During the exam i wont have the chance to use EES, how can i solve the problem manually? Thank you for the video.

thanks for everything, you really save my life

if i had these videos the first time i took the class, i would be graduated by now. great work thank you.

Sir cfa help to get into finance job in NZ

Thank you👍🏽

Correction: Not 120℉ and 140 ℉ linear interpolation The final state is superheated vapor and the internal energy at this state should be obtained by interpolation using 50 psia and 60 psia mini tables (100°F line) in Table A-13E.

My question is, how did you know where in the table to interpolate? I mean, how did you know to interpolate the values at 60 psia between 120 and 140 degrees Fahrenheit?

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. P2=1200 [kPa] T2=700 [C] T3=1200 [C] P3=P2 "1" v2=volume(steam_iapws, P=P2, T=T2) v1=v2 P1=pressure(steam_iapws, x=1, v=v1) "2" v3=volume(steam_iapws, P=P3, T=T3) w_b=P2*(v3-v2) "3" u1=intenergy(steam_iapws, x=1, v=v1) u3=intenergy(steam_iapws, P=P3, T=T3) q=u3-u1+w_b

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. R=0.3 [kJ/kg-K] c_v=0.7 [kJ/kg-K] DELTAT=100 [K] "(1)" c_p=R+c_v DELTAh=c_p*DELTAT "(2)" DELTAu=c_v*DELTAT "(5)" PV=R*DELTAT

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m_ball=5 [kg] T=100 [C] T1=25 [C] T2=100 [C] cp=1.8 [kJ/kg-C] Q=m_ball*cp*(T2-T1) Q=m_steam*h_fg h_f=ENTHALPY(Steam_IAPWS, x=0,T=T) h_g=ENTHALPY(Steam_IAPWS, x=1,T=T) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" Q=W1m_steam*h_g "Using h_g" Q=W2m_steam*cp*(T2-T1) "Using m*c*DeltaT = Q for water" Q=W3m_steam*h_f "Using h_f"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=6*0.355 [kg] T1=18 [C] T2=3 [C] c=4.18 [kJ/kg-K] DELTAT=T2-T1 "Applying energy balance E_in-E_out=dE_system and noting that dU_system=m*c*DELTAT gives" -Q_out=m*c*DELTAT "Some Wrong Solutions with Common Mistakes:" -W1_Qout=m*c*DELTAT/6 "Using one can only" -W2_Qout=m*c*(T1+T2) "Adding temperatures instead of subtracting" -W3_Qout=m*1.0*DELTAT "Using specific heat of air or forgetting specific heat"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=0.18 [kg] T1=17 [C] T2=5 [C] c=3.65 [kJ/kg-K] -Q_out = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) "Some Wrong Solutions with Common Mistakes:" -W1_Qout =c*(T2-T1) "Not using mass" -W2_Qout =m*c*(T2+T1) "adding temperatures

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=0.1 [kg] T1=5 [C] T2=95 [C] c=3.32 [kJ/kg-K] E_in = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Ein = m*c*T2 "Using T2 only" W2_Ein=m*(ENTHALPY(Steam_IAPWS,T=T2,x=1)-ENTHALPY(Steam_IAPWS,T=T2,x=0)) "Using h_fg

there are 15* people total. two are waiting besides the 13 machines being used.

