Math Problem Statement

take note this is the brief: A projectile requires a heat treatment operation in one of the furnace ovens. You have been asked by the metallurgist responsible for the heat treatment to demonstrate the use of differential calculus to monitor the process. The furnace is controlled using an electrical heater which must carefully regulate and monitor the projectile part temperature and the supply current.

The difference between the temperature of a body and its surroundings can be modelled using Newton’s Law of heating:

dT/dt=kT

T=F-P

Where T is the temperature difference in K and t is the time in seconds. A part initially at 295 K is placed inside a heat treatment furnace at a constant temperature of (F) K. The parts temperature rises to (P) after 1 minute. F=850 P=290

Solution

To demonstrate the use of differential calculus in this heat treatment operation, we will analyze the process step by step using Newton’s Law of Heating:

1. Newton’s Law of Heating

Newton’s law is given as:

dTdt=kT\frac{dT}{dt} = kT

where:

  • TT is the temperature difference between the body and its surroundings (T=FPT = F - P),
  • kk is the proportionality constant,
  • tt is time in seconds.

The solution to this differential equation is:

T(t)=T0ektT(t) = T_0 e^{kt}

where:

  • T0T_0 is the initial temperature difference at t=0t = 0.

2. Problem Setup

From the problem:

  • Initial temperature of the part: 295K295 \, \text{K},
  • Furnace temperature: F=850KF = 850 \, \text{K},
  • Temperature of the part after 1 minute: P=290KP = 290 \, \text{K}.

The initial temperature difference is:

T0=F295=850295=555K.T_0 = F - 295 = 850 - 295 = 555 \, \text{K}.

The temperature difference after 1 minute (60 seconds) is:

T(60)=FP=850290=560K.T(60) = F - P = 850 - 290 = 560 \, \text{K}.


3. Solving for the Constant kk

Substitute the general solution T(t)=T0ektT(t) = T_0 e^{kt} and the given conditions into the equation:

  1. At t=60t = 60 seconds: T(60)=560=555e60k.T(60) = 560 = 555 e^{60k}.

    Divide through by 555: 560555=e60k.\frac{560}{555} = e^{60k}.

    Simplify: 1.009=e60k.1.009 = e^{60k}.

  2. Take the natural logarithm on both sides: ln(1.009)=60k.\ln(1.009) = 60k.

  3. Solve for kk: k=ln(1.009)60.k = \frac{\ln(1.009)}{60}.

    Using ln(1.009)0.00896\ln(1.009) \approx 0.00896: k0.00896600.000149.k \approx \frac{0.00896}{60} \approx 0.000149.


4. Final Temperature Equation

The final equation for the temperature difference is:

T(t)=555e0.000149t.T(t) = 555 e^{0.000149t}.

This equation can now be used to monitor the temperature difference over time. If you wish to know the part's temperature at any time, substitute T(t)=FP(t)T(t) = F - P(t) and solve for P(t)P(t).


5. Example: Temperature after 120 seconds

To find the part's temperature after 120 seconds:

  1. Find T(120)T(120): T(120)=555e0.000149120.T(120) = 555 e^{0.000149 \cdot 120}.

    Calculate: T(120)555e0.017885551.018=565.0K.T(120) \approx 555 e^{0.01788} \approx 555 \cdot 1.018 = 565.0 \, \text{K}.

  2. Part temperature P(120)P(120): P(120)=FT(120)=850565.0=285.0K.P(120) = F - T(120) = 850 - 565.0 = 285.0 \, \text{K}.


Do you want to dive deeper into any part of this? Let me know!
Here are 5 related questions you could explore further:

  1. How can this process be extended to include non-constant furnace temperatures (FF)?
  2. What happens if the furnace temperature FF is increased or decreased dynamically?
  3. How does the proportionality constant kk affect the rate of heating?
  4. What other physical processes can be modeled using Newton's law of heating?
  5. How does the inclusion of heat loss change the differential equation?

Tip: Always double-check the units and initial conditions when solving differential equations for real-world problems.

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Math Problem Analysis

Mathematical Concepts

Differential Calculus
Newton's Law of Heating
Exponential Functions

Formulas

dT/dt = kT
T(t) = T0 * e^(kt)
k = ln(T(t)/T0)/t

Theorems

Newton's Law of Cooling (Heating)

Suitable Grade Level

Grades 11-12 and Undergraduate Level