Solve Heat Exchange Problem: Steel from 25°C to 60°C

In summary, the given task involves heating a block of steel from 25 degrees Celsius to 110 degrees Celsius using oil at 180 degrees Celsius and then cooling it down to 60 degrees Celsius using water at 15 degrees Celsius. The steel has a dimension of 230mm*230mm*40mm and a density of 0.00785 gram/cubic millimeters. The oil and water are flowing through channels with different diameters and lengths, and with different pressure and density. The specific heat capacity of the steel, oil, and water are also given. The calculation involves finding the flow rates of the oil and water, as well as the time it takes for the steel to reach the target temperature. A sample calculation is provided
  • #1
alex0919
2
0

Homework Statement



A block of steel dimension is 230mm*230mm*40mm=2,116,000cubic millimeters
0.00785 gram/cubic millimeters is the given steel density

Using 180 degree celsius oil (density given is 0.9) to heat up the steel from 25 degree celsius to 110 degree celsius in a 10mm diameter and 428.5mm long channel with oil pressure at 10 bar from the inlet. How long will it take to heat up the steel to the target temperature? What's the oil flow rate?

Then
Using 15 degree celsius water (density given is 1) to cool down the steel from 110 degree celsius to 60 degree celsius in a 8mm diameter and 489.5mm long channel with water pressure at 10 bar from the inlet. How long will it take to cool down the steel tempeerature to 60 degree celsius? What's the water flow rate?

Known S.H.C. of steel is 0.112 Kcal/Kg/per celsius degree
oil is 0.5 Kcal/Kg/per celsius degree
Water is 1 Kcal/Kg/Per celsius degree

Homework Equations



1atm = 14.7psi =101,325Pa
1kg/cm^2 =14.2psi= 97,878.57Pa
2kg/cm^2 =195,757Pa
10kg/cm^2 =978,785Pa
u^2 = 978,785/1,000
u^2/gc = P/d
gc=1

u = 31.29 m/s (Water with density of 1)
u = 32.98 m/s (Oil with density of 0.9)

The Attempt at a Solution



Oil flow rate=32.98m/sec*3.14*0.004m*0.004m=0.001657cubic meters/sec=1.66 Liter/sec.
Water flow rate=31.29m/sec*3.14*0.005m*0.005m=0.002456cubic meters/sec=2.46Liter/sec.
Is my calculation on the velocity right? It looks to me there is something wrong compare with the result we got in a practice similar to this.


(110-25)*0.112*16.61Kg(steel weight)=158.1272Kcal
(180-110)*0.5=140Kcal
158.1272/140=1.12948Kg of required Oil /0.9(Oil density)=1.2550 Liter
1.2550Liter/Oil flow rate at 1.66L/Sec=0.76 sec.


I am not quite sure if my calculation is right because practically, I don't think we can achieve the target in that short period of time. I think there must me something wrong on my calculation. Can anybody point out my mistake? Thank You!
 
Last edited:
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  • #2
Hi alex, welcome to PF.

Just as a pointer when posting homework questions you should always point out where you need help so that people may help you. Simply posting the question and calculating an answer means you posted the question and calculated an answer...nothing else.
 
  • #3
thanks! I had added more detail.
 

1. What is heat exchange?

Heat exchange refers to the transfer of thermal energy from one object or substance to another. This can occur through various processes such as conduction, convection, and radiation.

2. How does heat exchange affect steel?

Heat exchange can affect steel by causing it to expand or contract, depending on the direction of the temperature change. This can lead to changes in the dimensions and properties of the steel, which can impact its performance and structural integrity.

3. How is heat exchange problem solved for steel from 25°C to 60°C?

The heat exchange problem for steel can be solved by using the formula Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the steel, c is the specific heat capacity of steel, and ΔT is the change in temperature (60°C - 25°C = 35°C in this case).

4. What factors can affect the heat exchange of steel?

The heat exchange of steel can be affected by various factors such as the temperature difference between the steel and its surroundings, the material properties of the steel, the presence of any insulating materials, and the surface area of the steel exposed to heat.

5. How can the heat exchange problem be optimized for steel?

The heat exchange problem for steel can be optimized by using materials with higher thermal conductivity, minimizing the temperature difference between the steel and its surroundings, and increasing the surface area of the steel exposed to heat. Additionally, insulating materials can be used to reduce heat loss and improve the efficiency of heat exchange.

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