22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2018
Question 3 of 8: Air-standard Diesel cycle & fuel rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — first-law cycle analysis, wet-region steam properties, the ideal regenerative Rankine cycle with a closed feedwater heater, the air-standard Diesel cycle, and isentropic-efficiency compressor analysis; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady conduction with internal generation in a composite cylinder, internal-flow convection at constant wall temperature, combined convection–radiation surface balances, and the LMTD method for counterflow heat exchangers. Water/steam properties are IAPWS-consistent; air is treated as an ideal gas with constant specific heats.
Given. Air-standard Diesel cycle: compression ratio $r=14$, cutoff ratio $r_c=2.0$, isentropic compression and expansion, cold-air properties ($k=1.4$, $c_p=1.005$, $c_v=0.718\ \text{kJ/kg}\cdot\text{K}$).
Quantity
Value
Inlet, state 1
$T_1=20\ ^\circ\text{C}=293.15$ K, $P_1=1$ atm
Compression ratio, $r=v_1/v_2$
14
Cutoff ratio, $r_c=v_3/v_2$
2.0
Power output (part b)
3 750 kW
Air/fuel ratio (part b)
16 : 1
Find. (a) specific work output and thermal efficiency; (b) fuel consumption rate for the stated power.
Figure 3 — Air-standard Diesel cycle on the $T$–$s$ plane: isentropic compression 1→2, constant-pressure heat addition 2→3 (up to the cutoff), isentropic expansion 3→4, and constant-volume heat rejection 4→1.
Approach. March around the four processes with the isentropic and constant-property relations to get the corner temperatures, then form $q_\text{in}$, $q_\text{out}$, $w_\text{net}$ and $\eta$; in (b) convert power to an air-flow rate and then a fuel rate through the air/fuel ratio.
Constant-pressure heat addition 2→3. At constant pressure $T\propto v$, so $T_3=r_c T_2=2.0(842.4)=1684.9\ \text{K}$, and
$$q_\text{in}=c_p(T_3-T_2)=1.005(1684.9-842.4)=846.7\ \text{kJ/kg}$$
Expansion 3→4 (isentropic, back to $v_1$). With $v_4/v_3=v_1/v_3=r/r_c$:
$$T_4=T_3\left(\frac{r_c}{r}\right)^{k-1}=1684.9\left(\frac{2}{14}\right)^{0.4}=773.6\ \text{K}$$
Heat rejection 4→1 and net work.
$$q_\text{out}=c_v(T_4-T_1)=0.718(773.6-293.15)=345.0\ \text{kJ/kg}$$
$$w_\text{net}=q_\text{in}-q_\text{out}=846.7-345.0=501.7\ \text{kJ/kg}$$
$w_\text{net}\approx 502$ kJ/kg
Thermal efficiency. Either directly or from the closed-form Diesel relation:
$$\eta=1-\frac{q_\text{out}}{q_\text{in}}=1-\frac{1}{r^{\,k-1}}\,\frac{r_c^{\,k}-1}{k(r_c-1)}=59.3\%$$
$\eta\approx 59.3\%$
Part (b) — fuel consumption rate. The cycle delivers $w_\text{net}=501.7$ kJ per kg of air, so for $\dot W=3750$ kW the air flow is
$$\dot m_\text{air}=\frac{\dot W}{w_\text{net}}=\frac{3750}{501.7}=7.48\ \text{kg/s}$$
and with an air/fuel ratio of 16:
$$\dot m_\text{fuel}=\frac{\dot m_\text{air}}{16}=\frac{7.48}{16}=0.467\ \text{kg/s}\approx 1.68\times10^{3}\ \text{kg/h}$$
$\dot m_\text{fuel}\approx 0.47$ kg/s ($\approx 1680$ kg/h)
Check — power magnitude and heating value.
The printed "3.750 kW" is read as 3 750 kW (period as a thousands separator), consistent with a "large diesel engine"; a literal 3.75 kW would scale the fuel rate down by 1000. The stated heating value "4.65 kJ/kg" is physically implausible for a fuel (diesel is $\approx 42$–$46\times10^{3}$ kJ/kg) and is also inconsistent with the air-standard heat input already fixed by the cutoff ratio ($q_\text{in}=847$ kJ/kg-air $\Rightarrow 13.5\times10^{3}$ kJ/kg-fuel). The self-consistent route above (cycle work $\rightarrow$ air flow $\rightarrow$ fuel flow via the A/F ratio) is therefore used. For reference, an efficiency-based estimate with a realistic $HV=46.5\times10^{3}$ kJ/kg gives $\dot m_\text{fuel}=\dot W/(\eta\,HV)=0.14$ kg/s, but this ignores the A/F constraint and is not self-consistent with the specified cutoff ratio.
