Need a unique essay?
Order now

Questions, Problems and Answers on Heating and Temperature - Paper Example

Date:  2021-06-07 23:42:18
7 pages  (1753 words)
Categories: 
Back to list
logo_disclaimer
This essay has been submitted by a student.
This is not an example of the work written by our professional essay writers.

Part I - Site

If this sample essay on"Questions, Problems and Answers on Heating and Temperature - Paper Example" doesn’t help,
our writers will!

Calculate the number of trenches that can be accommodated in the proposed site, the length of pipe per trench (including the 2 m length to the next trench), and, hence, the total length of pipe available for heat absorption.

Number of Trenches

Length of trench=20m

Each trench occupies 106.88m; therefore, the total will be 1066.8m

=1067106.68

=10 trenches in total(per trench = 106.68 m, total = 1066.8 m)

Part II Ground Coil

In order to undertake an analysis of the potential rate of heat extraction from the above proposed site, an estimate of the likely operating inlet and outlet temperatures is necessarily made to allow the working fluid thermophysical properties to be determined. Thereafter, if, through calculation, the fluid outlet temperature is found to be significantly different from the initial estimate, properties need to be re-evaluated at the average fluid temperature, and the analysis reworked.

Considering the use of a 20% by mass glycol solution, with respective fluid inlet and outlet temperatures of 3 C and 7 C, using the Excel sheet provided Eth-Glyc Properties.xlsx, determine the appropriate thermophysical properties for use in the forthcoming heat transfer analyses.

Specific heat loss = 500 W/K

The site details:

Dimensions= 16m by 12m

Soil type: Clay, properties, -k= 1.2 W/mk

P= 1780kg/m3

Cp= 1.2Kj/kg.K

Vertical slinky parameters

Radius= 1m

Overlap=1m

Trenches=2m apart

Top of coil=1m below ground level

Polypropylene parameters

Thermal conductivity= 0.45W/Mk

Internal surface roughness= 310-6m

Internal diameter=27mm

Wall thickness=2mm

Circulating pump electrical rating parameters

100W or 150W

Efficiency=75%

Volumetric flow rate

Between 0.41/s and 0.61/s

For an initial assumed volumetric flow rate of 0.32 l/s, for the given pipe dimensions and relevant soil properties, determine the Reynolds number, the Nusselt number and internal surface heat transfer coefficient, and hence the rate of heat absorption per unit length of pipe, in W/m.K.

Re L = u L = 0.1m s 1m = 1161

n 86.1 10 6 m 2 s =55461447

=3.75 W/m.K

(ReD =5546, NuD =77.35, h = 1447 W/m2.K, UA/L = 1.26 W/m.K )

As discussed in class, as the internal fluid temperature increases along the pipe, the temperature differential and, hence, also, the rate of heat transfer, will decrease along the pipe length. To obtain a sufficiently accurate estimation of the fluid temperature at outlet from the pipe, we must consider heat transfer, and associated temperature rise, in incremental stages along the pipe length.

If you are confident that you fully understand this concept and that you can readily undertake the necessary analysis you are not obliged to undertake the following exercise

Exercise:

To consolidate your understanding of this concept, for a ground temperature of 9 C, and a fluid inlet temperature of 3 C, determine the rate of heat transfer for the first trench, considering the whole pipe length in that trench as 1 incremental step. Determine the resultant fluid outlet temperature.

Heat Transfer Analysis

The rate of heat absorption unit length=1.260 W/m.k

Internal surface heat transfer coefficient=1447 W/m2.K

The thermal conductivity of soil will be calculated as

The heat Absorption of the Pipe

Inlet temp= 3C

Outlet temp=6.93C

With the LMTD being 3.70 and the Q =4966.1 W,

Total Kw= 4.97kW

The change in temperature 6 C

The rate of heat transfer (1477 1.8334)

= 805.6 W

(T = 6 C, Rate of Heat transfer = 805.6 W, Tout = 3.64 C)

To calculate the rate of heat absorption in the second trench, we would use the above calculated outlet temperature as inlet, providing a T of 5.32 C. The rate of heat transfer would therefore be lower than for the first trench.

Again, considering heat transfer in the first trench, follow the same analysis as that above for 5m incremental steps, up to the full trench length of 106.68 m. Clearly, the calculations are highly repetitive. You may wish to set up a spreadsheet to do the calculation for you, using the outlet temperature from the previous 5 m section as inlet for the following section. To obtain a valid comparison with the analysis for the whole trench as one step it is necessary to maintain a high degree of accuracy in the temperature calculations, operating to 3 d.p. We can then summate the individual 5 m incremental steps rates of heat absorption to find the trench total rate.

Results from the analysis are given in the table to the left. It can be seen that, as fluid temperature rises along the pipe length, so heat transfer decreases. Considering the whole pipe length in the first trench as one step implies a rate of heat absorption of 805.6 W, while considering absorption based on 5 m incremental steps implies a decrease of 38.7W being approximately 5 %.

There is minimal benefit in accuracy obtained by reducing the incremental step size further.

Moving on.

Through continuing the thermal analysis as detailed above, for 5m incremental steps, the pipe end fluid outlet temperature is calculated as 6.93 C. (This is sufficiently close to the initial assumed outlet temperature of 7 C as to negate any requirement to re-evaluate fluid properties and rework the analysis)

For a ground temperature of 9 C, and respective fluid inlet and outlet temperatures of 3 C and 6.93 C respectively, determine the Log Mean Temperature Difference, and hence the total rate of heat absorption for the pipe.

