راهروی گرم خرس بزرگ در منطقه کوهستانی فاون اسکین در جنگل ملی سن برناردینو واقع شده است. یک خانه سفارشی در ساحل شمالی دریاچه خرس بزرگ ساخته شد و مسیر رانندگی در ماه های زمستان زیر سایه خانه بود. به دلیل بارش برف و حرکت وسایل نقلیه بر روی برف، مسیر عبور و مرور با یخبندان به شدت خطرناک شد. صاحبان خانه تصمیم گرفتند برای از بین بردن برف و یخ در ماه های زمستان یک راهروی گرم نصب کنند. 2010 تکمیل شد.
طرح
The heated drive way has 2 closed loops. One loop has water the other loop has glycol (antifreeze type stuff that has a very very low freezing point). The water loop goes through a water heater or boiler and a small pump of1/2 HP pushes the hot water through a heat exchanger. Another small pump pushes the glycol through the heat exchanger but the water and glycol do not mix because they are separated by tubes or plates creating chambers. As the hot water passes through the exchanger it warms the glycol and the water gets cooled. The water and glycol continue the heating and cooling exchange until they reach the desired temperature. The warm glycol exchanges its heat with the icy concrete driveway which warms and melts the snow.
- Sample gallery
Image 1. This photo shows the 8 separate loops that flow directly to the heat exchanger. Then it goes to the ½ HP pump to go back to the driveway. It also shows the pipes coming from the water heater that flows to the heat exchanger and back through the ½ HP pump. Underneath the evaporator is the plate heat exchanger.
Image 2. This is a 20 plate heat exchanger made of steel. It is 12"X4 ¼" X 2" X9 ¾" x 2 1/8". The max pressure is 435 psi. its max Temperature 437 °F. and minimum temperature of -256 °F.
Image 3.The water in the tankless water heater is pumped from this box to the heat exchanger then the cold water get pumped back to the water heater creating a cycle.
Image 4. Another view
Issues
The major issue with this project was the installation of 800 ft. tube under the driveway concrete which the glycol would travel through. The issue with too much tube in one continuous loop is that it causes friction. There are two way to fix the issue, one is to install a larger pump and the other is to break the tubes into smaller 8 loops about 100 ft. long each. The project went with 8 smaller 100 ft. long loops because the home owner didn't want a larger and louder pump. Think about using your mouth as a small pump blowing water thru a garden hose. Easy to blow water out of a 25 ft hose, harder to blow water thru a 100 ft hose, impossible to blow water through 800 ft of hose! So we intercepted the tubes along one side of the driveway and broke the one loop into 8 loops of approximately 100 ft each. The 8 loops come together into a manifold just before they go into the heat exchanger.
به دلیل تلفات اصطکاکی، می توان آن را تقریباً با یک نسخه بازآرایی معادله Hazen-Williams مدل کرد: f=0.2083*(100/C)1.852*(Q1.852/d) که در آن f = کاهش پا در هد در هر 100 فوت لوله Q نرخ جریان بر حسب گالن در دقیقه، C ضریب زبری لوله و d قطر لوله بر حسب اینچ است.
بحث
این سیستم فعلی برای تامین برق به شبکه متصل است. گام بعدی برای کارآمدتر بودن این سیستم، اتصال آبگرمکن به انرژی خورشیدی یا به پمپ حرارتی برای هزینه عملیات کمتر است.
