Duty cycle and calculated results工况与主要结论

This installment evaluates a repeated-output requirement: a 1 MW electrical input at the drivetrain DC port for 3 s, repeated ten times at start-to-start intervals of 20 s or 60 s. Initial and ambient temperatures are 25°C; the selected temperature limit is 45°C. The analysis checks when that limit is exceeded, whether extra thermal capacity or cooling resolves it, and how long gas replenishment takes if thrust assistance is used.本期评估一项重复输出要求：电驱直流端口输入 1 MW，每次持续 3 秒，分别按 20 秒和 60 秒的起点间隔重复十次。初始温度和环境温度均为 25°C，温度上限取 45°C。计算回答三个问题：何时超温，增加热容或散热能否满足要求，以及使用喷气辅助后需要多久补气。

Calculated results: at 20 s intervals, the baseline exceeds 45°C on pulse 6; at 60 s, it remains below the limit. Doubling thermal capacity passes ten pulses at 20 s but fails the steady-state requirement. In the separate gas case, refill takes about 348 s at 20 kW after gas conditioning.计算结果：基准方案按 20 秒周期运行，第 6 次超过 45°C；改为 60 秒周期则不超限。热容翻倍后可以通过十次、20 秒周期的输出，但稳态温度仍超限。独立储气算例中，完成气体调温后，20 kW 补气还需约 348 秒。

The parameters are illustrative design inputs, not Roadster specifications. Tesla’s product page does not provide the thermal parameters, compressor map or repeated-output test protocol needed to calculate these limits for the actual car. Part 5 addressed force allocation; this installment addresses the conditions required before the next output cycle.这些参数用于设计分析，不是 Roadster 规格。Tesla 产品页面未提供计算实车边界所需的热参数、压缩机性能曲线与重复输出测试协议。第五期讨论输出时的受力分配，本期讨论下一次输出前必须恢复的条件。

Heat generated by each power pulse每次输出产生多少热量

At an assumed constant DC-to-wheel efficiency of 90%, each 3 s pulse supplies 3 MJ of electrical energy and delivers 2.7 MJ of wheel work. The remaining 300 kJ is drivetrain loss. Deposit that loss in one effective thermal node for the calculations below. The imposed power profile does not define vehicle speed, available traction or a 0–100 km/h time.假定直流端口到轮端的效率恒为 90%，每次 3 秒脉冲输入电能 3 MJ，输出轮端功 2.7 MJ，电驱损耗为 300 kJ。后续计算将这 300 kJ 全部计入一个等效热节点。该功率曲线没有规定车速与抓地条件，不能据此计算 0–100 km/h 用时。

The 300 kJ loss is part of the 3 MJ input; adding it again would double-count it. Battery internal losses, fan and pump consumption, and compressor input require separate entries. Battery resistive heating can be approximated by integrating I²R, with a suitable electrical model; pack temperature, state of charge and electrochemical effects must also be considered. NREL’s BLAST model provides a reference for treating electrical losses, thermal state and cooling loads separately.300 kJ 损耗已经包含在 3 MJ 电输入中，不能重复计入电耗。电池内损、风扇和水泵功耗、压缩机输入需要单独计算。电池电阻热可在合适的电模型中用 I²R 的时间积分近似，还需考虑温度、荷电状态和电化学过程。NREL 的 BLAST 模型可用于参考电损耗、热状态与冷却负荷的分别建模方式。

At 20 s intervals, pulse 6 exceeds the limit20 秒周期：第六次超温

The baseline thermal model uses heat capacity C = 60 kJ/K and conductance G = 0.5 kW/K to a fixed 25°C ambient. With an initial temperature of 25°C and a selected limit of 45°C, the available temperature rise is 20 K. The thermal time constant is τ = C/G = 120 s. Here θ denotes temperature rise above ambient; 45°C is a calculation limit, not a component rating.基准热模型取热容 C = 60 kJ/K，到固定环境的等效热导 G = 0.5 kW/K。环境和初始温度均为 25°C，上限取 45°C，可用温升为 20 K。热时间常数 τ = C/G = 120 秒。θ 表示相对环境的温升；45°C 仅为计算上限，不代表部件额定温度。

