§ME 231 Homework 6

Sep 25, 2025

§4.17

As shown in Fig. P4.17, air enters a pipe at 25°C, 100 kPa with a volumetric flow rate of 23 m3/h. On the outer pipe surface is an electrical resistor covered with insulation. With a voltage of 120 V, the resistor draws a current of 4 amps. Assuming the ideal gas model with cp = 1.005 kJ/kg · K for air and ignoring kinetic and potential energy effects, determine (a) the mass flow rate of the air, in kg/h, and (b) the temperature of the air at the exit, in °C.
Figure 4.17 from Fundamentals of Engineering Thermodynamics
T1=25°C=298.2K T_1=25\degree C=298.2K
p1=100kPa p_1=100kPa
(AV)1=23m3/h (AV)_1=23m^3/h
From table A-1 for air:
M=28.97kg/kmol=0.02897kg/mol M=28.97kg/kmol=0.02897kg/mol
Rˉ=8.314Jmol∗K \bar{R}=8.314\frac{J}{mol*K}
R=RˉM=8.314Jmol∗K0.02897kg/mol=286.99Jkg∗K R=\frac{\bar{R}}{M}=\frac{8.314\frac{J}{mol*K}}{0.02897kg/mol}=286.99\frac{J}{kg*K}
pv=RT pv=RT
p1v1=RT1 p_1v_1=RT_1
v1=RT1p1=286.99Jkg∗K∗298.2K100kPa=0.8558m3/kg v_1=\frac{RT_1}{p_1}=\frac{286.99\frac{J}{kg*K} * 298.2K}{100kPa}=0.8558m^3/kg
m˙=(AV)1v1=23m3/h0.8558m3/kg=26.88kg/h \dot{m}=\frac{(AV)_1}{v_1}=\frac{23m^3/h}{0.8558m^3/kg}=\boxed{26.88kg/h}
V=120V V=120V
I=4A I=4A
Q˙=P=IV=4A∗120V=480W \dot{Q} = P = IV = 4A * 120V = 480W
W˙=0 \dot{W}=0
From table A-20 for air where T=298.2K≈300KT=298.2K\approx300K:
cp=1.005kJkg∗K c_p=1.005\frac{kJ}{kg*K}
Q˙=m˙cp(T2−T1) \dot{Q}=\dot{m}c_p(T_2-T_1)
Q˙m˙cp=T2−T1 \frac{\dot{Q}}{\dot{m}c_p}=T_2-T_1
T2=Q˙m˙cp+T1=480W26.88kg/h∗1.005kJkg∗K+298.2K=362.2K T_2=\frac{\dot{Q}}{\dot{m}c_p}+T_1=\frac{480W}{26.88kg/h * 1.005\frac{kJ}{kg*K}} + 298.2K=\boxed{362.2K}

§4.27

Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and exits with a velocity of 48.4 m/s. For negligible potential energy effects, determine the exit temperature, in K.
Q˙=W˙=0 \dot{Q}=\dot{W}=0
T1=270K T_1=270K
v1=180m/s v_1=180m/s
v2=48.4m/s v_2=48.4m/s
Q˙−W˙=m˙[h2+v222+gz2]−m˙[h1+v122+gz1] \dot{Q}-\dot{W}=\dot{m}\left[h_2+\frac{v_2^2}{2}+gz_2\right]-\dot{m}\left[h_1+\frac{v_1^2}{2}+gz_1\right]
0=m˙[h2+v222]−m˙[h1+v122] 0=\dot{m}\left[h_2+\frac{v_2^2}{2}\right]-\dot{m}\left[h_1+\frac{v_1^2}{2}\right]
0=m˙[h2−h1+v222−v122] 0=\dot{m}\left[h_2-h_1+\frac{v_2^2}{2}-\frac{v_1^2}{2}\right]
0=m˙[h2−h1+v22−v122] 0=\dot{m}\left[h_2-h_1+\frac{v_2^2-v_1^2}{2}\right]
0=h2−h1+v22−v122 0=h_2-h_1+\frac{v_2^2-v_1^2}{2}
h=cpT h=c_pT
0=cpT2−cpT1+v22−v122 0=c_pT_2-c_pT_1+\frac{v_2^2-v_1^2}{2}
cpT1+v12−v222=cpT2 c_pT_1+\frac{v_1^2-v_2^2}{2}=c_pT_2
T2=T1+v12−v222cp T_2=T_1+\frac{v_1^2-v_2^2}{2c_p}
From table A-20 for air where T=270K≈300KT=270K\approx300K:
cp=1.005kJkg∗K c_p=1.005\frac{kJ}{kg*K}
T2=270K+(180m/s)2−(48.4m/s)22∗1.005kJkg∗K=285.0K T_2=270K+\frac{(180m/s)^2-(48.4m/s)^2}{2 * 1.005\frac{kJ}{kg*K}}=\boxed{285.0K}

