2. HEAT TRANSFER THROUGH FORCED CONVECTION
AIM:
To find the heat transfer coefficient of horizontal tube losing heat by conduction, to determine the surface temperature distribution along the length of tube. APPARATUS REQUIRED:
1. Ammeter 2. Voltmeter 3. Temperature indicator 4. Forced convection apparatus THEORY:
Transfer of heat from one region to another due to macroscopic movement in a fluid or gas in addition to energy transfer by conduction is called heat transfer by convection. If fluid motion is caused by an external agency such as a blower or a pump, situation is to be forced convection. In other words, if convection heat transfer occurs due to the dynamic force of an external agency, then it is known as forced convection heat transfer. Newton’s law law of cooling governs the heat transfer. (i.e) Q = h A T Where h = heat transfer coefficient and is a function of density, diameter of tube (D), absolute viscosity, velocity (V), specific heat and thermal conductivity (K). The dependence dependence of h on all the above parameter is generally expressed in terms of dimensionless number. 1. Nusselt number, NU
=
h D/k
2. Prantle number, Pr
=
Cp μ/k
3. Reynolds number, Re
=
ρ V D/μ
Reynolds number plays on important role in forced convection heat transfer. TECHINICAL SPECIFICATION:
1. diameter of the orifice d = 20 mm 2. Inner Diameter, Di = 25mm
3. Length of the section = 400mm
FORMULA USED:
1. Density of air
ρa
=
P/RT
= hw.
ρw/ρa
2. Manometer difference of air,
ha
3. Discharge,
Q =3600 Cd a (2g)
4. Heat supplied through air,
q
=
m c p T/3600, kW
5. Heat transfer coefficient,
h
=
q / A (TS-Ta) W/m K
in (m) 0.5
ha in m3 /hr 2
Where,
ρw hw a Cd m T -
A TS Ta -
3
density of water in kg/m . difference in manometer level. Area of discharge = πd 2 /4, coefficient of discharge = 0.64 mass flow rate = Q ρa kg/hr Temperature difference at two extreme points T = T7 – T1°C 2 Surface area in m = πDiL., Surface température = (T2+T3+T4+T5+ T6) / 5 in °C Temperature of air = (T1 + T7) / 2 in °C
PROCEDURE:
1. Switch on the supply and select the range of voltmeter. 2. Adjust the dimmer stat say 50W, 60W and start heating test section. 3. Start the blower and adjust the flow by means of valve to some decide difference in manometer level, say 5 cm. 4. Wait till steady state is reached. 5. Note down the voltmeter, ammeter and thermocouple T1 to T6 readings. 6. Change the heat input to test sections and repeat the experiment. 7. Calculate the heat transfer coefficient by two methods.
MODEL CALCULATION:
Coefficient of discharge of orifice meter, Cd =0.64 Pressure of air at N.T.P, P = 1.01325 bar Specific heat of air,
Cp = 1.005kj/kg k
Density of water
ρw = 1000 kg/m
Density of air ρa
=
Discharge Q
=
Heat supplied q = m C p t q Heat transfer coefficient h = ----------A (Ts-Ta)
RESULT:
3
Thus the Heat transfer coefficient was calculated by varying the flow of air and results were tabulated.
TABULATION: S.No
Dimmer stat Reading VxI W
5
Voltage
Ammeter
V
I
V
A
Difference in Manometer x10-2
m
Thermometer Reading
Discharge Q
T1
T2
T3
T4
T5
T6
°c
°c
°c
°c
°c
°c
Heat supplied
Surface Temp
m3 /hr
kW
Average Temp
°c
Heat transfer coefficient
°c
W/m2K
6