The energy balance assumes a uniform node temperature and constant C and G. NREL’s thermal-design material describes this lumped approach and its distinction from models that resolve internal temperature gradients. Applying it to this drivetrain gives an aggregate temperature. Winding, semiconductor-junction and cell hotspots need separate temperature states and limits.能量平衡假定节点温度均匀，C 与 G 保持不变。NREL 热设计资料说明了这种集总方法及其与内部温度梯度模型的区别。本例计算的是电驱系统的等效温度；绕组、半导体结温和电芯热点需要分别建模、分别限温。

The 3 s pulse is short relative to the 120 s time constant. Approximate its heat input as an instantaneous Q/C = 5 K rise at the pulse end, followed by exponential cooling. Δ is the start-to-start interval and, for equal pulse durations, also the end-to-end interval. The peak rise after pulse n is θn+. The next peak equals the residual rise from the previous pulse plus the new 5 K increment:脉冲持续 3 秒，远小于 120 秒的热时间常数。将热输入近似为脉冲结束时 Q/C = 5 K 的瞬时温升，随后按指数规律冷却。Δ 为起点到起点的间隔；等时长脉冲的终点间隔也为 Δ。θn+ 表示第 n 次脉冲后的峰值温升。下一次峰值由上一轮残余温升与本轮新增的 5 K 相加得到：

Repeated thermal pulses: at 20 second intervals the sixth peak exceeds 45 degrees Celsius; at 60 seconds the peaks remain below the selected limit.重复热脉冲：20秒间隔的第六次峰温超过45摄氏度；60秒间隔的峰温保持在选定上限以内。
Figure 1 · Calculated peak temperatures for identical heat pulses at 20 s and 60 s intervals, starting at 25°C. Curves above 45°C continue the calculation without derating. A controller enforcing this limit must reduce or reject the requested output.图 1 · 初始温度为 25°C，按 20 秒与 60 秒间隔施加相同热输入，计算各次峰温。超过 45°C 的部分按未降额输出继续计算；控制器若要满足这一限制，必须降低或拒绝相应输出。

At 20 s intervals, peak 6 reaches 45.6°C and exceeds the limit; peak 10 would reach 51.4°C without derating. At 60 s, peak 10 is 37.6°C. A separate finite-pulse calculation applies 100 kW of heat for 3 s. Its peak temperature rises are 1.24% lower, and the first exceedance at 20 s remains pulse 6. The impulse approximation therefore preserves this comparison, but does not resolve hotspots within the pulse.20 秒周期下，第 6 次峰温达到 45.6°C，首次超限；若继续维持原输出，第 10 次将达到 51.4°C。60 秒周期下，第 10 次为 37.6°C。另以持续 3 秒、发热功率 100 kW 的有限脉冲校核，峰值温升比瞬时近似低 1.24%，20 秒周期仍在第 6 次首次超限。瞬时近似不改变本次对照结论，但不能解析脉冲内的局部热点。

Choosing thermal capacity or heat rejection热容与散热的选型依据

A ten-pulse test and a sustained-duty requirement have different acceptance criteria. Sustained thermal operation requires the periodic steady peak to stay within the limit. Solving the recurrence gives the peak rise and the minimum start-to-start interval below. A finite interval requires 0 < Q/C < θlim, where θlim is the allowed rise above ambient. The baseline requires about 35 s by the impulse model, or about 34 s with the finite pulse.十次输出与持续运行的验收条件不同。持续运行要求周期稳态峰温也不超过上限。对递推式求稳态，可得到峰值温升与最小起点间隔。有限间隔解要求 0 < Q/C < θlim，其中 θlim 为相对环境的允许温升。基准方案按瞬时脉冲计算需要约 35 秒间隔，按有限脉冲计算约为 34 秒。

