Question 1

ṁi=ṁe

ṁ1= ṁ2 + ṁ3

V 1. A 1 (30)(0.2)
ṁi = = =10.35 kg/ s
v1 0.5796
ṁi = ṁ2 + ṁ3

= ṁi =2ṁ2

V 2. A 2 V 3. A3
ṁi = = = ṁ3
v2 v3
10.35
ṁ2 = ṁ3 = =5.175 kg /s
2

Question 2

given

k=1.67

pv= ṁRT

T1=5500R

V1=150 ft/s

T2=4000R

P2=40PSI

A2=0.0085 ft2

Solution

Using energy balance equation to find V2

AV
ṁ=
v
P2v2=RT2

RT 2
V2=
P2
Ṝ 1545
R= = =386.25 FT2/S2 .OR
M 4
(386.25)(400)
v2= =26.822 ft 2 /lbm
(40)(144 )
finding the velocity

V 2=√ 2 ( Cp ( T 1−T 2 ) ) + v ₁ ²
Cp=KR/K-1

Cp=1.67(0.496)/(1.67-1)

Cp=1.236 Btu/lboR
2
V 2=√ 2 ( 1.236 ( 550−400 ) ) +150 = √ 2 ( 1.236 ( 778.17 )( 32.2)(550−400) ) +150 ²

V2=3051.83 ft/s

(0.0085)(3051.83)
Mass flowrate =
26.822
=0.967lbn/s
Question 3

Given

P1=25bar

V1=61 m/s

T1=?

Mass flowrate =50kg/s

W=28750kw

X2=1

P2=0.06 bar

V2=130m/s

61²−130² 1
0=w+ m ( 28750 ) + ( 50 ) [ h 1−2567.4 ] + ( )]
2 1000
H1=3148.99 kj/kg

Pressure (P) Temperature (T) (°C) Specific Enthalpy (h) (kJ/kg)

20(bar)
20 320 3148.99
360 3159.3
T20T_{20}T20 = 355.4
30 360 3158.7
400 3230.9
T30T_{30}T30 = 364.46

At h=3148.99 kJ/kg.99kJ/kg:

Pressure (P) Temperature (T) (°C)

(bar)
20 355.4
25 T25
30 364.46

The interpolated temperature at P=25 T1=359.93∘C.

Question 4

Given

T1=60OC

P1=14.2psi

1200 ft3/min

T2=500OF

P2=120 psi

V2?

Solution

w=m(h 1−h 2)
W=mCp(T1-T2)

P1v1=mRT

R=1545/28.013 = 55.15 ft.lbf/lbmoR

144
m=(14.2)( )(1200)/(55.15)(520)
1
=85.56lbm/min

W=(85.56)(0.249) x (520-960)

W=-9374.22

W=-221 hp

P1V1/T1=P2V2/T2

=(142)(1200)/520=(120)(144)(V2)/960

V2=262.15 ft3/min
QUESTION 5

Given Data:

• Volumetric flow rate, Q = 0.75 m³/min = 0.0125 m³/s

• Water temperature, T = 15°C
- From steam tables, the specific volume at this temperature is vf(T) = 0.0010009 m³/kg
• Gravitational acceleration, g = 9.81 m/s²
• Height difference, z2 - z1 = 15 m

We apply the Conservation of Mass principle to a control volume that includes the pump and the
delivery pipe:

dMcv/dt = Σṁin - Σṁout = 0 (steady-state)

ṁin = ṁout

Applying First Law of Thermodynamics

dEcv/dt = Σṁin(h + V²/2 + gz)in - Σṁout(h + V²/2 + gz)out + Q̇ - Ẇ

Since there is negligible heat transfer and no significant change in kinetic energy, this equation
simplifies to:

Ẇpump = ṁg(z2 - z1)

Mass Flow Rate:

ṁ = Q/vf(T) = 0.0125/0.0010009 = 12.49 kg/s

Pump Power Requirement:

Using the simplified energy equation:

Ẇpump = ṁg(z2 - z1)

Substitute the known values:

Ẇpump = 12.49 × 9.81 × 15 = 1,838.5 W

Thus, the required pump power is:

Ẇpump = 1.84 kW
Question 6

energy balance on the condenser:

Q_in = Q_out

where:
• Q_in is the heat transferred to the cooling water (Btu/h)
• Q_out is the heat removed from the refrigerant (Btu/h)

The heat transferred to the cooling water can be calculated using the following equation:

Q_in = ṁ_w * c_w * ΔT_w

where:
• ṁ_w is the mass flow rate of the cooling water (lb/h)
• c_w is the specific heat of the cooling water (Btu/lb°F)
• ΔT_w is the temperature change of the cooling water (°F)

The heat removed from the refrigerant can be calculated using the following equation:

Q_out = ṁ_r * (h_1 - h_2)

where:
• ṁ_r is the mass flow rate of the refrigerant (lb/h)
• h_1 is the specific enthalpy of the refrigerant at the inlet (Btu/lb)
• h_2 is the specific enthalpy of the refrigerant at the outlet (Btu/lb)

Solution
We are given the following:

• ṁ_r = 3100 lb/h

• ΔT_w = 20°F
• c_w = 1 Btu/lb°F

From refrigerant tables for Refrigerant 134a, we find:

• h_1 = 127.2 Btu/lb (at 70 lbf/in², 160°F)
• h_2 = 41.47 Btu/lb (at 60 lbf/in², saturated liquid)