مدلی از مبدل حرارتی ساده
یک مبدل حرارتی ساده ممکن است به عنوان دو لوله مستقیم با جریان سیال در نظر گرفته شود که از نظر حرارتی به هم متصل هستند. بگذارید لوله ها با طول L برابر باشند و سیالاتی با ظرفیت گرما را حمل کنندسیمن{\displaystyle C_{i}}
Temperature profiles for the pipes are T1(x){\displaystyle T_{1}(x)}
- du1dt=γ(T2−T1){\displaystyle {\frac {du_{1}}{dt}}=\gamma (T_{2}-T_{1})}
- du2dt=γ(T1−T2){\displaystyle {\frac {du_{2}}{dt}}=\gamma (T_{1}-T_{2})}
where ui(x){\displaystyle u_{i}(x)}
- du1dt=J1dT1dx{\displaystyle {\frac {du_{1}}{dt}}=J_{1}{\frac {dT_{1}}{dx}}}
- du2dt=J2dT2dx{\displaystyle {\frac {du_{2}}{dt}}=J_{2}{\frac {dT_{2}}{dx}}}
where Ji=Ciji{\displaystyle J_{i}={C_{i}}{j_{i}}}
- J1∂T1∂x=γ(T2−T1){\displaystyle J_{1}{\frac {\partial T_{1}}{\partial x}}=\gamma (T_{2}-T_{1})}
- J2∂T2∂x=γ(T1−T2).{\displaystyle J_{2}{\frac {\partial T_{2}}{\partial x}}=\gamma (T_{1}-T_{2}).}
Note that, since the system is in a steady state, there are no partial derivatives of temperature with respect to time, and since there is no heat transfer along the pipe, there are no second derivatives in x as is found in the heat equation. These two coupled first-order differential equations may be solved to yield:
- T1=A−Bk1ke−kx{\displaystyle T_{1}=A-{\frac {Bk_{1}}{k}}\,e^{-kx}}
- T2=A+Bk2ke−kx{\displaystyle T_{2}=A+{\frac {Bk_{2}}{k}}\,e^{-kx}}
where k1=γ/J1{\displaystyle k_{1}=\gamma /J_{1}}
- T¯1=1L∫0LT1(x)dx{\displaystyle {\overline {T}}_{1}={\frac {1}{L}}\int _{0}^{L}T_{1}(x)dx}
- T¯2=1L∫0LT2(x)dx.{\displaystyle {\overline {T}}_{2}={\frac {1}{L}}\int _{0}^{L}T_{2}(x)dx.}
Using the solutions above, these temperatures are:
T10=A−Bk1k{\displaystyle T_{10}=A-{\frac {Bk_{1}}{k}}} | T20=A+Bk2k{\displaystyle T_{20}=A+{\frac {Bk_{2}}{k}}} |
T1L=A−Bk1ke−kL{\displaystyle T_{1L}=A-{\frac {Bk_{1}}{k}}e^{-kL}} | T2L=A+Bk2ke−kL{\displaystyle T_{2L}=A+{\frac {Bk_{2}}{k}}e^{-kL}} |
T¯1=A−Bk1k2L(1−e−kL){\displaystyle {\overline {T}}_{1}=A-{\frac {Bk_{1}}{k^{2}L}}(1-e^{-kL})} | T¯2=A+Bk2k2L(1−e−kL).{\displaystyle {\overline {T}}_{2}=A+{\frac {Bk_{2}}{k^{2}L}}(1-e^{-kL}).} |
Choosing any two of the temperatures above eliminates the constants of integration, letting us find the other four temperatures. We find the total energy transferred by integrating the expressions for the time rate of change of internal energy per unit length:
- dU1dt=∫0Ldu1dtdx=J1(T1L−T10)=γL(T¯2−T¯1){\displaystyle {\frac {dU_{1}}{dt}}=\int _{0}^{L}{\frac {du_{1}}{dt}}\,dx=J_{1}(T_{1L}-T_{10})=\gamma L({\overline {T}}_{2}-{\overline {T}}_{1})}
- dU2dt=∫0Ldu2dtdx=J2(T2L−T20)=γL(T¯1−T¯2).{\displaystyle {\frac {dU_{2}}{dt}}=\int _{0}^{L}{\frac {du_{2}}{dt}}\,dx=J_{2}(T_{2L}-T_{20})=\gamma L({\overline {T}}_{1}-{\overline {T}}_{2}).}
By the conservation of energy, the sum of the two energies is zero. The quantity T¯2−T¯1{\displaystyle {\overline {T}}_{2}-{\overline {T}}_{1}}
نتیجه گیری
پیمانکاری استخدام کنید که می داند چه کار می کند. پیمانکاری که برای اولین بار این سیستم را نصب کرد تکالیف خود را انجام نداد و لوله های زیادی را در زیر راهرو نصب کرد. و دومین پیمانکار آگاه تر باید بعد از ریختن بتن وارد می شد و مجبور بود بخشی از راهروی کاملاً جدید را برای دسترسی به لوله ها به نمایش بگذارد تا آنها را به طول مناسب و به 8 حلقه جدید برش دهد. سپس مجبور به تعمیر راهروی بتنی شد.
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