Homework 6 Solution - Introduction to Thermodynamics | ME 205, Study notes of Thermodynamics

Download Homework 6 Solution - Introduction to Thermodynamics | ME 205 and more Study notes Thermodynamics in PDF only on Docsity! PROBLEM 4.257 _ As shown in Fig, P4.25, air enters a pipe at 25°C, 100 kPa with a volumetric flow rate of 23 m'7h. On the outer pipe surface is an electrical resistor covered with insulation. With rollage of 120 V, the resistor draws a current of 4 amps. assuming the ideal gas model with ¢, = 1.005 kJ/kg > K for air and ignoring kinetig and potential energy effects, determine (a) the mass low rate of the air, in kg/h, and (b) the temperature of the alr al exit, in °C. Schematic § 6vEw DATA: CGR MODEL: Insulation — 1. The conte! vetume Shown we the sketch is at steady state. air 2 9, For the conto volwrek » SIE $= 1004 me? reat trance fan and daneie (Av), = 23 meh aud petenhor energy effect Fig. P4.25 Caw lee i here . AWALTSIS! Spree hut & cy we @O, eater | (oe) Loe 94 eg a Gq Ry (3 eee (248) . h pear SF (ey Radweing EG 4:20 vary Lacked asrump bene, 5 = fils Ney + AUR TAP yeas wlan yn ho p= Spe C1 -Te] on Sey am (vo te: Caveent Wey = > wx en) AWettfamp || kW a (novelty (t¢')| ae Watt _ — 0-45 KW Gtlectns Apults, and sulvons fre Ty 27, — Wy TL i nye — (-0.4%ew) aeons ) 4 KT 2 te ooo I (26.etke (ees Sc) an || dkw Hn 302k ( 84°C) &) Copyright © John Wiley & Sons, Inc. 2015 Do not copy or distribute without explicit permission from John Wiley & Sons, Inc PROBLEM ‘tal 4.31 Steam enters a nozzle operating at steady state at 20 bar, 280°C, with a velocity of 80 m/s. The exit pressure and temperature are 7 bar and 180°C, respectively. The mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energy, determine (a) the exit velocity, in m/s. (b) the inlet and exit flow areas, in cm’. KNOWN: Steam at a specified mass flow rate flows through a nozzle with known pressure, temperature, and velocity at the inlet and known pressure and temperature at the exit, FIND: Determine the exit velocity and the inlet and exit flow areas. SCHEMATIC AND GIVEN DATA: m= 1.5 kgs Pp, = 20 bar T, = 280°C V,=80 m/s ENGINEERING MODEL: 1. The control volume shown on the accompanying figure is at steady state. 2. Heat transfer and the change in potential energy from inlet to exit can be neglected. 3. W,, =0, ANALYSIS: State 1 is fixed by 7) and p; and is in the superheated vapor region. From Table A-4, hy = 2976.4 kJ/kg and v, = 0.1200 m’/kg. State 2 is fixed by 72 and p» and also is in the superheated vapor region. From Table A-4, hy = 2799.1 kJ/kg and vz = 0.2847 m/kg. (a) The steady-state, one-inlet, one-exit energy balance gives 0= Q,, —W,, + [Un — ha) +2 (VP ~ V2) + gi 22) Neglecting heat transfer and potential energy change, recognizing no work is associated with a nozzle, and dividing by the mass flow rate, the energy balance simplifies to = (hy Ie) £8 2 Vy 2 cg Copyright © John wiley sat incl23i5 2 (Ve Va) Solving for the ByKo¥ €4GFUSaistribute without explicit permission from John Wiley & Sons, Inc PROBLEM 4.47 ENOWN: Sfeady- state data are prov: deal for as tam teebrne, EUIND! Deternuna te Power de velped by the turbine. Scrematic & G/vew DATA ENGR. MOpEL: . toh = Wye OKI nom 1, As phews vn the tke teh 5 ee TA is Control Volume encloses the turbine. = Bloona/eg 2. The cee! vale Gg ot Steady stote, We Sevan, ea 3. 7 VBinss® e 4sm/5 hype zaooealey z= ASFA OF AOS CFP AWALTSIS: Mass rate balance at Steady stl, Mae ty, Sneey nats balance af steady shite, eet wm (CA ot Ca 5-20] Oz @ey- x, (,-2 > Whey = Ge + ¥ Eler bay MES ww +7 G2 Jaw, Sev a f Cunha VOB 5 etn] = aa =F 4 Gv. oped “= *t Vo Chyoha) = Gree 29001 3, = B0KZ, 4nI =: ik Y ee (Leen carn eal pan)? 70% AvT je 003 &3 Ton “4 Z sage OFEy=lete Caiteeting anottt J v a (eu + yoo - 0-56 + 0.03) We ® a = Bast A = ve (oes (79835 2) ee | I) 535 | a = 13R1 «WW SO 4. Enthalpy change ic clearly the dominant effect, Copyright © John Wiley & Sons, Inc. 2015 Do not copy or distribute without explicit permission from John Wiley & Sons, Inc 4.62. Air, modeled as an ideal gas, is compressed at steady state from 1 bar, 300 K, to 5 bar, 500 K, with 150 kW of power input. Heat transfer occurs at a rate of 20 kW from the air to cooling water circulating in a water jacket enclosing the compressor. Neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in kg/s. KNOWN: Air flows through a compressor. FEND: The mass flow rate of the air, in kg/s. SCHEMATIC AND GIVEN DATA: Air py = 1 bar T, = 300K ir | Ooy =20 KW (out) | We, = 150 kW (in) P,= 5 bar T, = 500K ENGINEERING MODEL: 1. The control volume shown with the schematic is at steady state. 2. For the control volume, Ake = 0 and Ape = 0. 3. Model air as an ideal gas. ANALYSIS: The energy rate balance 0= Oy —Wey +t —n) + VE AVP) + gaz) simplifies to 0= On —Woy + (iy — he) Solving for mass flow rate gives Qoy = Wey fy —hy Copyright © John Wiley & Sons, Inc. 2045 Do not copy or distribute without explicit permission from John Wiley & Sons, Inc Since heat transfer is from the control volume, Ory =—20kW. Since power is to the control volume, We, =—150 kW. Using inlet and exit temperatures, specific enthalpy values are obtained from Table A-22: A, = 300.10 kJ/kg and Ay = 503.02 kJ/kg. Substituting values and solving for mass flow rate give (3) iia {-20 kW) —(-150 kW) |\Us (s03.02-300.19) | SY kg rt = 0.64 kg/s Copyright © John Wiley & Sons, inc, 2045 Do not copy or distribute without explicif permission from John Wiley & Sons, Inc