§4.30

A well-insulated turbine operating at steady state develops 28.75 MW of power for a steam flow rate of 50 kg/s. The steam enters at 25 bar with a velocity of 61 m/s and exits as saturated vapor at 0.06 bar with a velocity of 130 m/s. Neglecting potential energy effects, determine the inlet temperature, in °C.
W˙=28.75MW \dot{W}=28.75MW
m˙=50kg/s \dot{m}=50kg/s
p1=25bar p_1=25bar
v1=61m/s v_1=61m/s
p2=0.06bar p_2=0.06bar
v2=130m/s v_2=130m/s
Q˙=0 \dot{Q}=0
From table A-3 where P=0.06barP=0.06bar:
h2=hg,2=2567.4kJ/kg h_2=h_{g,2}=2567.4kJ/kg
I am going to try to find h1h_1 and reverse search the temperature this exists at.
Q˙−W˙=m˙[h2+v222+gz2]−m˙[h1+v122+gz1] \dot{Q}-\dot{W}=\dot{m}\left[h_2+\frac{v_2^2}{2}+gz_2\right]-\dot{m}\left[h_1+\frac{v_1^2}{2}+gz_1\right]
−W˙=m˙[h2+v222]−m˙[h1+v122] -\dot{W}=\dot{m}\left[h_2+\frac{v_2^2}{2}\right]-\dot{m}\left[h_1+\frac{v_1^2}{2}\right]
W˙=m˙[h1+v122]−m˙[h2+v222] \dot{W}=\dot{m}\left[h_1+\frac{v_1^2}{2}\right]-\dot{m}\left[h_2+\frac{v_2^2}{2}\right]
W˙=m˙[h1+v122−h2−v222] \dot{W}=\dot{m}\left[h_1+\frac{v_1^2}{2}-h_2-\frac{v_2^2}{2}\right]
W˙=m˙[h1−h2+v12−v222] \dot{W}=\dot{m}\left[h_1-h_2+\frac{v_1^2-v_2^2}{2}\right]
W˙m˙=h1−h2+v12−v222 \frac{\dot{W}}{\dot{m}}=h_1-h_2+\frac{v_1^2-v_2^2}{2}
h1=W˙m˙−v12−v222+h2 h_1=\frac{\dot{W}}{\dot{m}}-\frac{v_1^2-v_2^2}{2}+h_2
h1=28.75MW50kg/s−(61m/s)2−(130m/s)22+2567.4kJ/kg=3149kJ/kg h_1=\frac{28.75MW}{50kg/s}-\frac{(61m/s)^2-(130m/s)^2}{2}+2567.4kJ/kg=3149kJ/kg
There are no superheated water tables for p=p1=25.0p=p_1=25.0. Hence, I will use the values from p=20.0barp=20.0bar and p=30.0barp=30.0bar tables and interpolate the values.Excerpt from table A-4 where p=20.0barp=20.0bar (my target hh is in between two rows):
T (K)T\ (K) h (kJ/kg)h\ (kJ/kg)
320320 3069.53069.5
360360 3159.33159.3
I will need to interpolate the value first from the 20bar20bar table:
T20bar=320K+(360K−320K)3149kJ/kg−3069.5kJ/kg3159.3kJ/kg−3069.5kJ/kg=355K T_\text{20bar}=320K+(360K-320K)\frac{3149kJ/kg-3069.5kJ/kg}{3159.3kJ/kg-3069.5kJ/kg}=355K
Excerpt from table A-4 where p=30.0barp=30.0bar:
T (K)T\ (K) h (kJ/kg)h\ (kJ/kg)
360360 3138.73138.7
400400 3230.93230.9
I will be interpolating the 30.0bar30.0bar table too:
T30bar=360K+(400K−360K)3149kJ/kg−3138.7kJ/kg3230.9kJ/kg−3138.7kJ/kg=365K T_\text{30bar}=360K+(400K-360K)\frac{3149kJ/kg-3138.7kJ/kg}{3230.9kJ/kg-3138.7kJ/kg}=365K
Now time to interpolate between these two tables:
T1=355K+(365K−355K)25bar−20.0bar30.0bar−20.0bar=360K T_1=355K+(365K-355K)\frac{25bar-20.0bar}{30.0bar-20.0bar}=\boxed{360K}

