### Heat Transfer Lab Report

Thermodynamics is a branch of physical science which deals with the process of heat and mass transfer in physical systems. The process of heat and mass transfer in physical system is govern by the laws of thermodynamics and according to the second law of thermodynamic it is not possible for a physical system to convert all the heat into work or work into heat. Some of the heat and work involve in the physical system is loss of the surrounding and within the system. The process of heat transfer in a physical system happened on basic of any one of the three modes of heat transfer, conduction, convection and radiation. Conduction is the mode of the transfer where heat it transferred between solids bodies in contact with each other or from one side of a solid body to the other side of the solid body. Conduction in Watt W depends on the material ability to transfer heat called conductive heat transfer coefficient of material K in watt per meter centigrade W/m C, thickness of the cross section area t in meter m. the area involve in heat transfer A in square meter m^2 and temperature difference across the cross section area delta T in centigrade C.
Heat=K/t*Area* delat T
Convection is the mode of the transfer where heat it transferred between solids body and fluid in contact with each other. Convection in Watt W depends on both the fluids and solid ability to transfer heat called convective heat transfer coefficient of h in watt per meter centigrade W/m C, the area involve in heat transfer A in square meter m^2 and temperature difference across the cross section area delta T in centigrade C.
Heat=h*Area* delat T
There is a special case of convective heat transfer called force convection where fluid is force to move at a specific speed against the metal surface. This is done to increase the flow of heat at the solid and fluid interface point. The force convection depend only mass flow rate of the fluid m in kilogram per second Kg/sec, specific heat capacity Cp of the fluid in joule per kilogram centigrade J/Kg C and the  temperature difference across the cross section area delta T in centigrade C.
Heat=m*Cp* delat T
Other than the mode of heat transfer the system in which the heat transfer is being done is also important. There are three types of the thermodynamic system, open, close and isolated. Open system is one in which heat and mass both moves in and out of the system boundaries, close system is one in which only heat can move in and out of the system boundaries and the isolated system is one in which heat and mass both cannot move in and out of the boundaries of the system. For any system the heat that move into the system as input and moves out as work done or output, there is some amount of it being wasted into the surrounding. To account this the energy balance of the thermodynamics system is analyzed where the heat provided should be converted into useful work, waste heat and system losses. Any imbalance in the system should be properly discussed.
Heat=Work+losses+Waste

Apparatus
The force convection apparatus consist of simple base frame which hold the centrifugal fan inside it and a complete assembly of rest of apparatus on top of it. The top assembly consists of copper tube around which are heating coil is winded to heat the copper tube as required. Coil heat electrically. In order to stop heat from moving out lagging is placed on top of heating coil winded copper tube. Centrifugal fan is connected to copper tube with the help of a u shaped pipe which allow air to enter the copper tube smoothly. An orifice plate is placed inside the copper tube to measure the flow rate of the air. Thirteen T type thermocouples are on different locations on the copper tubes and lagging to measure the temperature at that specific point. Inlet and outlet temperature of the air is measures separately.
Procedure

Following is the procedure to perform the experiment of force convection
• First step is the installation and setting of the required components to make the apparatus fit to use. Make sure apparatus is on the flat surface and all electrical connections are properly attached and tight.
• Start the apparatus and wait until a stead reading on the manometer of pilot tube and thermocouple is achieved
• After the stead state condition measure the input voltage and current of the fan
• Note the air inlet temperature using special thermocouple installed at the inlet of the tube
• Note all thirteen thermocouple reading and write them in the provided table
• Note the height of fluid of manometer of orifice plate to calculated the mass flow rate of air
• Without changing the parameters, change the location of pilot tube as per provided data and measure the note the temperature of outlet air
• Make necessary calculations
Measurements
Number Temperature C
1 57.7
2 61.9
3 64.2
4 67.1
5 71.3
6 66.2
7 65.4
8 59.7
10 66.6
12 84.1
9 37.5
11 39.2
13 45

Position on Scale mm T 14
141.5 40
149 46.9
156.5 40.1
164 43
171.5 50.4

Calculations
The following properties can be assumed for the purpose of calculation:
• Thermal conductivities of the copper pipe: 380 W/m· C.
• Thermal conductivities of the lagging: 0.04 W/m· C.
• Average specific heat capacity of air: 1005 J/kg· C.
• Average density of air: 1.177 kg/m^3
• Density of water: 1000 kg/m^3
• Discharge coefficient of orifice: 0.613.

Discussion
According to the law of the thermodynamics it is not possible to convert or transfer all the provided heat into the useful work. This is what happens here some of the heat provided by the electrically heat coil is used to heat the air moving through the copper tube and some of the energy is loss to the surrounding. Loss of heat from the heating area is happens even with the lagging material surrounding the heat source. The heat provided to the system by the electrically heated coil is around 1320 watt from which the 1172 watt of heat being transferred to air moving inside the heated copper tube. Some of the heat is loss to the surrounding environment for example the radial heat loss by the lagging material is about 34 Watt and longitudinal heat loss by the lagging material is about 0.00142 watt. Reason for this small heat loss by lagging material that it has very low thermal conductivity and this was the reason it was used for insulation purpose.
Energy balance equation shows that the system is imbalance in terms of the total heat transferred and useful heat plus the losses. The imbalance is about 112 watt which show that the 112 watt of heat is wasted somewhere in the system or 112 watt of energy is over calculated. In the case of heat wasted in the system the reason by 112 watt being wasted can be due to faulty apparatus means heat calculated is more but actual heat generated is less or thermocouple does not show faulty reading. It also can be due to human error where error in noting the reading from the apparatus.

Conclusion
Aim of this lab work was to understand the force convection happening inside the copper tube with air as the fluid and that has been completed successfully as the heat of 1172 watt has been transfer to the air from the 1320 watt actually generated. This means that most of the heat generated in transfer to the air where some amount of it has been wasted and loss to the surrounding. The energy balance shows that system in imbalance where the 112 watt of heat produces does not have any record. This can be due to the any fault in the apparatus or personal error in calculation or measurement.