EXAMPLE: CLOSED SYSTEM

A gas expands in a cylinder from 500 kPa

and 0.1 m3 to a final volume of 0.5 m3.
Determine the work done if the pressure
distribution is determined to be:
a. P=C
b. PV=C
where V is in m3 and P in kPa.
3
𝑘𝑁
𝑚 [ 2 ] = 𝑘𝑁 ∙ 𝑚 = 𝑘𝐽
𝑚

Given:
P1 = 500 kPa
V1 = 0.1 m3
V2 = 0.5 m3

Required:
a. WNF = ? kJ if P = C
 b. WNF = ? kJ if PV = C

Solution:
Solving for letter (a) WNF when P=C:
𝑉2

𝑊𝑁𝐹 = ∫ 𝑃𝑑𝑉
𝑉1

𝑤ℎ𝑒𝑟𝑒: 𝑃 = 𝐶

𝑉2

𝑊𝑁𝐹 = ∫ 𝐶𝑑𝑉
𝑉1
𝑉2

𝑊𝑁𝐹 = 𝐶 ∫ 𝑑𝑉
𝑉1

𝑊𝑁𝐹 = 𝐶 [𝑉2 − 𝑉1 ]

𝑏𝑢𝑡 𝑃 = 𝐶;
 𝑊𝑁𝐹 = 𝑃 [𝑉2 − 𝑉1 ]

𝑏𝑦 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛:
𝑘𝑁
𝑊𝑁𝐹 = [500 2 ] [0.5𝑚3 − 0.1𝑚3 ]
𝑚

𝑾𝑵𝑭 = 𝟐𝟎𝟎 𝒌𝑵 ∙ 𝒎 𝒐𝒓 𝒌𝑱 (𝑨𝑵𝑺𝑾𝑬𝑹)

Solving for letter (b) WNF when PV=C:

𝑉2

𝑊𝑁𝐹 = ∫ 𝑃𝑑𝑉
𝑉1

𝑤ℎ𝑒𝑟𝑒: 𝑃𝑉 = 𝐶
𝐶
𝑃=
𝑉

𝑏𝑦 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛:
 𝑉2
𝐶
𝑊𝑁𝐹 = ∫ 𝑑𝑉
𝑉
𝑉1
𝑉2
𝑑𝑉
𝑊𝑁𝐹 =𝐶 ∫ = 𝐶 [ln 𝑉 ]𝑓𝑟𝑜𝑚 𝑉1 𝑡𝑜 𝑉2
𝑉
𝑉1

𝑊𝑁𝐹 = 𝐶 [𝑙𝑛𝑉2 − 𝑙𝑛𝑉1 ]

𝑉2
𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔: 𝑊𝑁𝐹 = 𝐶 ln [ ]
𝑉1

𝑏𝑢𝑡: 𝑃𝑉 = 𝐶
𝑉2
𝑠𝑜: 𝑊𝑁𝐹 = 𝑃𝑉 ln [ ]
𝑉1

𝑊𝑒 𝑚𝑎𝑦 𝑢𝑠𝑒 𝑃1 𝑉1 𝑜𝑟 𝑃2 𝑉2 ; 𝑏𝑢𝑡 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑜𝑛𝑙𝑦 𝑔𝑖𝑣𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑠 𝑃1 , 𝑡ℎ𝑢𝑠:

𝑉2
𝑊𝑁𝐹 = 𝑃1 𝑉1 ln [ ]
𝑉1
 3
𝑘𝑁 0.5 𝑚
𝑊𝑁𝐹 = [500 2 ] [0.1 𝑚3 ] ln [ 3
]
𝑚 0.1 𝑚

𝑾𝑵𝑭 = 𝟖𝟎. 𝟒𝟕 𝒌𝑱 (𝑨𝑵𝑺𝑾𝑬𝑹)

EXAMPLE: OPEN SYSTEM

A steady flow system receives 4.56 kg m per minute of
a fluid where P1 = 137.90 kPa, ʋ1 = 0.0388 m3/kgm,
ῡ1 = 122 m/s, U1 = 17.16 kJ/kgm. The fluid leaves the
system at a boundary where P2 = 551.6 kPa, ʋ2 =
0.193 m3/kgm, ῡ2 = 183 m/s, and U2 = 52.80 kJ/kgm.
During passage through the system, the fluid receives
3000 J/s of heat. Determine the work in Watts.

WSF=?