Substituting these values into the energy balance equation:

ṁ_w * 1 Btu/lb°F * 20°F = 3100 lb/h * (127.2 Btu/lb - 41.47 Btu/lb)

Solving for ṁ_w:

ṁ_w = 12,900 lb/h

Volumetric Flow Rate

We can convert the mass flow rate to a volumetric flow rate using the specific volume of the
cooling water:

v̇ _w = ṁ_w * v_w

where:
• v̇ _w is the volumetric flow rate of the cooling water (ft³/h)
• v_w is the specific volume of the cooling water (ft³/lb)

Given:

• v_w = 0.0161 ft³/lb

Substituting the values:

v̇ _w = 12,900 lb/h * 0.0161 ft³/lb = 208.29 ft³/h

convert the volumetric flow rate to gal/min:

v̇ _w = 208.29 ft³/h * (7.48 gal/ft³) * (1 h/60 min) = 260.2 gal/min

Therefore, the volumetric flow rate of the entering cooling water is approximately 260.2
gal/min.

The rate of heat transfer to the cooling water can be calculated using the equation:

Q_in = ṁ_w * c_w * ΔT_w

Substituting the values:

Q_in = 12,900 lb/h * 1 Btu/lb°F * 20°F = 258,000 Btu/h

Therefore, the rate of heat transfer to the cooling water from the condensing refrigerant is
258,000 Btu/h.
Question 7

Determine enthalpy of water at inlet 1:

Liquid water enters at inlet 1 at 7 bar and 42°C.

Using steam tables, the enthalpy, h₁, of sub-cooled liquid water at these conditions is:

h₁ = 176.8 kJ/kg

Determine enthalpy of the two-phase mixture at inlet 2:

h₂ = h_f + x(h_fg)

Where:

h_f is the enthalpy of saturated liquid

h_fg is the enthalpy of vaporization

x is the quality (0.98 in this case)

From the steam tables at 7 bar:

h_f = 721.2 kJ/kg

h_fg = 2087.1 kJ/kg

Thus,

h₂ = 721.2 + 0.98(2087.1) = 2766.96 kJ/kg

Determining enthalpy at outlet 3:

Saturated liquid exits at outlet 3 at 7 bar.

Using steam tables, the enthalpy, h₃, at these conditions (corresponding to h_f at 7 bar) is:

h₃ = 721.2 kJ/kg

m₁h₁ + m₂h₂ = m₃h₃

Where:
m₁ is the mass flow rate at inlet 1

m₂ is the mass flow rate at inlet 2

m₃ is the total mass flow rate out of the heater (m₁ + m₂)

This equation can be further simplified to:

m₁h₁ + m₂h₂ = (m₁ + m₂)h₃

Since m₃ = m₁ + m₂, we can substitute and rearrange:

m₁h₁ + m₂h₂ = (m₁ + m₂)h₃

Expanding and simplifying:

m₁h₁ + m₂h₂ = m₁h₃ + m₂h₃

m₂h₂ - m₂h₃ = m₁h₃ - m₁h₁
m₂(h₂ - h₃) = m₁(h₃ - h₁)

Solve for m₂:

Dividing both sides by (h₂ - h₃):

m₂ = m₁(h₃ - h₁) / (h₂ - h₃)

Substituting these values into the equation:

m₂ = 70 kg/s * (721.2 - 176.8) / (2766.96 - 721.2)

m₂ ≈ 13.84 kg/s

Therefore, the mass flow rate at inlet 2, m₂, is approximately 13.84 kg/s.
Question 8

Power Output of the Turbine:

W_turbine = m (h1 - h2)

= 118.86 kg/s (3426.4 kJ/kg - 2335.96 kJ/kg)

= 127,537.04 kW

Power Input to the Pump:

W_pump = m (h4 - h3)

= 118.86 kg/s * (190.152 kJ/kg - 173.08 kJ/kg)

= 2009.58 kW

Net Power Output:

Net Power Output = W_turbine - W_pump

= 127,537.04 kW - 2009.58 kW

= 125,527.46 kW

Thermal Efficiency:
Thermal Efficiency = (Net Power Output) / (Heat Input to the Boiler)

= (125,527.46 kW) / (3400 kJ/kg * 118.86 kg/s)

= 0.332

= 33.2%

Isentropic Efficiency of the Turbine:

Isentropic Efficiency = (Actual Work Output) / (Isentropic Work Output)

= (h1 - h2) / (h1 - h2s)

= (3426.4 kJ/kg - 2335.96 kJ/kg) / (3426.4 kJ/kg - 2084.61 kJ/kg)

= 0.813

= 81.3%

Isentropic Efficiency of the Pump:

Isentropic Efficiency = (Isentropic Work Input) / (Actual Work Input)

= (h4s - h3) / (h4 - h3)

 = (183.92 kJ/kg - 173.08 kJ/kg) / (190.152 kJ/kg - 173.08 kJ/kg)

= 0.635

= 63.5%

Mass Flow Rate of Cooling Water:

mwater (h6 - h5) = m (h2 - h3)

mwater = (m(h2 - h3)) / (h6 - h5)

= (118.86 kg/s (2335.96 kJ/kg - 173.08 kJ/kg)) / (146.725 kJ/kg - 86.199 kJ/kg)

= 3900 kg/s