IDEAL GAS PROCESSES: ISOBARIC & ISOMETRIC - PROBLEM SOLVING (PART-1)

[@JohnCarloHerrera-t1m]

Sa problem 3 po ndi po ba gagawin na absolute temperature yung 200 at 500 F. Thank you engineer

[@EngrJomDeGuia]

No need na sir. Kung kkunin mo lang dn naman ung change nilang dalawa, same lang po kalalabasan.

[@nickoemmanuelvillacruz6441]

Thank you po Engr. more power po sa channel nyo!

[@irishmaeguerrero5442]

Same naman po ng input sa calc

[@JEROMEPANEZARIS]

Engr. need po ba talaga iconvert into BTU kahit di hinihingi sa problem?

[@EngrJomDeGuia]

Pwde dn namang hindi na. Para less pagod heheheh

[@JEROMEPANEZARIS]

@@EngrJomDeGuia thank you po engineer!

[@kennethjamero4075]

sir, hindi na po ba need i-convert yung F to R? problem 3 isobaric

[@EngrJomDeGuia]

Change in Fahrenheit is equal to change in rankine kaya ok lng kht d na magconvert

[@kennethjamero4075]

@@EngrJomDeGuia tnx sir

[@JeckyDelasolidad-dg8qy]

engr. bakit po sa delta U, delta S, and Q ay Cv ang ginagamit and sa enthalpy naman po ay Cp. Ito po ay sa Isometric

[@EngrJomDeGuia]

Actually we can use Cp naman din. Convertible din naman ung Cv sa Cp.
So its up to you if gagamit ka ng Cp or Cv syempre maximize the given

[@hannahrosco2611]

diba po engr amg heat transferred sa isobaric process is mcpdelta t bakit po naging Q=change in enthalpy + work non flow sa isothermal po diba yan? thank you engr digs 🙌

[@EngrJomDeGuia]

Kapag isothermal un po talaga formula ♥️

[@adrianbicoy8959]

Thank you Engineer

[@kevinroi3631]

Hello engr. nalilito parin ako sa may bandang Enthalpy pano po naging k(∆u) pwde pa elaborate?

[@jillchristiannim6791]

Oo nga asan galing po yung 1.4 na K

[@EngrJomDeGuia]

Ung 1.4 na K constant ng dry air un diba.

[@irishmaeguerrero5442]

Sa problem no. 1 po bakit 1200 pi po lumalabas ty

[@EngrJomDeGuia]

Press S-D button of your scical para maging decimal yung sagot mu

[@JmsLlane-pz4ov]

Sa no. 3 pano po naging Cp = KR/K-1

[@EngrJomDeGuia]

Formula po yan derived

[@raulestimo8497]

35m²/min of air flows steadily through a compressor. It undergoes an isentropic change from 320C and 0.85 bar to 1.7 bar. The measured change in kinetic energy. is 2.5kJ/kg. Neglect changes in potential energy and determine (a) the change in
internal energy; (b) the compressor power requirement.

[@jadedig6512]

kailan po nag aassume ng 1lbm?

[@EngrJomDeGuia]

Pag walang bnanggit sa problem

[@jadedig6512]

kung metric naman po, 1 kgm po ba gagamitin?

[@EngrJomDeGuia]

Yes since ang pnakagnagamit naman pag SI and Metric is Kgm

[@jadedig6512]

okay thank you po Engineer!

[@EngrJomDeGuia]

Welcome poo