Given:

P1 = 137.90 kPa P2 = 551.6 kPa

ʋ1 = 0.0388 m3/kgm ʋ2 = 0.193 m3/kgm

SYSTEM
ῡ1 = 122 m/s ῡ2 = 183 m/s

U1 = 17.16 kJ/kgm U2 = 52.80 kJ/kgm

ḿ = 4.56 kgm/min

Q = 3000 J/s
Required:
Work: WSF = ? W

Solution:
The system is an open system. Let’s do energy
balance in the system.
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑊𝑓1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑓2 + 𝑊𝑆𝐹
𝑄 = (𝑃𝐸2 − 𝑃𝐸1 ) + (𝐾𝐸2 − 𝐾𝐸1 ) + (𝑈2 − 𝑈1 ) + (𝑊𝑓2 − 𝑊𝑓1 ) + 𝑊𝑆𝐹
𝑄 = ∆𝑃𝐸 + ∆𝐾𝐸 + ∆𝑈 + ∆𝑊𝑓 + 𝑊𝑆𝐹

We are looking for WSF, so:

𝑊𝑆𝐹 = 𝑄 − ∆𝑃𝐸 − ∆𝐾𝐸 − ∆𝑈 − ∆𝑊𝑓

Let’s do this one by one. In order to determine WSF, we need to

know the value of Q, ΔPE, ΔKE, ΔU and ΔWf. Remember, the
required unit is Watts.
Let’s start with Q (heat).
𝐽
𝑄 = 3000
𝑠
𝐽
𝑁𝑂𝑇𝐸: = 𝑊𝑎𝑡𝑡
𝑠
𝑠𝑜: 𝑸 = 𝟑𝟎𝟎𝟎 𝑾
Second, let’s compute for ΔPE.
∆𝑃𝐸 = 𝑃𝐸2 − 𝑃𝐸1
ḿ𝑔𝑧
𝑤ℎ𝑒𝑟𝑒: 𝑃𝐸 =
𝑘
ḿ2 𝑔𝑧2 ḿ1 𝑔𝑧1
∆𝑃𝐸 = −
𝑘 𝑘

𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑠𝑠:

ḿ1 = ḿ2 = ḿ
ḿ𝑔𝑧2 ḿ𝑔𝑧1
𝑡ℎ𝑢𝑠: ∆𝑃𝐸 = −
𝑘 𝑘

𝑏𝑢𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑔𝑖𝑣𝑒𝑛 𝑧 (𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛);

𝑠𝑜 𝑤𝑒 𝑚𝑎𝑦 𝑎𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 1 𝑖𝑠 𝑎𝑡 𝑠𝑎𝑚𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑝𝑜𝑖𝑛𝑡 2;
𝑧1 = 𝑧2 = 𝑧
ḿ𝑔𝑧 ḿ𝑔𝑧
𝑠𝑜: ∆𝑃𝐸 = −
𝑘 𝑘

𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:
∆𝑷𝑬 = 𝟎

Third, let’s solve for ΔKE.

∆𝐾𝐸 = 𝐾𝐸2 − 𝐾𝐸1
 ḿῡ2
𝑤ℎ𝑒𝑟𝑒: 𝐾𝐸 =
2𝑘

𝑙𝑒𝑡 ′ 𝑠 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑓𝑜𝑟 𝐾𝐸1 :

ḿῡ1 2
𝐾𝐸1 =
2𝑘
𝑘𝑔𝑚 𝑚 2 1 𝑚𝑖𝑛
[4.56 ] [122 ] [ ]
𝑚𝑖𝑛 𝑠 60 𝑠
𝐾𝐸1 =
𝑘𝑔𝑚 ∙ 𝑚
2 [1 ]
𝑁 ∙ 𝑠2
𝑁∙𝑚
𝐾𝐸1 = 565.592 𝑜𝑟 𝑊
𝑠

𝑁∙𝑚
𝑊𝑎𝑖𝑡. 𝐷𝑖𝑑 𝑦𝑜𝑢 𝑔𝑒𝑡 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑚𝑒 𝑓𝑟𝑜𝑚?
𝑠
𝐿𝑒𝑡 ′ 𝑠 𝑑𝑜 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠:
𝑘𝑔𝑚 2
[ ]2 𝑚 1 𝑚𝑖𝑛
[4.56 ] 122 [ 2 ] [ ]
𝑚𝑖𝑛 𝑠 60 𝑠
𝑘𝑔𝑚 ∙ 𝑚
2 [1 ]
𝑁 ∙ 𝑠2
𝐶𝑜𝑙𝑜𝑟 𝑅𝑒𝑑: 𝑘𝑔𝑚 𝑖𝑠 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑒𝑑
𝐶𝑜𝑙𝑜𝑟 𝑂𝑟𝑎𝑛𝑔𝑒: min 𝑖𝑠 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑒𝑑
𝐶𝑜𝑙𝑜𝑟 𝐺𝑟𝑒𝑒𝑛: 𝑠 2 𝑖𝑠 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑒𝑑
𝐶𝑜𝑙𝑜𝑟 𝑃𝑖𝑛𝑘: 𝑚 𝑖𝑠 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑒𝑑, 𝑏𝑢𝑡 𝑜𝑛𝑒 𝑚 𝑤𝑖𝑙𝑙 𝑠𝑡𝑎𝑦 𝑜𝑛 𝑡ℎ𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑠𝑖𝑛𝑐𝑒 𝑖𝑡 𝑖𝑠 𝑠𝑞𝑢𝑎𝑟𝑒𝑑
𝑇ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑢𝑛𝑖𝑡𝑠 𝑎𝑟𝑒:
 𝑚
565.592
𝐾𝐸1 = 𝑠
1
𝑁
𝑡𝑜 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦:
𝑁∙𝑚
𝐾𝐸1 = 565.592 𝑜𝑟 𝑊
𝑠