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=3.6 [kg] T1=120 [C] x1=1 P2=500 [kPa] T2=300 [C] u1=INTENERGY(Steam_IAPWS,T=T1,x=x1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) "Noting that Eout=0 and dU_system=m*(u2-u1), applying energy balance E_in-E_out=dE_system gives" E_out=0 E_in=m*(u2-u1) "Some Wrong Solutions with Common Mistakes:" cp_steam=1.8723 [kJ/kg-K] cv_steam=1.4108 [kJ/kg-K] W1_Ein=m*cp_Steam*(T2-T1) "Assuming ideal gas and using cp" W2_Ein=m*cv_steam*(T2-T1) "Assuming ideal gas and using cv" W3_Ein=u2-u1 "Not using mass" h1=ENTHALPY(Steam_IAPWS,T=T1,x=x1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) W4_Ein=m*(h2-h1) "Using enthalpy"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=1.5 [kg] T1=12 [C] T2=95 [C] Q_loss=0 [kJ] W_e=0.8 [kJ/s] c=4.18 [kJ/kg-K] time*W_e-Q_loss = dU_system "time in minutes" "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) time_min=time/60 "Some Wrong Solutions with Common Mistakes:" W1_time*60*W_e-Q_loss = m*c*(T2+T1) "Adding temperatures instead of subtracting" W2_time*60*W_e-Q_loss = c*(T2-T1) "Not using mass"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=75 [kg] time=15*60 [s] W_e=2 [kJ/s] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] time*W_e=m*cv*DELTAT "from energy balance E_in-E_out=dE_system" "Some Wrong Solutions with Common Mistakes:" cp=1.005 [kJ/kg-K] time*W_e=m*cp*W1_DELTAT "Using cp instead of cv" time*W_e/60=m*cv*W2_DELTAT "Using min for time instead of s"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=5 [kg] Q_loss=300 [kJ] time=10*60 [s] W_e=2 [kJ/s] c=4.18 [kJ/kg-K] time*W_e-Q_loss = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*DELTAT "Some Wrong Solutions with Common Mistakes:" time*W_e = m*c*W1_T "Ignoring heat loss" time*W_e+Q_loss = m*c*W2_T "Adding heat loss instead of subtracting" time*W_e-Q_loss = m*1.0*W3_T "Using specific heat of air or not using specific heat"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m_w=0.32 [kg] T1=20 [C] T2=0 [C] c=4.18 [kJ/kg-K] h_melting=334 [kJ/kg-K] DELTAT=T2-T1 "Noting that there is no energy transfer with the surroundings and the latent heat of melting of ice is transferred form the water, and applying energy balance E_in-E_out=dE_system to ice+water gives" dE_ice+dE_w=0 dE_ice=m_ice*h_melting dE_w=m_w*c*DELTAT "Some Wrong Solutions with Common Mistakes:" W1_mice*h_melting*(T1-T2)+m_w*c*DELTAT=0 "Multiplying h_latent by temperature difference" W2_mice=m_w "taking mass of water to be equal to the mass of ice

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=5 [kg] P_1=400 [kPa] T=30 [C] Wout_b=15 [kJ] Win_pw=3 [kJ] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] Q_in+Win_pw-Wout_b=0 "From energy balance E_in-E_out=dE_system, T=constant and thus dE_system=0" "Some Wrong Solutions with Common Mistakes:" W1_Qin=Q_in/cv "Dividing by cv" W2_Qin=Win_pw+Wout_b "Adding both quantities" W3_Qin=Win_pw "Setting it equal to paddle-wheel work" W4_Qin=Wout_b "Setting it equal to boundaru work"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. N=3 [kmol] T_change=10 [C] Ru=8.314 [kJ/kmol-K] Q_diff=N*Ru*T_change "cp-cv=R, and thus Q_diff=m*R*dT=N*Ru*dT" "Some Wrong Solutions with Common Mistakes:" W1_Qdiff=0 "Assuming they are the same" W2_Qdiff=Ru*T_change "Not using mole numbers" W3_Qdiff=Ru*T_change/N "Dividing by N instead of multiplying" W4_Qdiff=N*Rair*T_change; Rair=0.287 "using Ru instead of R"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=75 [kg] P_1=100 [kPa] T_1=15 [C] time=30*60 [s] W_ref=0.250 [kJ/s] W_TV=0.120 [kJ/s] W_heater=1.8 [kJ/s] W_fan=0.05 [kJ/s] R=0.287 [kJ/kg-K] "Applying energy balance E_in-E_out=dE_system gives E_out=E_in since T=constant and dE=0" E_gain=W_ref+W_TV+W_heater+W_fan Q_loss=E_gain*Convert(kJ/s, kJ/h) "Some Wrong Solutions with Common Mistakes:" E_gain1=-W_ref+W_TV+W_heater+W_fan "Subtracting Wrefrig instead of adding" W1_Qloss=E_gain1*3600 "kJ/h" E_gain2=W_ref+W_TV+W_heater-W_fan "Subtracting Wfan instead of adding" W2_Qloss=E_gain2*3600 "kJ/h" E_gain3=-W_ref+W_TV+W_heater-W_fan "Subtracting Wrefrig and Wfan instead of adding" W3_Qloss=E_gain3*3600 "kJ/h" E_gain4=W_ref+W_heater+W_fan "Ignoring the TV" W4_Qloss=E_gain4*3600 "kJ/h"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. m=60 [kg] P1=200 [kPa] T1=25 [C] Qsol=0.8 [kJ/s] time=30*60 [s] Wfan=0.12 [kJ/s] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] time*(Wfan+Qsol)=m*cv*(T2-T1) "from energy balance E_in-E_out=dE_system" "Some Wrong Solutions with Common Mistakes:" cp=1.005 [kJ/kg-K] time*(Wfan+Qsol)=m*cp*(W1_T2-T1) "Using cp instead of cv " time*(-Wfan+Qsol)=m*cv*(W2_T2-T1) "Subtracting Wfan instead of adding" time*Qsol=m*cv*(W3_T2-T1) "Ignoring Wfan" time*(Wfan+Qsol)/60=m*cv*(W4_T2-T1) "Using min for time instead of s"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. V1=0.5 [m^3] V2=0.2 [m^3] P1=600 [kPa] T1=300 [K] R=8.314/28 P1*V1=m*R*T1 W=m*R*T1* ln(V1/V2) "constant temperature process" "Some Wrong Solutions with Common Mistakes:" W1_W=R*T1* ln(V2/V1) "Forgetting m" W2_W=P1*(V1-V2) "Using V*DeltaP" P1*V1/T1=P2*V2/T1 W3_W=(V1-V2)*(P1+P2)/2 "Using P_ave*Delta V" W4_W=P1*V1-P2*V2 "Using W=P1V1-P2V2"