Doubling heat capacity passes ten pulses but the steady peak is 56.3 degrees Celsius; doubling conductance lowers the steady peak to 42.6 degrees Celsius.热容翻倍可以通过十次脉冲，但稳态峰温为56.3摄氏度；热导翻倍则将稳态峰温降至42.6摄氏度。
Figure 2 · At a fixed 20 s interval, double C or G separately. Increasing C passes ten pulses but exceeds 45°C at steady state; increasing G keeps the steady peak below 45°C. These comparisons do not hold hardware mass, auxiliary power or packaging equal.图 2 · 固定 20 秒周期，分别将 C 或 G 翻倍。增加 C 可通过十次输出，但稳态仍超过 45°C；增加 G 可使稳态峰温低于 45°C。两种改动的硬件质量、辅助功耗与安装空间未作等价比较。

At a 20 s interval, doubling C lowers peak 10 to 42.7°C, but the steady peak is still 56.3°C. Doubling G instead lowers the steady peak to 42.6°C. The average heat balance explains the difference: 300 kJ every 20 s is 15 kW, while baseline cooling at the temperature limit is only Gθlim = 10 kW. Extra heat capacity stores the accumulating energy for longer; no finite C eliminates this 5 kW deficit at the limit. At periodic steady state, the mean rise is Q/(GΔ), independent of C.固定 20 秒周期，只将 C 翻倍，第 10 次峰温降至 42.7°C，但稳态峰温仍达 56.3°C。只将 G 翻倍，稳态峰温降至 42.6°C。差别来自平均热收支：每 20 秒产生 300 kJ 热量，平均发热为 15 kW；基准方案在温度上限处的散热能力只有 Gθlim = 10 kW。增加热容可以延长蓄热时间，但任何有限热容都不能消除上限处这 5 kW 的缺口。周期稳态的平均温升为 Q/(GΔ)，与 C 无关。

These thermal limits assume a fixed heat sink and sufficient electrical energy. Recalculate them for the intended ambient and airflow: a hotter ambient reduces temperature margin, lower airflow can reduce G, and shared cooling loads can raise coolant temperature or consume available heat-rejection capacity. A thermal steady state does not restore battery charge; state of charge continues to fall unless energy is replenished.这些热边界以固定热沉和充足供能为前提。设计校核还需覆盖目标环境与气流：环境升温会减小温度余量，气流下降可能减小 G，共用冷却负荷会抬高冷却介质温度或占用散热能力。热稳态不代表电量恢复；若不补充电能，荷电状态仍会下降。

Gas recovery requires mass and temperature control储气恢复：质量与温度控制

For the gas-recovery calculation, specify 5 kN·s of useful impulse and assume a mean effective exhaust speed of 500 m/s. Gas consumption is then J/ve = 10 kg. Set the conditioned low and high tank states to 10 and 30 MPa absolute at 300 K. With ideal-air R = 287 J/(kg·K), the required internal volume is 43.05 L, containing 5 kg at the low state and 15 kg at the high state. This is a separate case from the 100 L tank in Part 4.储气恢复算例规定有效冲量为 5 kN·s，平均有效排气速度假定为 500 m/s，对应耗气量 J/ve = 10 kg。调温后的低、高状态分别取 10 MPa 和 30 MPa 绝对压力，温度均为 300 K。按理想空气 R = 287 J/(kg·K) 计算，所需内部容积为 43.05 L，低、高状态分别存有 5 kg 和 15 kg 气体。本例参数独立于第四期的 100 L 储罐。

The 43.05 L result sizes the inventory difference under the ideal-gas assumption. Vessel selection at 10–30 MPa requires real-gas properties. The gas temperature of 300 K is independent of the drivetrain model’s 25°C ambient.43.05 L 是理想气体假设下满足质量差的容积计算值。10–30 MPa 储罐选型需要改用真实气体物性。气体温度 300 K 与电驱模型的环境温度 25°C 是独立设定。

