A bottle of fresh orange juice is taken from a fridge at 4°C and left in a room with an ambient temperature of 20°C. The bottle is 70 mm diameter and 125 mm long, and you may assume that the juice has the same physical properties as water. The overall heat transfer coefficient for the can in the room is 15 W m² K-1. a) Derive a dynamic model (ie a first-order ODE) for the warming of the bottle using deviation variables; state clearly any assumptions you make. b) Use separation of variables to solve your model; estimate the temperature of the juice after 20 minutes.

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4. A bottle of fresh orange juice is taken from a fridge at 4°C and left in a room with an
ambient temperature of 20°C. The bottle is 70 mm diameter and 125 mm long, and
you may assume that the juice has the same physical properties as water. The
overall heat transfer coefficient for the can in the room is 15 W m-² K-1.
a) Derive a dynamic model (ie a first-order ODE) for the warming of the bottle using
deviation variables; state clearly any assumptions you make.
b) Use separation of variables to solve your model; estimate the temperature of the
juice after 20 minutes.
Transcribed Image Text:4. A bottle of fresh orange juice is taken from a fridge at 4°C and left in a room with an ambient temperature of 20°C. The bottle is 70 mm diameter and 125 mm long, and you may assume that the juice has the same physical properties as water. The overall heat transfer coefficient for the can in the room is 15 W m-² K-1. a) Derive a dynamic model (ie a first-order ODE) for the warming of the bottle using deviation variables; state clearly any assumptions you make. b) Use separation of variables to solve your model; estimate the temperature of the juice after 20 minutes.
Those who got this far seem to have noticed the similarity to problem 1. After taking
the balance, we obtain an expression in a similar general form. Notice that we are
doing a balance on energy and the energy of the contents of the bottle is mCp(T-Tref).
Taking the can out of the fridge is equivalent to a step change in the temperature of
the surroundings, i.e. Tsurr* = 16 °C.
The steady-state gain is 1, since we expect the ultimate change in temperature of the
juice to be equal to Tsurr*.
Always start towards a dynamic model by taking the balance:
Accum. In - Out + Gen - Cons.
We are given an overall heat transfer coefficient (i.e. U). This tells us about the rate
of heat loss per unit surface area i.e. it helps us with the "Out" term.
You are asked to assume that the orange juice has identical physical properties to
water. Density and specific heat capacity are relevant.
You can evaluate the surface area and volume of the vessel.
By working from there, we obtain,
UA
61-6 (-Mant)]
T* (t) = 16 1 exp
Thus, the juice temperature obeys
T(t) = 20 16exp
My estimate for the timescale:
U = 15Wm ²K-1
-
UA
MCw
M≈ 0.48kg
A≈ 0.035m²
UA
CwM
After 20 minutes = 1200s, T = 2016. exp(-0.3)~ 8°C
≈ 2.6 × 10-45-1
Cw = 4200 Jkg-¹K-¹
Transcribed Image Text:Those who got this far seem to have noticed the similarity to problem 1. After taking the balance, we obtain an expression in a similar general form. Notice that we are doing a balance on energy and the energy of the contents of the bottle is mCp(T-Tref). Taking the can out of the fridge is equivalent to a step change in the temperature of the surroundings, i.e. Tsurr* = 16 °C. The steady-state gain is 1, since we expect the ultimate change in temperature of the juice to be equal to Tsurr*. Always start towards a dynamic model by taking the balance: Accum. In - Out + Gen - Cons. We are given an overall heat transfer coefficient (i.e. U). This tells us about the rate of heat loss per unit surface area i.e. it helps us with the "Out" term. You are asked to assume that the orange juice has identical physical properties to water. Density and specific heat capacity are relevant. You can evaluate the surface area and volume of the vessel. By working from there, we obtain, UA 61-6 (-Mant)] T* (t) = 16 1 exp Thus, the juice temperature obeys T(t) = 20 16exp My estimate for the timescale: U = 15Wm ²K-1 - UA MCw M≈ 0.48kg A≈ 0.035m² UA CwM After 20 minutes = 1200s, T = 2016. exp(-0.3)~ 8°C ≈ 2.6 × 10-45-1 Cw = 4200 Jkg-¹K-¹
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