Great news! I passed my FE Mechanical exam! Just got the news today. I came back to this video bc I got a question just like this. I got the right answer too! It was listed so I assumed it was correct. Thanks so much for your help with these videos! Next stop... P.E.!
Hi Anthony! Awesome man! You seemed to be very committed to getting this thing over with and it was clearly shown by your interaction on here. I'm glad this video helped you answer a question on the FE exam and thank you for the kind words. Celebrate, as you will remember this day for a long time (I know I did when I passed lol). Keep it up and slowly start to put in these hours into studying for the PE. Well done!
Hi Samuel, A heat exchanger is specifically designed for the efficient heat transfer from one fluid to another fluid over a solid surface. For example, if we have a parallel flow type heat exchanger, the two fluids enter at the same end of the heat exchanger and flow in the same direction, parallel to one another. In this design, the temperature differences are large at the inlet, but the fluid temperatures will approach a similar value at the outlets. This is often the case when doing FE type problems, if not the temperatures at the outlet will be very close.
Hello Ayubir, it can be confusing in this case but here is a photo showing I conducted the mass energy balance: drive.google.com/file/d/1gqU9hbvACTYWRtHC5Q889wNfLAaZK9DG/view
Note: What was done is ---> Energy (IN) = Energy (Out). For the IN part (water enters from the left at 288K and oil enters from the top at 388 K). For the OUT part (water leaves at the right at 348 K and oil leaves at the bottom at 348 K). Let me know if this helps.
Im a bit lost with the last part. for m_dot oil T=oil temp T_inlet = 388k T_oulet = 348k Cp_oil= 2.20 kj/kgk m_dot(cp_oil)(T_inlet) = m_dot(cp_oil)(T_outlet) dont I have two unknowns? I am stuck on this step.
@@directhubfeexam thank you for helping but it still doesn’t make sense to me. If I don’t cancel the m_dot I will have two and those will both be unknown. m_dot(cp_oil)(T_inlet) = m_dot(cp_oil)(T_outlet) So I have everything but the m_dots. How can I solve for it? I can’t divide since it will cancel out the m_dots.
@@danithaman4610 sorry, it’s hard texting this on phone. But, I think we forgot m_dot for the water which is known to be 65 kg/min. So you have water entering at the inlet at a temperature of 288K and water leaving at a temperature of 348 k. Let me know if you get an answer.
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Great news! I passed my FE Mechanical exam! Just got the news today. I came back to this video bc I got a question just like this. I got the right answer too! It was listed so I assumed it was correct. Thanks so much for your help with these videos! Next stop... P.E.!
Hi Anthony! Awesome man! You seemed to be very committed to getting this thing over with and it was clearly shown by your interaction on here. I'm glad this video helped you answer a question on the FE exam and thank you for the kind words. Celebrate, as you will remember this day for a long time (I know I did when I passed lol). Keep it up and slowly start to put in these hours into studying for the PE. Well done!
Hey Anthony,
Can you see my other comment on this video? Any idea?
Also, thank you for the stuff! I appreciate it!
I believe the answer to the last question is 3.09 kg/s (B)
Why is the exit temperature of oil and water are the same 348k?
Hi Samuel,
A heat exchanger is specifically designed for the efficient heat transfer from one fluid to another fluid over a solid surface.
For example, if we have a parallel flow type heat exchanger, the two fluids enter at the same end of the heat exchanger and flow in the same direction, parallel to one another. In this design, the temperature differences are large at the inlet, but the fluid temperatures will approach a similar value at the outlets.
This is often the case when doing FE type problems, if not the temperatures at the outlet will be very close.
for energy balance is it m_dot cp (delta T) of water = m_dot cp delta T of oil?
Hello Ayubir, it can be confusing in this case but here is a photo showing I conducted the mass energy balance: drive.google.com/file/d/1gqU9hbvACTYWRtHC5Q889wNfLAaZK9DG/view
Note: What was done is ---> Energy (IN) = Energy (Out). For the IN part (water enters from the left at 288K and oil enters from the top at 388 K). For the OUT part (water leaves at the right at 348 K and oil leaves at the bottom at 348 K). Let me know if this helps.
@@directhubfeexam This helps, thanks
@@ayubirfahim281 Nice!
Im a bit lost with the last part.
for m_dot oil
T=oil temp
T_inlet = 388k
T_oulet = 348k
Cp_oil= 2.20 kj/kgk
m_dot(cp_oil)(T_inlet) = m_dot(cp_oil)(T_outlet)
dont I have two unknowns? I am stuck on this step.
I think the mass flow rate cancels since it will not change. The mass flow rate at the inlet is the same as the outlet.
@@directhubfeexam but then what am I solving for? I’ll just have cp(T_inlet) = cp(T_outlet)
@@danithaman4610 sorry m_dot does cancel , that will be your unknown you would need to solve for.
@@directhubfeexam thank you for helping but it still doesn’t make sense to me.
If I don’t cancel the m_dot I will have two and those will both be unknown.
m_dot(cp_oil)(T_inlet) = m_dot(cp_oil)(T_outlet)
So I have everything but the m_dots. How can I solve for it? I can’t divide since it will cancel out the m_dots.
@@danithaman4610 sorry, it’s hard texting this on phone. But, I think we forgot m_dot for the water which is known to be 65 kg/min. So you have water entering at the inlet at a temperature of 288K and water leaving at a temperature of 348 k. Let me know if you get an answer.