Pressure alone does not establish recovery. With ideal adiabatic discharge from 15 kg to 5 kg, the remaining gas would be about 193 K at 6.44 MPa. Warming the same 5 kg to 300 K raises pressure to 10 MPa without replenishing any gas. The low state used here is therefore defined after conditioning to 300 K. The assumed mean exhaust speed does not determine the discharge history or verify delivery of the specified impulse.仅检查压力无法判断是否恢复。若按理想绝热模型从 15 kg 放气至 5 kg，残余气体约为 193 K、6.44 MPa。将这 5 kg 气体加热至 300 K，压力会回升到 10 MPa，气量却没有增加。因此，本例的低压状态定义在气体调温至 300 K 之后。假定的平均排气速度不能确定放气过程，也不能验证规定冲量是否已经实现。

The refill estimate starts from that conditioned state. Intake air is at 100 kPa absolute and 300 K. Assume reversible isothermal compression to the instantaneous tank pressure and hold the tank at 300 K throughout filling. Reversible work per added kilogram is RTg ln(p/pa); substitute dm = V dp/(RTg) and integrate from low to high pressure. Convert reversible work to electrical input using ηc = Wrev/Efill = 0.65, an assumed overall ratio rather than a measured compressor efficiency.补气从这个调温后的状态开始。进气条件取绝对压力 100 kPa、温度 300 K，假定气体可逆等温压缩至瞬时罐压，储罐在整个补气过程中保持 300 K。每千克新增气体的可逆功为 RTg ln(p/pa)；代入 dm = V dp/(RTg)，从低压积分到高压。再按 ηc = Wrev/Efill = 0.65 换算电输入。这里的 ηc 是假定的总比值，不是压缩机实测效率。

A 43.05 litre tank is refilled from a conditioned 5 kg state to 15 kg; reversible work is 4.52 MJ and estimated electrical input is 6.96 MJ.43.05升储罐从调温后的5千克补至15千克，可逆功为4.52兆焦，估算电输入为6.96兆焦。
Figure 3 · Tank inventory at 300 K and the isothermal refill estimate. Maintaining 300 K requires heat exchange throughout filling. Post-discharge conditioning time is additional and remains unknown.图 3 · 300 K 下的储气质量与等温补气估算。补气期间维持 300 K 需要持续换热；放气后的调温时间尚未确定，需另行计入。

The calculation gives 4.52 MJ of reversible work and 6.96 MJ of electrical input, or 1.93 kWh. At a constant 20 kW input, refill takes about 348 s after conditioning. To achieve that time, the compressor must supply the required flow over the full pressure range, and cooling must maintain the assumed temperatures. Compressor selection therefore needs a pressure–flow map and a cooling specification, not just a power rating. The DOE compressed-air sourcebook covers these relationships between cooling, pressure, flow and storage.计算得到可逆功 4.52 MJ，电输入 6.96 MJ，即 1.93 kWh。按恒定 20 kW 电输入，调温完成后的补气时间约为 348 秒。达到这一时间，要求压缩机在整个压力范围内提供所需流量，冷却系统维持假定温度。因此，选型必须核对压力—流量曲线和冷却要求，不能只看额定功率。DOE 压缩空气系统手册讨论了冷却、压力、流量与储气之间的关系。

Recovery time sets the repeat interval恢复时间决定重复周期

For the reference schedule, a 3 s burst is followed by gas conditioning and then the specified isothermal refill. The conditioning time tcond is unknown. Drivetrain cooling runs in parallel, provided gas conditioning and compression do not reduce the electrical or cooling capacity assigned to the drivetrain model. Under these assumptions, the longer requirement sets the start-to-start interval:参考时序规定：完成 3 秒脉冲后，先调节气体温度，再按上述等温过程补气。调温时间 tcond 尚未确定。电驱冷却并行进行，但前提是调温和压缩不占用电驱模型所需的供电与冷却能力。在这些条件下，起点间隔由较长的一项要求决定：

