T_C}$$

Based on the general definition of heat pump COP, the heat pump power input requirement can be determined as

$$\dot W_{HP} = \dfrac{\dot Q_p}{COP_{HP}}$$

Finally, the heat input requirement for the heat engine can be determined based on the efficiency and total power output requirement for option B.

$$\dot Q_{H,B} = \frac{\dot W_{p}+\dot W_{hp}}{\eta_{th}}$$

For option B, the total heat transfer requirement is the heat transfer to the heat engine from the high temperature reservoir or

$$\dot Q_{total,B} = \dot Q_{H,B}$$
In [10]:
COP_HP = T_P/(T_P-T_C)
W_dot_HP = Q_dot_P/COP_HP
Q_dot_total_B = (W_dot_P + W_dot_HP)/eta_th

print('COP_HP = ',round(COP_HP,3))
print('W_dot_HP = ',round(W_dot_HP,1),'MW')
print('Q_dot_total_B = ',round(Q_dot_total_B,1),'MW')

COP_HP =  2.49
W_dot_HP =  12.0 MW
Q_dot_total_B =  47.6 MW

The required heat transfer rate from the high temperature reservoir is significantly less for option B than option A. Thus option B would consume less fuel to generate this heat transfer rate.

$$\dot Q_{total,B}<\dot Q_{total,A}$$

Option B has no irreversibilities associated with delivering the required process power and heat transfer rate for drying because the heat engine and heat pump are reversible. In contrast, option A has a highly irreversible heat transfer occurring between the high temperature source at 1000°C and the process at 200°C. Heat transfer across a finite temperature difference is inherently irreversible. Later we will discovery that option has significant entropy generation, while option B has no entropy generation. Option B would undoubtedly be better than option A even if the heat engine and heat pump were not considered to be reversible.