Thermal conductivity in a heating system

[astri_lfc]

Homework Statement

In a domestic heating system, a room is warmed by a 'radiator' through which water passes at a rate of 0.12kg s-1. The steady-state difference between the inlet and outlet temperatures of the water is 6.0 K.

The radiator is made of iron of thermal conductivity 80 W m-1 K-1 and has an effective surface area of 1.5 m2 with walls 2.0 mm thick.
(a) At what rate is heat supplied to the room?
(b) What is the mean temperature difference between the inner and outer surfaces of the radiator walls?
[Specific heat capacity of water = 4.2 x 103 J kg-1 K-1]

Homework Equations

a)q/t = k a ΔT / L
is it true use this equations?

b) q = m c ΔT

The Attempt at a Solution

a)
i put all the data to the formula, but the answer is 360000 w
my teacher give me the answer is 3024 w
which one is true?

 

[Chestermiller]

Your teacher is right. Show us your work.

Chet

 

[astri_lfc]

k = 80
A = 1.5 m2
L = 2 x 10^-3
delta T = 6 K

q/t = 80 x 1.5 x 6 / (2x10^-3)

so q/t = 360000

what's wrong with thiss? :'(

 

[Chestermiller]

What's wrong is that the 6 C is the temperature rise of the water in passing through the radiator, not the temperature difference across the thickness of the radiator iron. How much heat does the water lose per second in passing through the radiator?

Chet

 

[astri_lfc]

oh i see
so shall i use q/t= m c deltaT ?

 

[Chestermiller]

oh i see
so shall i use q/t= m c deltaT ?

Yes, if m is the mass flow rate and q/t is the rate of heat transfer.

Chet

 

[CWatters]

There are two parts to the question.

For part..

a) Apply conservation of energy. In steady state Power going into the rad = Power coming out. Power going in is (m/t)cΔT where (m/t) is the flow rate in kg/s^-1, c = specific heat capacity, and ΔT the temperature difference between input and output pipes (the output pipe isn't at absolute zero so some power is returned to the furnace/boiler)

b) A different equation will apply. Have a go first.

 

[CWatters]

Actually there is a better way to explain equation in part a...

Power emitted to the room = Power into the rad - Power returned to the furnace

= mcT_flow - mcT_return
= mcΔT

 

[astri_lfc]

okay thanks a lot! i have found the answer