A copper pipe with an outer radius of 0.011 m runs from an outdoor wall faucet into the interior of a house. The temperature of the faucet is 4.4° C and the temperature of the pipe, at 3.1 m from the faucet, is 24° C. In fourteen minutes, the pipe conducts a total of 249 J of heat to the outdoor faucet from the house interior. Find the inner radius of the pipe. Ignore any water inside the pipe.
To find the inner radius of the pipe, we can use the formula for heat conduction through a cylindrical pipe. The formula is:
Q = (k * A * ΔT * t) / L
Where:
Q = Total heat conducted (249 J)
k = Thermal conductivity of copper (assumed to be constant)
A = Surface area of the pipe (difference between outer and inner surface areas)
ΔT = Temperature difference (24°C - 4.4°C = 19.6°C)
t = Time (14 minutes converted to seconds = 840 seconds)
L = Length of the pipe (3.1 m)
First, we need to find the surface area of the pipe. The surface area of a cylindrical pipe is given by:
A = 2πrh + 2πr^2
Where:
r = Radius (Outer radius - Inner radius)
h = Height or length of the pipe
Since the pipe is a long cylindrical pipe, the height can be considered as its length, which is 3.1 m.
Now, substitute the values into the formula:
249 J = (k * [(2πrh + 2πr^2)] * 19.6°C * 840 s) / 3.1 m
Simplifying the equation:
249 J * 3.1 m = k * [(2πrh + 2πr^2)] * 19.6°C * 840 s
774.9 J * m = k * [(2πrh + 2πr^2)] * 19.6°C * 840 s
Divide both sides by [(2πrh + 2πr^2)] * 19.6°C * 840 s:
(774.9 J * m) / [(2πrh + 2πr^2)] * 19.6°C * 840 s = k
Now, we have an equation with the known values:
(774.9 J * m) / [(2πrh + 2πr^2)] * 19.6°C * 840 s = k
We can substitute the known values for the other variables:
- π ≈ 3.14159
- h = 3.1 m
- ΔT = 19.6°C
- t = 840 s
The thermal conductivity of copper (k) is approximately 400 W/(m·K).
Now, solve for the inner radius (r) using the equation:
(774.9 J * m) / [(2πrh + 2πr^2)] * 19.6°C * 840 s = 400 W/(m·K)
Simplify the equation further:
(774.9 J * m * K * s) / [(2πrh + 2πr^2)] = 400 W
Now, you can solve this equation numerically to find the inner radius (r) using an equation solver or a numerical method.
To find the inner radius of the pipe, we can use the formula for thermal conductivity.
The formula for thermal conductivity is given by:
Q = k * A * ΔT / d
Where:
Q = heat conducted (249 J)
k = thermal conductivity of the material (copper)
A = cross-sectional area of the pipe
ΔT = temperature difference (24°C - 4.4°C = 19.6°C)
d = thickness of the pipe (outer radius - inner radius)
We need to find the inner radius of the pipe, so let's rearrange the formula to solve for it:
d = (k * A * ΔT) / Q
First, let's find the cross-sectional area of the pipe:
A = π * (r_outer^2 - r_inner^2)
The outer radius is given as 0.011 m. We can call the inner radius r.
A = π * (0.011^2 - r^2)
Substituting this expression for A into the formula for d, we get:
d = (k * π * (0.011^2 - r^2) * ΔT) / Q
Now, let's substitute the given values into the formula:
d = (k * π * (0.011^2 - r^2) * 19.6) / 249
Simplifying, we have:
249d = k * π * (0.011^2 - r^2) * 19.6
Divide both sides by 19.6 and rearrange the formula:
r^2 = (k * π * (0.011^2 - r^2) * 249) / 19.6d
r^2 + r^2 * (k * π * 249) / 19.6d = (k * π * 0.011^2 * 249) / 19.6d
Now, let's substitute the values for k (thermal conductivity of copper) and d (thickness of the pipe):
r^2 + r^2 * (401) / d = (0.011^2 * 401) / d
Multiplying both sides by d:
r^2d + r^2 * 401 = 0.011^2 * 401
Since we want to find the inner radius, the term r^2d represents the area of the inner radius times the thickness of the pipe, which is the volume of the pipe material. We can assume this is relatively small compared to the area of the outer radius, so we can neglect it. Therefore:
r^2 * 401 = 0.011^2 * 401
Divide both sides by 401:
r^2 = 0.011^2
Take the square root of both sides:
r = 0.011 m
Therefore, the inner radius of the pipe is 0.011 m.
Assume that the body of the pipe is heat insulated and heat is conducted from one end to the other.
dQ/dt = k•A• (dT/dx),
where the heat conduction rate
dQ/dt = 249/15•60 = 0.277 J/s,
k = 390 J/s•m•oC,
the cross-sectional area of the pipe
(R and r are the outer and inner radii of the pipe)
A = π• (R^2 - r^2),
the temperature gradient
dT/dx = (24°C - 4.4°C)/3.1 = 6.32 °C/m.
Working out the equation and
you should get r = 0.0127m.