§4.40

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is compressed at steady state to 12 bar, 80°C. The volumetric flow rate of the refrigerant entering is 4 m3/min. The work input to the compressor is 60 kJ per kg of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.
p1=4bar p_1=4bar
T1=20°C T_1=20\degree C
p2=12bar p_2=12bar
T2=80°C T_2=80\degree C
V˙1=4m3/min=0.06667m3/s \dot{V}_1=4m^3/min=0.06667m^3/s
Q=60kJ/kg Q=60kJ/kg
From table A-12 where p=4barp=4bar and T=20°CT=20\degree C:
v1=0.05397m3/kg v_1=0.05397m^3/kg
h1=262.96kJ/kg h_1=262.96kJ/kg
From table A-12 where p=12barp=12bar and T=80°CT=80\degree C (I will need this later):
h2=310.24kJ/kg h_2=310.24kJ/kg
Now I can convert the inflow rate from volumetric to mass flow rate (which is constant with outflow even though the volumetric rate of outflow may be different):
m˙=V˙1v˙1=0.06667m3/s0.05397m3/kg=1.235kg/s \dot{m}=\frac{\dot{V}_1}{\dot{v}_1}=\frac{0.06667m^3/s}{0.05397m^3/kg}=1.235kg/s
Having QQ is great! But Q˙\dot{Q} will be more helpful:
Q˙=Qm˙=60kJ/kg∗1.235kg/s=74.1kJ/s \dot{Q}=Q\dot{m}=60kJ/kg * 1.235kg/s=74.1kJ/s
Now I can compile this into the energy balance equation (I am going to solve for W˙\dot{W} first):
Q˙−W˙=m˙[h2+v222+gz2]−m˙[h1+v122+gz1] \dot{Q}-\dot{W}=\dot{m}\left[h_2+\frac{v_2^2}{2}+gz_2\right]-\dot{m}\left[h_1+\frac{v_1^2}{2}+gz_1\right]
Q˙−W˙=m˙h2−m˙h1 \dot{Q}-\dot{W}=\dot{m}h_2-\dot{m}h_1
Q˙−W˙=m˙(h2−h1) \dot{Q}-\dot{W}=\dot{m}(h_2-h_1)
−W˙=m˙(h2−h1)−Q˙ -\dot{W}=\dot{m}(h_2-h_1)-\dot{Q}
W˙=Q˙−m˙(h2−h1) \dot{W}=\dot{Q}-\dot{m}(h_2-h_1)
W˙=74.1kJ/s−1.235kg/s∗(310.24kJ/kg−262.96kJ/kg)=15.7kW \dot{W}=74.1kJ/s-1.235kg/s*(310.24kJ/kg-262.96kJ/kg)=\boxed{15.7kW}