An electric heater is used to heat a slab, and the following model has been d the slab temperature: dT = Q(t) – a(T* – T,“) dt where T is the slab temperature in °R, Q(t) is the rate of heat input in Btu/h variable, C= 250 Btu/ºR, Ts = 530°R and a = 5x10-8 Btu/h-°R“.
An electric heater is used to heat a slab, and the following model has been d the slab temperature: dT = Q(t) – a(T* – T,“) dt where T is the slab temperature in °R, Q(t) is the rate of heat input in Btu/h variable, C= 250 Btu/ºR, Ts = 530°R and a = 5x10-8 Btu/h-°R“.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![**Modeling Slab Heating with an Electric Heater**
An electric heater is used to heat a slab, and the following model has been derived to predict the slab temperature:
\[ C \frac{dT}{dt} = Q(t) - \alpha (T^4 - T_s^4) \]
Where:
- \( T \) is the slab temperature in degrees Rankine (°R),
- \( Q(t) \) is the rate of heat input in Btu/h, which is an input variable,
- \( C = 250 \) Btu/°R,
- \( T_s = 530 \) °R,
- \( \alpha = 5 \times 10^{-8} \) Btu/h·°R⁴.
### Tasks
(a) **Linearized Model Derivation:**
Obtain a linearized model around a slab steady-state temperature of 650°R.
(b) **Transfer Function Development:**
Obtain the transfer function for the process relating the slab temperature to the heating rate. Determine the time constant and steady-state gain of the linearized model.
### Explanation
This model represents the dynamics of slab heating via an electric heater. The equation captures the change in temperature over time (\( \frac{dT}{dt} \)) as a balance between the heat input and the radiation losses/gains (represented by \( \alpha (T^4 - T_s^4) \)). The variables and constants given describe the system's physical properties and how it responds to heat input.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ed819a1-7cc7-4b35-b605-cdaaef8761fd%2Fc28c17a9-cf1d-45d4-8900-ba70b4deb75f%2Fd9xh29h_processed.png&w=3840&q=75)
Transcribed Image Text:**Modeling Slab Heating with an Electric Heater**
An electric heater is used to heat a slab, and the following model has been derived to predict the slab temperature:
\[ C \frac{dT}{dt} = Q(t) - \alpha (T^4 - T_s^4) \]
Where:
- \( T \) is the slab temperature in degrees Rankine (°R),
- \( Q(t) \) is the rate of heat input in Btu/h, which is an input variable,
- \( C = 250 \) Btu/°R,
- \( T_s = 530 \) °R,
- \( \alpha = 5 \times 10^{-8} \) Btu/h·°R⁴.
### Tasks
(a) **Linearized Model Derivation:**
Obtain a linearized model around a slab steady-state temperature of 650°R.
(b) **Transfer Function Development:**
Obtain the transfer function for the process relating the slab temperature to the heating rate. Determine the time constant and steady-state gain of the linearized model.
### Explanation
This model represents the dynamics of slab heating via an electric heater. The equation captures the change in temperature over time (\( \frac{dT}{dt} \)) as a balance between the heat input and the radiation losses/gains (represented by \( \alpha (T^4 - T_s^4) \)). The variables and constants given describe the system's physical properties and how it responds to heat input.
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