Question

Question1 You must solve the following problem: An automobile heater is required to provide 30m/min air at 30°C to the passengers when the outside temperature is -10°C. If a coolant is available at 85°C, determine the mass flow of coolant needed to heat the air if the coolant cannot leave the heater at a temperature lower than 45°C. (a) Would you need transport equations or equilibrium thermodynamics assumptions to solve this problem? Briefly give your reasons that justify your election (b) Solve the problem 120 marksl
0 0
Add a comment Improve this question Transcribed image text
Answer #1

ven con sideina density *f oi1 : . mass flow Yate of a m in m c Temperature of inside aiv- Tia: 300 Temperatue f dutside cir:If water is used as Coolant, I- must be Svpplird at the

Add a comment
Know the answer?
Add Answer to:
Question1 You must solve the following problem: An automobile heater is required to provide 30m/min air...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Problem #3 (Total marks-14) Given: An engine has the following operating conditions per air-standard Otto cycle:...

    Problem #3 (Total marks-14) Given: An engine has the following operating conditions per air-standard Otto cycle: Gasolina (SI) TYCO engine Pressure at beginning of compression process (Pal state 1 Temperature at beginning of compression process state 1) Volumetric compression ratio , Specific heat added during combustion process ( k g) (process 273) 300 1400 Data used for constant Specific Heat (standard cold-air assumption) Method: Ic ... 1.005 kJ/kg. K(C) = 0.718 kJ/kg.K. k=1.400, R -0.287 kJkgK Required: Draw a clear...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT