3 replies to this topic

[Kristen]

Hello,

I'm working with a single-pass shell and tube heat exchanger (108 tubes, 14.1 sq. ft heat transfer area). I've been tasked with experimentally calculating heat losses from the exchanger to the surroundings and then comparing the experimental loss with the predicted amount of heat loss based on the size of the exchanger, temperature of the exchanger surface, and temperature of surroudings.

I ran hot water through the tubes and cold water through the shell.

My method for experimentally calculating the heat lost to the surroundings is to find the difference between the rate of heat transfer on the shell and tube sides. In mathematical terms:

q_tube - q_shell = q lost to surroundings

q_tube = (mass flow rate of tube water)*(Cp water)*(T_{entering tube} - T_{exiting tube})

q_shell = (mass flow rate of shell water)*(Cp water)*(T_{entering shell} - T_{exiting shell})

When I do this calculation for the data I collected, I find that q_tube = -26,054 Watts and q_shell = 24,223 Watts. That gives a difference of 1,831 Watts. As I understand it, q_tube is the heat lost from the hot tube side. Since the shell side only has 24,000 Watts, I'm assuming the rest of the heat was lost to the surroundings. Is that assumption correct, or could there be 1800 Watts accumulated somewhere in the heat exchanger?

To calculate the expected heat lost to the surroundings I used the following equation:

q = (Natural convection individual heat transfer coefficient, h_L)(Surface Area of Exchanger)(T_surroundings - T_{exchanger surface})

For natural convection in air at room temperature, the individual heat transfer coefficient varies from 1-100 W/m²K, so at the very most, my expected heat loss to the surrounding should be:

(100 W/m^2K)(.2 m^2)(4 K) = 80 Watts

I feel more confident in this theoretical answer than the one I got from my experimental data. Does anyone see an error in my reasoning?

Thanks in advance!

Edited by Kristen, 12 November 2009 - 06:33 PM.

[rana680]

everything seems logical to me. recalculate your results.

[ankur2061]

Kristen,

The fundamental error in your experimental run is that you are considering that heat is lost from the tubes to the surrounding.

The heat lost to surroundings is only from the surface of the STHE shell.

Theoretically you could use Fourier's law of conductivity for a cylindrical surface for heat loss which in it's most simplified form would be as follows:

q/l = 2*PI*K_m*delta T / ln(D_o/D_i)

where:

q= heat lost to the surroundings, W
l= length of your shell, m
K_m = Thermal conductivity of shell metal @ avg. temperature ((T1 +T2)/2), W/m-K
delta T = T1 - T2
T1 = surface temperature of the hot shell, K
T2 = ambient temperature under controlled conditions, K
D_o = Outside Diameter of Shell, m (= D_i + 2*t)
D_i = Inside Diameter of shell, m
t= thickness of shell, m

This method should give you a fairly conservative result for your heat loss to the surroundings.

Hope, I have been of some help.

Regards,
Ankur.

[S.AHMAD]

Hello,

I'm working with a single-pass shell and tube heat exchanger (108 tubes, 14.1 sq. ft heat transfer area). I've been tasked with experimentally calculating heat losses from the exchanger to the surroundings and then comparing the experimental loss with the predicted amount of heat loss based on the size of the exchanger, temperature of the exchanger surface, and temperature of surroudings.

I ran hot water through the tubes and cold water through the shell.

My method for experimentally calculating the heat lost to the surroundings is to find the difference between the rate of heat transfer on the shell and tube sides. In mathematical terms:

q_tube - q_shell = q lost to surroundings

q_tube = (mass flow rate of tube water)*(Cp water)*(T_{entering tube} - T_{exiting tube})

q_shell = (mass flow rate of shell water)*(Cp water)*(T_{entering shell} - T_{exiting shell})

When I do this calculation for the data I collected, I find that q_tube = -26,054 Watts and q_shell = 24,223 Watts. That gives a difference of 1,831 Watts. As I understand it, q_tube is the heat lost from the hot tube side. Since the shell side only has 24,000 Watts, I'm assuming the rest of the heat was lost to the surroundings. Is that assumption correct, or could there be 1800 Watts accumulated somewhere in the heat exchanger?

To calculate the expected heat lost to the surroundings I used the following equation:

q = (Natural convection individual heat transfer coefficient, h_L)(Surface Area of Exchanger)(T_surroundings - T_{exchanger surface})

For natural convection in air at room temperature, the individual heat transfer coefficient varies from 1-100 W/m²K, so at the very most, my expected heat loss to the surrounding should be:

(100 W/m^2K)(.2 m^2)(4 K) = 80 Watts

I feel more confident in this theoretical answer than the one I got from my experimental data. Does anyone see an error in my reasoning?

Thanks in advance!

Hi,
Re-calculate the heat loss using the equation given by ankur. That equation is being used industrially.

The two results should be close enough but it will never be the same - in the order of 2%. However, if the result is far apart, then suspect the experimental data in error. There two types of error in measurement. Standard error and random error. Standard error can be minimized by calibration and for random error collect a lot of data and take the average.You can do heat balance reconciliation if you want to by minimizing the error of measurement.