𝑁𝑒𝑥𝑡, 𝑙𝑒𝑡 ′ 𝑠 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝐾𝐸2 :

ḿῡ2 2
𝐾𝐸2 =
2𝑘
𝑘𝑔𝑚 𝑚 2 1 𝑚𝑖𝑛
[4.56 ] [183 ] [ ]
𝑚𝑖𝑛 𝑠 60 𝑠
𝐾𝐸2 =
𝑘𝑔𝑚 ∙ 𝑚
2 [1 ]
𝑁 ∙ 𝑠2
𝑁∙𝑚
𝐾𝐸2 = 1272.582 𝑜𝑟 𝑊
𝑠

𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:
∆𝐾𝐸 = 𝐾𝐸2 − 𝐾𝐸1
∆𝐾𝐸 = 1272.582 𝑊 − 565.592 𝑊
∆𝑲𝑬 = 𝟕𝟎𝟔. 𝟗𝟗 𝑾

Fourth, compute for ΔU.

∆𝑈 = 𝑈2 − 𝑈1
 𝑤ℎ𝑒𝑟𝑒:
𝑘𝐽
𝑈1 = 17.16
𝑘𝑔𝑚
𝑘𝐽
𝑈2 = 52.80
𝑘𝑔𝑚

𝑘𝐽 𝑘𝐽
∆𝑈 = 52.80 − 17.16
𝑘𝑔𝑚 𝑘𝑔𝑚
𝑘𝐽
∆𝑈 = 35.64
𝑘𝑔𝑚

𝑘𝐽
𝑏𝑢𝑡 𝑡𝑎𝑘𝑒 𝑛𝑜𝑡𝑒 𝑜𝑓 𝑡ℎ𝑖𝑠, 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 ∆𝑈 𝑖𝑠 ;
𝑘𝑔𝑚
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑖𝑛 𝑢𝑛𝑖𝑡 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒎𝒂𝒔𝒔;
𝑖𝑛 𝑜𝑡ℎ𝑒𝑟 𝑤𝑜𝑟𝑑𝑠, 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑎𝑙𝑙 𝑖𝑡 𝑎𝑠 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚:
𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒎𝒂𝒔𝒔

𝑆𝑜, 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠:

𝑘𝑔𝑚
𝑤ℎ𝑒𝑟𝑒: ḿ = 4.56
𝑚𝑖𝑛
𝑘𝐽 𝑘𝑔𝑚
∆𝑈 = 35.64 [4.56 ]
𝑘𝑔𝑚 𝑚𝑖𝑛
 𝑘𝐽
∆𝑈 = 162.5184
𝑚𝑖𝑛

𝑘𝐽
𝑛𝑜𝑤, 𝑜𝑢𝑟 𝑔𝑜𝑎𝑙 𝑖𝑠 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑊𝑎𝑡𝑡𝑠:
𝑚𝑖𝑛
𝑘𝐽 1000 𝐽 1 𝑚𝑖𝑛
∆𝑈 = 162.5184 [ ][ ]
𝑚𝑖𝑛 1 𝑘𝐽 60 𝑠
𝑱
∆𝑼 = 𝟐𝟕𝟎𝟖. 𝟔𝟒 𝒐𝒓 𝑾
𝒔

Fifth, determine ΔWf.

∆𝑊𝑓 = 𝑊𝑓2 − 𝑊𝑓1
𝑤ℎ𝑒𝑟𝑒:
𝑊𝑓 = 𝑃Ṿ

𝑙𝑒𝑡 ′ 𝑠 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑊𝑓1 :

𝑊𝑓1 = 𝑃1 Ṿ1

𝑤ℎ𝑒𝑟𝑒:
𝑃1 = 137.90 𝑘𝑃𝑎
Ṿ1 = ?

𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛, 𝑏𝑢𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑣𝑜𝑙𝑢𝑚𝑒:

 𝑚3
ʋ1 = 0.0388
𝑘𝑔𝑚

𝐿𝑒𝑡 ′ 𝑠 𝑟𝑒𝑐𝑎𝑙𝑙 𝑓𝑟𝑜𝑚 𝑀𝑜𝑑𝑢𝑙𝑒 1:

𝑉𝑜𝑙𝑢𝑚𝑒 𝑉
ʋ= =
𝑀𝑎𝑠𝑠 𝑚
𝑠𝑜: 𝑉 = 𝑚 ∙ ʋ
𝐻𝑜𝑤 𝑐𝑎𝑛 𝑤𝑒 𝑟𝑒𝑙𝑎𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑛𝑑 𝑚𝑎𝑠𝑠 𝑡𝑜 𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑛𝑑 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦?

𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡:
𝑉 𝑚
Ṿ= 𝑎𝑛𝑑 ḿ =
𝑡 𝑡

𝑡ℎ𝑢𝑠, 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑛𝑑 𝑚𝑎𝑠𝑠:

𝑉 = Ṿ ∙ 𝑡 𝑎𝑛𝑑 𝑚 = ḿ ∙ 𝑡

𝑏𝑦 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛:
𝑓𝑟𝑜𝑚: 𝑉 = 𝑚 ∙ ʋ
[Ṿ ∙ 𝑡] = [ḿ ∙ 𝑡] ∙ ʋ

𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔: Ṿ = ḿ ∙ ʋ

𝑠𝑜, 𝑤𝑒 𝑐𝑎𝑛 𝑛𝑜𝑤 𝑔𝑒𝑡 Ṿ1 :

Ṿ1 = ḿ ∙ ʋ1
 𝑘𝑔𝑚 𝑚3
Ṿ1 = [4.56 ] [0.0388 ]
𝑚𝑖𝑛 𝑘𝑔𝑚
𝑚3
Ṿ1 = 0.176928
𝑚𝑖𝑛

𝐴𝑛𝑑, 𝑔𝑜𝑖𝑛𝑔 𝑏𝑎𝑐𝑘 𝑡𝑜 𝑊𝑓1 :

𝑊𝑓1 = 𝑃1 Ṿ1
𝑘𝑁 𝑚3 1000 𝑁 1 𝑚𝑖𝑛
𝑊𝑓1 = [137.90 2 ] [0.176928 ][ ][ ]
𝑚 𝑚𝑖𝑛 1 𝑘𝑁 60 𝑠
𝑁∙𝑚
𝑊𝑓1 = 406.63952 𝑜𝑟 𝑊
𝑠

𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑊𝑓2 :
𝑊𝑓2 = 𝑃2 Ṿ2
𝑤ℎ𝑒𝑟𝑒:
𝑃2 = 551.6 𝑘𝑃𝑎
Ṿ2 = ḿ ∙ ʋ2

𝐿𝑒𝑡 ′ 𝑠 𝑓𝑖𝑛𝑑 Ṿ2 :
𝑘𝑔𝑚 𝑚3
Ṿ2 = [4.56 ] [0.193 ]
𝑚𝑖𝑛 𝑘𝑔𝑚
𝑚3
Ṿ2 = 0.88008
𝑚𝑖𝑛
 𝐶𝑜𝑚𝑝𝑢𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑊𝑓2 :
𝑊𝑓2 = 𝑃2 Ṿ2
𝑘𝑁 𝑚3 1000 𝑁 1 𝑚𝑖𝑛
𝑊𝑓2 = [551.6 2 ] [0.88008 ][ ][ ]
𝑚 𝑚𝑖𝑛 1 𝑘𝑁 60 𝑠
𝑁∙𝑚
𝑊𝑓2 = 8090.8688 𝑜𝑟 𝑊
𝑠

𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 ∆𝑊𝑓 :

∆𝑊𝑓 = 𝑊𝑓2 − 𝑊𝑓1
∆𝑊𝑓 = 8090.8688 𝑊 − 406.63952 𝑊
∆𝑾𝒇 = 𝟕𝟔𝟖𝟒. 𝟐𝟐𝟗𝟐𝟖 𝑾

Lastly, we may now compute for WSF.

𝑊𝑆𝐹 = 𝑄 − ∆𝑃𝐸 − ∆𝐾𝐸 − ∆𝑈 − ∆𝑊𝑓
𝑤ℎ𝑒𝑟𝑒:
𝑄 = 3000 𝑊
∆𝑃𝐸 = 0
∆𝐾𝐸 = 706.99 𝑊
∆𝑈 = 2708.64 𝑊
∆𝑊𝑓 = 7684.22928 𝑊

𝑊𝑆𝐹 = 3000 𝑊 − 0 − 706.99 𝑊 − 2708.64 𝑊 − 7684.22928 𝑊

𝑾𝑺𝑭 = −𝟖𝟎𝟗𝟗. 𝟖𝟔 𝑾 (𝑨𝑵𝑺𝑾𝑬𝑹)
Work is done on the system. (-)
Work is done by the system. (+)