FLUID MACHINERY

INDUSTRIAL BLOWERS Wherein,

✓ is a fan used to force air under pressure, that is, ∆𝑈 + ∆𝑊𝑓 = ∆𝐻

the resistance to gas flow is imposed primarily ∆𝐻 = 𝑚𝑐𝑝 (𝑇2 − 𝑇1 )
upon the discharge.
Substituting in the equation,
Fans Blowers Compressors
Presssure 𝑟𝑝 ≤ 1.11 1.11 < 𝑟𝑝 ≤ 1.2 𝑟𝑝 > 1.2 𝑊𝐵 = 𝑚𝑐𝑝 (𝑇2 − 𝑇1 )
Ratio
Blower usually have pressures and volume given in the
Energy Equation problem instead of temperatures and mass.
Rearranging the equation and express it using the
other thermodynamic properties,

𝑊𝐵 = 𝑐𝑝 [𝑚(𝑇2 − 𝑇1 )]

𝑇1
𝑊𝐵 = 𝑐𝑝 [𝑚(𝑇2 − 𝑇1 )] ×
𝑇1

𝑇2 − 𝑇1
For Steady State Energy System (∆𝐸𝑆 = 0) 𝑊𝐵 = 𝑐𝑝 [𝑚𝑇1 ( )]
𝑇1
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝐸1 + 𝑊𝐵 = 𝐸2 + 𝑄𝐿 𝑇2
𝑊𝐵 = 𝑐𝑝 𝑚𝑇1 [( ) − 1]
𝑇1
Where,
𝐸1 = 𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑊𝑓1
From,
𝐸2 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑓2 𝑃𝑉
𝑊𝐵 = 𝑏𝑙𝑜𝑤𝑒𝑟 𝑤𝑜𝑟𝑘 𝑃𝑉 = 𝑚𝑅𝑇 → 𝑚𝑇 =
𝑅
𝑄𝐿 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑠𝑒𝑠
And,
Therefore,
𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑊𝑓1 + 𝑊𝐵 𝑘𝑅
= 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑓2 + 𝑄𝐿 𝑐𝑝 =
𝑘−1

And, The equation will be,

𝑊𝐵 = ∆𝑃𝐸 + ∆𝐾𝐸 + ∆𝑈 + ∆𝑊𝑓 + 𝑄𝐿
𝑘𝑅 𝑃1 𝑉1 𝑇2
𝑊𝐵 = ( ) [( ) − 1]
Considering, T ≠ C or 𝑇1 ≠ 𝑇2, 𝑘−1 𝑅 𝑇1

∆𝑈 ≠ 0 𝑘 𝑇2
𝑊𝐵 = 𝑃1 𝑉1 [( ) − 1]
𝑘−1 𝑇1
And for blowers, ∆𝑃𝐸 & ∆𝐾𝐸 are negligible,
And for an isentropic process,
∆𝑃𝐸 ≈ 0 ; ∆𝐾𝐸 ≈ 0
𝑘−1
𝑇2 𝑃2 𝑘
We also assume that, =( )
𝑄𝐿 = 0 𝑇1 𝑃1

𝑊𝐵 = ∆𝑃𝐸 + ∆𝐾𝐸 + ∆𝑈 + ∆𝑊𝑓 + 𝑄𝐿 Therefore,

𝑘−1
So, 𝑘 𝑃2 𝑘
𝑊𝐵 = ∆𝑈 + ∆𝑊𝑓 𝑊𝐵 = 𝑃 𝑉 [( ) − 1]
𝑘 − 1 1 1 𝑃1

Prepared by: Engr. FJRG

 FLUID MACHINERY

Blower Work (WB) for 3 Different Processes: Blower Head (HB) for 3 Different Processes:

Isentropic Process: 𝑾𝑩
𝑯𝑩 =
𝒎𝒈
𝒌−𝟏
𝒌 𝑷𝟐 𝒌
𝑾𝑩 = 𝑷𝑽 [( ) − 𝟏] Isentropic Process:
𝒌−𝟏 𝑷𝟏
𝒌−𝟏
𝒌 𝑹𝑻 𝑷𝟐 𝒌
𝒌−𝟏 𝑯𝑩 = ( ) [( ) − 𝟏]
𝒌 𝑷𝟐 𝒌 𝒌−𝟏 𝒈 𝑷𝟏
𝑾𝑩 = 𝒎𝑹𝑻 [( ) − 𝟏]
𝒌−𝟏 𝑷𝟏
Polytropic Process:
Polytropic Process:
𝒏−𝟏
𝒏 𝑹𝑻 𝑷𝟐 𝒏
𝒏−𝟏 𝑯𝑩 = ( ) [( ) − 𝟏]
𝒏 𝑷𝟐 𝒏 𝒏−𝟏 𝒈 𝑷𝟏
𝑾𝑩 = 𝑷𝑽 [( ) − 𝟏]
𝒏−𝟏 𝑷𝟏
Isothermal Process:
𝒏−𝟏
𝒏 𝑷𝟐 𝒏 𝑹𝑻 𝑷𝟐
𝑾𝑩 = 𝒎𝑹𝑻 [( ) − 𝟏] 𝑯𝑩 = 𝐥𝐧 ( )
𝒏−𝟏 𝑷𝟏 𝒈 𝑷𝟏

Isothermal Process:

𝑷𝟐
𝑾𝑩 = 𝑷𝑽 𝐥𝐧 ( )
𝑷𝟏

𝑷𝟐
𝑾𝑩 = 𝒎𝑹𝑻 𝐥𝐧 ( )
𝑷𝟏

Prepared by: Engr. FJRG