Recovery comparison: thermal cadence is about 35 seconds, while 20 kilowatt refill plus a three second burst needs about 351 seconds before adding gas conditioning.恢复对照：热周期约35秒，20千瓦补气加3秒脉冲约需351秒，尚未计入气体调温。
Figure 4 · Start-to-start interval with independent recovery resources and zero conditioning time. At 20 kW, 3 s of output plus about 348 s of refill gives about 351 s. This is an optimistic schedule; actual conditioning, cooling and flow limits require further evaluation.图 4 · 恢复资源独立、调温时间取零时的起点间隔。20 kW 下，3 秒输出加约 348 秒补气，共约 351 秒。这是乐观参考时序，实际调温、冷却与流量限制仍需评估。

With tcond set to zero, the required interval is about 351 s, or 5.85 minutes. Gas refill sets this reference interval; additional drivetrain cooling alone provides no reduction. Actual conditioning time must be added. If post-discharge conditioning overlaps with filling, or the subsystems share radiators, current limits or fan capacity, replace the independent timing calculation with a coupled mass-and-energy model. Temperature-dependent compressor derating must also be included.将 tcond 暂取为零，起点间隔约为 351 秒，即 5.85 分钟。该参考工况由补气限制，单独增强电驱散热不能缩短间隔；实际调温时间还需加上。若放气后的调温与补气重叠，或各子系统共用散热器、受同一电流上限和风扇能力约束，应采用质量—能量耦合模型重新计算，并计入压缩机随温度降额的影响。

The combined electrical input is 3 + 6.96 ≈ 9.96 MJ per cycle, excluding battery internal losses and auxiliaries. Over 351 s, that averages about 28.4 kW. The 300 kJ drivetrain loss is already included. Allocate heat to the relevant components: compressor input is not all battery heating, and compressor cooling duty is not determined solely by the electrical inefficiency fraction. Compression and tank-temperature control also require heat rejection. Without energy replenishment, state of charge falls each cycle.组合算例每周期电输入为 3 + 6.96 ≈ 9.96 MJ，不含电池内损与辅助功耗；按 351 秒计算，平均约为 28.4 kW。电驱的 300 kJ 损耗已计入。热负荷应按部件分配：压缩机电输入不能全部算作电池发热，压缩系统的散热需求也不能只按电效率损失计算，压缩过程和罐温控制同样需要排热。不补充电能时，荷电状态会逐周期下降。

Upgrade priorities and integration costs改进优先级与集成代价

For ten pulses at 20 s, doubling thermal capacity meets the selected temperature limit. For sustained thermal operation at 20 s, it does not; doubling conductance meets the limit in this model, while reducing per-pulse losses is another route to evaluate. For the gas-assisted reference cycle, prioritize gas conditioning and refill capability. Increasing the motor’s peak rating alone does not guarantee a better heat balance or a shorter refill time. The table separates each change’s benefit from its integration cost.若要求是按 20 秒周期完成十次输出，热容翻倍可满足本例温度限制。若要求在 20 秒周期下持续满足温度限制，热容翻倍不满足要求，热导翻倍则可满足；降低单次损耗也是需要评估的方案。对喷气辅助参考循环，应优先处理调温和补气能力。只提高电机峰值额定功率，不能保证改善热收支或缩短补气时间。各项改动的收益与集成代价如下。

Change改动Useful effect能解决的问题Cost or remaining limit代价与剩余约束
Increase thermal capacity增加热容Useful effect能解决的问题Reduces the temperature increment of one pulse and extends a finite burst sequence.降低单次温升，延长有限次连续输出。Cost or remaining limit代价与剩余约束Adds thermal storage, not heat rejection. Assess added mass, gradients and the steady heat balance.增加的是蓄热能力，不是散热能力；仍需评估质量、温度梯度与稳态热平衡。
Increase heat rejection增强散热Useful effect能解决的问题Lowers steady temperature at the specified airflow and ambient.在给定气流与环境下，降低稳态温度。Cost or remaining limit代价与剩余约束Needs radiator, coolant and fan capacity. Recheck auxiliary draw, drag and shared-loop loads.需要散热器、冷却液与风扇能力；重新核算辅助功耗、阻力及共用回路负荷。
Increase gas storage增加储气Useful effect能解决的问题Supports more pulses before refill at the same usable mass per pulse.在每次耗气相同的条件下，增加补气前可用的脉冲次数。Cost or remaining limit代价与剩余约束Does not reduce the gas needed per pulse. Adds vessel mass and refill inventory; installed mass changes the vehicle duty.不减少每次耗气；增加储罐质量和待补气量，装车增重也会改变整车任务。
Increase refill capability提高补气能力Useful effect能解决的问题Can shorten refill if pressure-dependent flow, conditioning and cooling all support it.在流量、调温与冷却均支持时，可以缩短补气。Cost or remaining limit代价与剩余约束Electrical and thermal loads rise. An external supply transfers these requirements to infrastructure.电负荷与热负荷增加；外接气源则将相应要求转移到基础设施。