Revised LMTD and Rate of Heat Absorption

LMTD =3.61C

UA/L =1.259 W/m.K

L=1066.8M

Q=4844.48W

Rate = 4844.481100

=4.4040 Kw

Determine required pump head

Pump head=head loss+ Static lift

=2.2+2.9

=5.1m

The pump output

Output= density gravity pump head V

=1014 105.10.05

=2585.7watts

=2.58 kW

(LMTD = 3.70 C, Q = 4966.1 W = 4.97 kW)

[ Note: summating the individual 5 m incremental steps rates of heat transfer for the whole pipe length implies an overall rate of heat transfer of 4978.1 W, a difference of 0.2 %]

Part III Head Loss and Circulating Pump Selection

The above analysis was based on an initial assumed volumetric flow rate of 0.32 l/s.

Considering the specified pipe internal diameter of 27mm, and a surface roughness of 0.003 mm, calculate the friction factor, the pressure loss ( either in m or Pa), and hence the pump hydraulic output required to develop the assumed flow rate.

hf = f (L/D) x (v^2/2g) where: hf = head loss (m) f = friction factor L = length of pipe work (m) d = inner diameter of pipe work (m) v = velocity of fluid (m/s) g = acceleration due to gravity (m/s2) or: hf = head loss (ft) f = friction factor L = length of pipe work (ft) d = inner diameter of pipe work (ft) v = velocity of fluid (ft/s) g = acceleration due to gravity (ft/s2)

hf = 0.002083 x L x (100/C)^1.85 x (gpm^1.85 / d^4.8655)

(Relative roughness = 0.00011, from Chart or Swamee and Jain eqn, f = 0.0368, h = 23.16 m, Phyd = 75.2 W)

Assuming that from the range of circulating pumps you have available, the preferred choice is a 100 W pump (electrical rating) with an overall electrical to hydraulic output efficiency of 65 %, determine a more realistic flow rate for the system when operating with this pump. You can either do this through manual calculation and an iterative process, or by developing an appropriate spreadsheet in Excel and using the Goal seek function.

RELATIVE ROUGHNESS = 310-6m (3010-3)

0.00003 0.03

= 0.001

PUMP. USE 150 W PUMPS, WITH AN EFFICIENCY OF 65%

Hydraulic Output= 0.65 150

= 97.5 W

= 22/7 (0.03/2)2 (2 16)

= 22/7 (0.015)2 (2 16)

= 0.393 I/s

(Revised flow rate = 0.303 l/s)

The reduced flow rate implies a reduced Reynolds number. The impact on the rate of heat absorption per m of pipe, in W/m.K, is, however, minimal.

(Revised analysis: ReD = 5252, h = 1385 W/m2.K, UA/L = 1.259 W/m.K)

With UA/L effectively unaltered, the reduced flow rate, however, also implies a greater temperature increase over the pipe length, reducing the LMTD and hence the rate at which heat can be abstracted from the ground. Using the same incremental method as detailed above, the pipe outlet temperature is found to be 7.05 C (previously 6.93 C). Determine the new LMTD, and hence the total rate of heat absorption for the pipe.

The rate of heat absorption unit length=1.260 W/m.k

Internal surface heat transfer coefficient=1447 W/m2.K

The thermal conductivity of soil will be calculated as

The heat Absorption of the Pipe

Inlet temp= 3C

Outlet temp=6.93C

With the LMTD being 3.70 and the Q =4966.1 W,

Total Kw= 4.97kW

(LMTD = 3.61 C, Q= 4.84 kW)

If there was a significant divergence from the initial assumed flow rate, leading to a marked drop in potential rate of heat gain, a significantly larger pump may be required. For example, a 200 W pump with 65% efficiency would imply a flow rate of 0.392 l/s, and an increase in heat absorption to 5.40 kW.

This increase in the rate of heat absorption would lower the ground temperature more rapidly, hence having a greater impact on the Coefficient of Performance (CoP) of the heat pump. Additionally, the system outlet temperature, in response to the increase in mass flow rate, is reduced, promoting a higher LMTD. While this is favourable with regard to heat abstraction, the lower outlet temperature potentially has implications for the rate at which heat can be transferred from this fluid to the refrigerant in the heat pump evaporator.

Part IV CoP and Ground Temperature

The heating CoP is a measure of the ratio of heat delivered to the compressor work input. The compressor work input, in the form of electrical energy to drive the motor, is also, along with that abstracted from the ground, delivered as heat to the building. A heating CoP of 3 implies that for every 2 units of heat removed from the ground, 3 will be delivered to the building. (The cooling CoP is always 1 less than the heating CoP). The CoP depends primarily on the temperature/pressure differential between the evaporator and condenser in the refrigeration cycle, and the isentropic efficiency of the compression process. Both of these factors will be considered in more detail in forthcoming lectures and practical sessions. If the temperature differential increases, the CoP decreases. If the isentropic efficiency decreases, the CoP decreases. For a given rate of heat delivery a lower CoP implies a greater running cost.

Considering typical operating temperature differentials and isentropic efficiency, a heating CoP of 3 may be taken for initial consideration. However, through the action of a GSHP, the temperature of the soil adjacent to the coil decreases as heat is drawn from the ground. The rate at which heat is replaced depends on the solar availability (varies seasonally/monthly) and on the soil thermophysical properties and far field temperatures. While we have been considering an effective radius of collection of 1m and a fixed ground temperature at this distance from the coil, in practice, the thermal influence of the system will be greater than this. Simulation of detailed temperature profiles requires significant computational effort.

Our thermal analysis thus far has shown that, for the initial assumed operating conditions, heat can be abstracted at a rate of 4.84 kW from the ground. For a CoP of 3, the system has the potential to deliver heat at a rate of:

Ground Availability + [...

If you are the original author of this essay and no longer wish to have it published on the midtermguru.com website, please click below to request its removal: