0% found this document useful (0 votes)
324 views4 pages

HW 10

Uploaded by

Selena Gomez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
0% found this document useful (0 votes)
324 views4 pages

HW 10

Uploaded by

Selena Gomez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
You are on page 1/ 4
9.22. Consider a cold air-standard Diesel cycle. At the beginning of compression, p = 14.0 Ibffin. and 7; = 520°R. ‘The mass of air is 0.145 Ib and the compression ratio is 17. The ‘maximum temperature in the cycle is 4000°R. Determine (a) the heat addition, in Bru. (b) the thermal efficiency, (c) the cutoff ratio, KNOWN: An air-standard Diesel cycle has a known compression ratio and a specified state at the beginning of compression. The mass and the maximum cycle temperature are given. FIND: Determine (a) the heat addition, (b) the thermal efficiency, and (c) the cutoff ratio. SCHEMATIC AND GIVEN DATA: m=0.145 Ib eee = 140 Ibffin? 7, = 520°R r=VlVa=17 ENGINEERING MODEL: See Example 9.2. ANALYSIS: Begin by fixing each principal state of the cycle (Table A-22E), State 1: 7) = 520°R + m = 88.62 Btullb, va = 158.58 State 2: For the isentropic compression va = (Val i)-va = (1/17)(158.58) = 9.3282 Thus, interpolating in the table: 72 = 1534.5°R, fy = 378,32 Buullb State 3: 7 = 4000°R — hy = 1088.3 Btu/lb, v3 = 0.4518 State 4: For the isentropic expansion Problem 9.22 (Continued) ~ Page 2 (Vdb¥V3) = (Vila) (Vals) = Wl Vay T/T) * (17-(1534.5/4000) = 6.522 and va = (ViVi = 2.9466 Thus, interpolating in the table: 7, = 2253.7°R, 14 = 421.25 Btw/lb (a) ‘The heat addition is determined from an energy balance on Process 2-3, as follows. Q2s = (us — ua) + Way = mus — un) + mpa(vy ~ v2) = mths ~ hn) Inserting values Qa = (0.145 1b)(1088.3 ~ 378,32) Bru/lb = 102.9 Btu (b) To determine the thermal efficiency, first evaluate the net work of the cycle. Weyce = Qeyste = Ors ~ Our = Qas ~ (us — m1) = 102.9 — (0.145)(421.25 — 88.62) = 54.67 Btu Thus, the thermal efficiency is 1 = WeyalQas = 54,67/102.9 = 0.531 (53.1%) (c) Since p:= ps, the cutoff ratio is re= ValVy = 5/7, = 4000/1534.5 = 2.61 9.26 Consider an air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. ‘The states are numbered as in Fig. 9.5. Determine (a) the cut-off ratio. (b) the heat addition per unit mass, in Btu/Ib. (c) the net work per unit mass, in Buw/lb. (d) the thermal efficiency. State TCR) p(ibiin.”) wu (Bulb) fh (Btu/lby 1 320 142 88.62_ 124.27 2 1502.5 657.8 266.84 369.84 3 3000 657.8 585.04 790.68 4 1527.1 418 271.66 376.36 29.26 KNOWN: An air-standard Diesel cycle operates with property data given at principal states. FIND: Determine (a) the cut-off ratio, (b) the heat addition per unit mass, (c) the net work per unit mass, and (d) the thermal efficiency. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: 1. Air, modeled as an ideal gas, is the system. 2. The compression and expansion processes are adiabatic, 3. Kinetic and potential energy effects are negligible. ANALYSIS: (a) The cut-off ratio can be determined as follows. For the constant pressure process, p2 = ps. Noting that for an ideal gas, p = RTIv Problem 9.26 (Continued) ~ Page 2 Since r. = vy/vy 3000°R__ 9 49 a | hase = (b) The heat addition occurs during process 2-3. Thus, noting that I= mp(vs— v2) mus — ta) = Ors — Was = Qos — mp v3 ~ v2) Thus Qasim = (us ~ Ua) + ps ~ V2) = ha In Inserting values Ge 5, -h, = 750.68 369. sabe 420.84 Btu/tb, (©) The net work per unit mass can be determined from the net heat transfer per unit mass Weyce Oz On Applying the closed system energy balance to process 4-1 tg, = 271 oR 98 ot 183.04 Btuilb Solving for net work per unit mass gives #208451 s018 237.80 Bru/lb (d) Thermal efficiency is W. Qr5 1m tm

You might also like