Derating is an available control measure when hardware limits cannot meet the requested duty. Tesla’s Model 3 Track Mode manual describes reduced drive power for more consistent multi-lap performance and continued cooling between sessions. It also states that the endurance setting does not guarantee protection against overheating. This documents a production-vehicle tradeoff, not Roadster calibration.当硬件无法满足规定工况时，控制策略可以降低输出功率。Tesla 的 Model 3 Track Mode 手册说明，可降低驱动功率以提高多圈表现的一致性，并在两段驾驶之间继续冷却；耐久设置仍不保证避免过热。这是量产车中的功率取舍，不能据此推断 Roadster 标定。

Vehicle tests and acceptance criteria实车测试与验收指标

Acceptance testing must cover the complete preparation, acceleration, braking and recovery cycle. Wheel work changes kinetic energy and overcomes road resistance. Braking returns some energy through regeneration and dissipates the rest in brakes and other losses. Battery charge acceptance and temperature constrain regeneration. Include the return-to-start segment when calculating the initial state for the next run.验收试验应覆盖准备、加速、制动与恢复的完整循环。轮端功用于改变动能并克服道路阻力；制动时，一部分能量通过再生返回电池，其余进入制动器及其他损耗。再生能力受电池可接受功率和温度限制。计算下一次运行的初态时，应计入回到起始状态的全过程。

Test boundary测试边界Record同步记录Decision判断依据
Repeat the same duty重复同一任务Record同步记录Initial charge, component temperatures, ambient and airflow; actual wheel work, impulse and complete start-to-start time.初始电量、部件温度、环境与气流；实际轮端功、冲量及完整起点间隔。Decision判断依据Set output tolerances and allowed initial-state ranges before testing. Report every attempt, including curtailed runs.测试前规定输出容差与允许初态范围，报告全部尝试，包括被限功率的次数。
Check recovery and hotspots检查恢复与热点Record同步记录Winding, inverter and battery temperatures; gas mass estimate, pressure and temperature; refill energy and time.绕组、逆变器和电池温度；气体质量估计、压力与温度；补气电耗和时间。Decision判断依据Validate the model against repeated cycles and the worst relevant conditions, rather than fitting only the first pulse.用连续循环及相关最不利工况验证模型，不能只拟合第一次脉冲。
Check durability and degradation检查耐久与退化Record同步记录Pressure-cycle count, leakage, valve response, insulation and cooling performance before and after a specified test programme.在规定试验计划前后记录压力循环数、泄漏、阀门响应、绝缘与冷却性能。Decision判断依据Ten successful pulses establish only that test. Lifetime and production variation need separate qualification.通过十次脉冲只能证明该项测试；寿命与制造差异需要独立验证。

The model parameters and results include checks of energy and mass balances, pulse recurrence, independent thermal integration, steady-state limits and refill-work integration. Validation is limited to these calculations. No Roadster measurements, coupled vehicle/thermal/gas simulation, hardware tests or lifetime qualification were performed. Sources checked on 25 October 2026.模型参数与计算结果包含能量和质量平衡、脉冲递推、独立热积分、稳态边界与补气功积分的校验。验证范围限于这些计算，尚未进行 Roadster 实测、整车—热—气路耦合仿真、实物试验或寿命验证。资料核对日期：2026 年 10 月 25 日。