The document discusses the first law of thermodynamics as it applies to control volumes. It defines control volumes and explains that mass can cross their boundaries, so the net mass transfer into and out of the control volume must be tracked. The conservation of mass principle for control volumes is presented, along with equations relating the rate of change of mass within the control volume to the mass flow rates entering and exiting. The first law of thermodynamics is also expressed in rate equation form for a control volume with fixed mass.

1. BITS Pilani
Pilani Campus
Lecture 10 – First Law for Control Mass

2. Problem 1
A piston/cylinder contains 50 kg
of water at 200 kPa with a
volume of 0.1 m3. Stops in the
cylinder restricts the enclosed
y
volume to 0.5 m3, as shown in
fig. The water is now heated to
200oC Find the final pressure
C. pressure,
volume and the work done by
the water.
BITS Pilani, Pilani Campus

3. Example 2
A piston-cylinder arrangement has a linear
spring and the outside atmosphere acting
g g
on the piston. It contains water at 3 MPa
and 400oC with a volume of 0.1 m3. If the
piston is at the bottom the spring exerts a
bottom,
force such that a pressure of 200 kPa
inside is required to balance the forces. The
system now cools until th
t l til the pressure
reaches 1 MPa. Find the heat transfer for
the process.
BITSPilani, Pilani Campus

4. Example 2:Solution
Initial state determined by P and v, it is superheated. Final
state fixed by P (given) and v again. How do we find v?
P = 200 kPa + ksV/A2, where ks is the spring constant, and A
the area of cross-section. This is since the pressure to just
balance when piston is at bottom is given as 200 kPa Since
kPa.
P1 = 3 MPa and V1 = 0.1 m3 given, ks/A2 = 2.8x104 kJ/m3, so
V2 = 0.02857 m3, v2 = 0.02840 m3/kg, saturated mixture with
x2 = 0.14107
State P (kPa) T (C) V (m3) v (m3/kg) Mass M (kg) x u (kJ/kg) U(kJ)
1 3000 400 0.1 0.09936 1.006 ‐ 2950
2 1000 143.6 0.02857 0.02840 1.006 0.14107 1019
1W2 = ∫PdV = 200 (∆V) + 2 8x104(V22 – V12)/2 = -143 kJ
2.8x10 143
1Q2 = ∆U + 1W2 = -1925 -143 = -2068 kJ
BITSPilani, Pilani Campus

5. Problem 3
An insulated piston cylinder device
contains 5 L of saturated liquid
water at a constant pressure of
175kPa. Water is stirred by a peddle
wheel while current of 8 A flows for
45 min through a resistor place in
the water. If 50% of liquid (by mass)
q ( y )
is evaporated during this constant
pressure process and the peddle
work amounts to 300kJ determine
300kJ,
the voltage of the source. Also show
the on P-v diagram.
BITS Pilani, Pilani Campus

6. Fig 4 2 Example of work crossing th b
Fi 4.2 E l f k i the boundary of a system
d f t
because of electric current flow across the system boundary
BITS Pilani, Pilani Campus

7. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
Solution:Assumptions
1 The cylinder is stationary and thus the kinetic and potential
energy changes are zero.
2 The cylinder is well-insulated and thus heat transfer is
well insulated
negligible.
3 The compression or expansion process is quasi-equilibrium.
BITS Pilani, Pilani Campus

8. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
• We take the contents of the cylinder as the system.
• This is a closed system since no mass enters or leaves
leaves.
• The energy balance for this stationary closed system can be
expressed as
m(V2 − V1 )
2 2
1 Q2 −1W2 = U 2 − U 1 + + mg ( Z 2 − Z1 )
2
− 1W2 = U 2 − U1
BITS Pilani, Pilani Campus

9. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
We,,in + Wpw,,in − Wb,,out = ∆U
p
We,in + Wpw,in = m( h2 − h1 )
( VI∆t ) + Wp iin = m( h2 − h1 )
pw,
P1 = 175 kPa ⎫ h1 = h f @175 kPa = 486.97 kJ/kg
⎬v = v = 0.001057 m 3 /
/kg
sat.liquid ⎭ 1 f @175 kPa
P2 = 175 kPa, x2 = 0.5
h2 = h f + x2h fg = 486.97 + (0.5 × 2213.57) = 1593.755 kJ/kg
V1 0.005 m 3
m= = 3
= 4 7304 k
4.7304 kg
v 1 0.001057 m /kg
BITS Pilani, Pilani Campus

10. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
SUBSTITUTING
VI∆t + (300kJ) = (4.7304 kg)(1593.755 − 486.97)kJ/kg
VI∆t = 4935.536 kJ
P 4935.536 kJ ⎛ 1000 VA ⎞
V= ⎜ ⎟
⎜ 1 kJ/s ⎟ = 228.497 V
(8 A)(45 × 60 s) ⎝ ⎠
1 2
v
BITS Pilani, Pilani Campus

11. First law as a rate equation
∆U + ∆KE + ∆PE = δQ − δW
∆U ∆KE ∆PE δQ δW
+ + = −
δt δt δt δt δt
∆U dU
δt
lim 0 =
δt dt
∆ ( KE ) d ( KE )
δt
lim 0 =
δt dt
∆ ( PE ) d ( PE )
δt
lim =
0
δt dt
BITS Pilani, Pilani Campus

12. First law as a rate equation
δQ &
δt
lim 0 =Q
δt
δW &
δt
lim 0 =W
δt
dU d ( KE ) d ( PE ) & &
+ + = Q −W
dt dt dt
dE & &
= Q −W
dt
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13. BITS Pilani
Pilani Campus
FIRST-LAW
FIRST LAW ANALYSIS FOR A
CONTROL VOLUME

14. FIRST-LAW ANALYSIS FOR A
CONTROL VOLUME
BITS Pilani, Pilani Campus

15. Conservation of mass and
Control Volume
Control volumes:
Mass can cross the boundaries, and so we must keep track
of the amount of mass entering and leaving the control
volume.
Conservation of Mass Principle
Net mass transfer to or from a system during a process is
equal to the net change in total mass of the system
during that process.
BITS Pilani, Pilani Campus

16. Conservation of mass and
Control Volume
dmC .V .
= ∑ mi − ∑ me
& &
dt
Total mass inside the control volume
mC .V . = ∫ ρ dV = ∫ (1 / v) dV = m A + mB + mC + .......
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17. Conservation of mass and
Control Volume
The l
Th volume fl
flow rate is
t i
V = ∨ A = ∫ ∨ local dA
&
The mass flow rate becomes
V& ∨ local ∨A
m = ρ avgV = = ∫ (
& & )dA =
v v v
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18. FIRST-LAW ANALYSIS FOR A
CONTROL VOLUME
For a Fixed mass
E2 − E1 = 1 Q2 − W2
1
The instantaneous rate equation is
= Q − W
&
dE C . M .
dt
&
BITS Pilani, Pilani Campus