I am having a hard time solving this problem. The formula given is:
T=Tin - (Tin - Tinit.)e(-v/v)t
If the initial temperature Tinit is 115 (degrees F), the cold water temperature is 35 (degrees F) (1.7 degrees C), and the volume and volumetric flow rate are 3,000 liters and 30 liters per minute, respectively,
(a) Calculate the expected water temperature at 5-minute intervals for the first 60 seconds after the flow of cold water is established;
You really don't need an excel program to do this.
heat lost by warm water (3000L) + heat gained by cool water (30L/min) = 0
Five minutes into the operation you have 3000L of 115 C water(convert to C and call that Tinitial for the hot water.That part of the equation is
[mass hot water x specific heat H2O x (Tfinal-Tinitial)]
At 30L/min x 5 min = 150L so that part of the equation is
[mass cool water x specific heat x (Tfinal-Tinitial)] = 0
Tfinal is what you solve for.
Tinitial for warm water is 115F (but converted to C). Tinitial for cold water is 1.7C and Tfinal is what you solve for. Put all that together and you have your first point into the 5-min operation. Then just redo the whole thing for the next 5 min cycle. The whole thing looks like this for the first cycle.
[3E6g x sp.h. x (Tf-about 46)] + [150E3g x sp.h. x (Tf-1.7)] = 0
Solve for Tf.
To calculate the expected water temperature at 5-minute intervals for the first 60 seconds after the flow of cold water is established, we need to use the given formula:
T = Tin - (Tin - Tinit)e(-v/v)t
Given Values:
Tin = Initial Temperature = 115 degrees F
Tinit = Cold Water Temperature = 35 degrees F
v = Volumetric Flow Rate = 30 liters/min
t = Time in minutes (or fractions of a minute)
Since we need to calculate at 5-minute intervals for the first 60 seconds (1 minute), we will calculate for t = 0, 0.083 (5 minutes/60 minutes), 0.167 (10 minutes/60 minutes), 0.25 (15 minutes/60 minutes), 0.333 (20 minutes/60 minutes), and 0.417 (25 minutes/60 minutes).
Let's calculate the expected water temperature at these intervals:
For t = 0 (initial):
T(0) = 115 - (115 - 35)e(-30/3000)x0
= 115 - 80e^0
= 115 - 80
= 35 degrees F
For t = 0.083 (5 minutes):
T(0.083) = 115 - (115 - 35)e(-30/3000)x0.083
= 115 - 80e(-30/3000)(0.083)
= 115 - 80e^(-0.001) ~ 114.999
For t = 0.167 (10 minutes):
T(0.167) = 115 - (115 - 35)e(-30/3000)x0.167
= 115 - 80e(-30/3000)(0.167)
= 115 - 80e^(-0.00167) ~ 114.999
For t = 0.25 (15 minutes):
T(0.25) = 115 - (115 - 35)e(-30/3000)x0.25
= 115 - 80e(-30/3000)(0.25)
= 115 - 80e^(-0.0025) ~ 114.999
For t = 0.333 (20 minutes):
T(0.333) = 115 - (115 - 35)e(-30/3000)x0.333
= 115 - 80e(-30/3000)(0.333)
= 115 - 80e^(-0.00333) ~ 114.999
For t = 0.417 (25 minutes):
T(0.417) = 115 - (115 - 35)e(-30/3000)x0.417
= 115 - 80e(-30/3000)(0.417)
= 115 - 80e^(-0.00417) ~ 114.999
We can see that for the first 60 seconds after the flow of cold water is established, the expected water temperature remains approximately constant at around 114.999 degrees Fahrenheit.
To calculate the expected water temperature at 5-minute intervals for the first 60 seconds, we will use the formula:
T = Tin - (Tin - Tinit) * e^(-v/v * t)
Given:
Tin = initial temperature = 115 degrees F
Tinit = cold water temperature = 35 degrees F
v = volumetric flow rate = 30 liters per minute
t = time elapsed (in minutes)
Since we want to calculate the temperature at 5-minute intervals for the first 60 seconds, we will calculate the temperature at t = 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, and 55 seconds.
First, let's convert the cold water temperature from degrees Fahrenheit to degrees Celsius:
Tinit = 35 degrees F = 1.7 degrees C
Now, we can substitute the values into the formula and calculate the expected water temperature for each time interval:
At t = 0 seconds:
T = Tin - (Tin - Tinit) * e^(-v/v * t)
= 115 - (115 - 1.7) * e^(-30/30 * 0)
= 115 - 113.3 * 1
= 1.7 degrees C
At t = 5 seconds:
T = Tin - (Tin - Tinit) * e^(-v/v * t)
= 115 - (115 - 1.7) * e^(-30/30 * 5/60)
= 115 - 113.3 * e^(-1/12)
≈ 56.9 degrees C
At t = 10 seconds:
T = Tin - (Tin - Tinit) * e^(-v/v * t)
= 115 - (115 - 1.7) * e^(-30/30 * 10/60)
= 115 - 113.3 * e^(-1/6)
≈ 75.5 degrees C
Repeat the above steps for the remaining time intervals.
By following this process, you can calculate the expected water temperature at 5-minute intervals for the first 60 seconds after the flow of cold water is established.