Disclaimer the code is has been uploaded to anti cheat software such as Turnitin. Use it to check your answer. V=3 [m^3] P1=500 [kPa] T1=300 [K] P2=800 [kPa] W=0 "since constant volume" "Some Wrong Solutions with Common Mistakes:" R=0.297 W1_W=V*(P2-P1) "Using W=V*DELTAP" W2_W=V*P1 W3_W=V*P2 W4_W=R*T1*ln(P1/P2)

Thanks

Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Vol_A=0.2 [m^3] P_A[1]=400 [kPa] x_A[1]=0.8 T_B[1]=250 [C] P_B[1]=200 [kPa] Vol_B=0.5 [m^3] T_final=25 [C] E_in-E_out=DELTAE E_in=0 E_out=Q_out DELTAE=m_A*(u_A[2]-u_A[1])+m_B*(u_B[2]-u_B[1]) m_A=Vol_A/v_A[1] m_B=Vol_B/v_B[1] Fluid$='Steam_IAPWS' u_A[1]=INTENERGY(Fluid$,P=P_A[1], x=x_A[1]) v_A[1]=volume(Fluid$,P=P_A[1], x=x_A[1]) T_A[1]=temperature(Fluid$,P=P_A[1], x=x_A[1]) u_B[1]=INTENERGY(Fluid$,P=P_B[1],T=T_B[1]) v_B[1]=volume(Fluid$,P=P_B[1],T=T_B[1]) u_B[2]=u_final u_A[2]=u_final m_final=m_A+m_B Vol_final=Vol_A+Vol_B v_final=Vol_final/m_final u_final=INTENERGY(Fluid$,T=T_final, v=v_final) P_final=pressure(Fluid$,T=T_final, v=v_final)

which edition of the book are you using ? thanks for sharing ❤

Disclaimer the code has been uploaded to anti cheat software such as Turnitin. Change it up if you plan on using it. V = 0.3 [L] T_1_ice = 0 [C] T_1 = 20 [C] T_2 = 5 [C] c_ice = 2.11 [kJ/kg-C] c_water = 4.18 [kJ/kg-C] h_if = 333.7 [kJ/kg] T_1_ColdWater = 0 [C] rho_water = 1 [kg/L] m_water = rho_water*V E_in - E_out = DELTAE_sys E_in = 0 E_out = 0 DELTAE_sys = DELTAU_water+DELTAU_ice "The system is water+ice" DELTAU_water = m_water*C_water*(T_2 - T_1) T_melting = 0 [C] DELTAU_ice = DELTAU_solid_ice+DELTAU_melted_ice DELTAU_solid_ice =m_ice*c_ice*(T_melting-T_1_ice) + m_ice*h_if DELTAU_melted_ice=m_ice*c_water*(T_2 - T_melting) m_ice_grams=m_ice*convert(kg,g) DELTAE_sys = DELTAU_water+DELTAU_ColdWater "The system is water+cold water" DELTAU_water = m_water*c_water*(T_2_ColdWater - T_1) DELTAU_ColdWater = m_ColdWater*c_water*(T_2_ColdWater - T_1_ColdWater) m_ColdWater_grams=m_ColdWater*convert(kg,g)

how can you just assume the boundary work is 0? that was the only part that didn't make sense to me.

I hope you much successful as an educator your videos and explanations are amazing I use them to studying for thermodynamics exams and have gotten good grades thank you

But the plate is turning??

the equation is change in internal energy but why you use change in enthalpy???can you explain

Thank you very much!

what is the name of this book ? ?

Where is the file pls attach it

Sir is it good moving to newzealand for chemical engineers . Will they have good job opportunities or any good destination

Be blessed. YOU ARE HELPING SOMEONE, TRULYY!!

Im thinking about migratung go newzealand to live and work from Trinidad

At 6:15 how did you know to use the internal energy values in the middle column? Aren't the left and right columns also encompassing the the range of 3040?

Frequently Asked Questions. Why is it V_1 - V_2 not V_2 - V_1 ( V_2 < V_1 ), then ( V_1 - V_2 ) will be a positive number. However, in thermodynamics, the sign convention for work is such that work done by the system on the surroundings is considered positive, and work done on the system by the surroundings is considered negative. So, when a gas is compressed (like in your piston-cylinder device), the volume decreases (( V_2 < V_1 )), and the surroundings do work on the system. According to the convention, this work is negative, which is why we use ( P(V_1 - V_2) ) and not ( P(V_2 - V_1) ). The formula ( W = P(V_1 - V_2) ) inherently accounts for this sign convention by giving a negative result for compression work. To put it simply, the formula ( W = P(V_1 - V_2) ) ensures that the work is negative for compression because ( V_1 ) is greater than ( V_2 ), and the work is positive for expansion because ( V_1 ) would be less than ( V_2 ). This sign convention is consistent with the first law of thermodynamics, which in one form can be written as: ΔU=Q-W If work is done on the system (compression), ( W ) is negative, and the internal energy ( U ) increases. Conversely, if the system does work on the surroundings (expansion), ( W ) is positive, and the internal energy decreases. Part B Why do we use 1000 kPa not 500 kPa The reason we use 1000 kPa in the work calculation for part (b) is because the work done is calculated up to the point where the piston hits the stops, not the final state of the system. In this problem, the stops are set at 40% of the initial volume. This means that the piston can only compress the steam up to this volume before it hits the stops and can no longer do work on the system. At this point, the pressure inside the cylinder is still 1000 kPa, which is the initial pressure. After the piston hits the stops, the system continues to cool down until it reaches the final pressure of 500 kPa. However, during this cooling process, the volume of the system remains constant because the piston is not moving. Therefore, no work is done on the system during this stage. So, the work calculated is the work done during the compression process up to the point where the piston hits the stops, which is why we use the initial pressure of 1000 kPa in the calculation. The final pressure of 500 kPa is not used in the work calculation because no work is done on the system after the piston hits the stops.

sry but really bad explaining, no stetps

Why we use mP(v1-v2) instead of mP(v2-v1)

Why we use mP(v1-v2) instead of mP(v2-v1)

Where are you getting the heights for h in the Ptop and Pbot equations?

Plz sir i want to join the Newzland

what text book is this from

Hi, I think I found out what you did wrong on row 2 of the table. The formula for finding the quality of a liquid-vapor region is actually x= (h-h_f)/(h_fg). When using the given value h=1800, you'll get x = (1800 - 589.16)/2144.3 This will get you x ~ 0.565

How can i get a QS job in NZ from India, please suggest.

When calculating the pressure difference, why was 10.3 used and not 15.4