Find. (a) number of welds $n$; (b) total welding time; (c) total electricity cost.
Approach. Size the weld count from the parent-metal shear/tensile capacity of one weld nugget; get the electrical energy needed per weld from the casting-style heat balance (heat the nugget from room temperature to melting, including latent heat, then divide by the 65% efficiency); get the weld time from $H=I^2Rt$ with $R$ from the assumed resistivity and geometry; get the cost from the total electrical energy at the given electricity rate.
(a) Number of welds. Each weld's load capacity is set by the parent-metal yield strength acting over the electrode contact area:
$$A_{weld}=\frac{\pi d^2}{4}=\frac{\pi(6\ \text{mm})^2}{4}=28.3\ \text{mm}^2\qquad F_{weld}=\sigma_y A_{weld}=90\ \text{MPa}\times 28.3\ \text{mm}^2=2545\ \text{N}$$
$$n=\left\lceil\frac{F_{total}}{F_{weld}}\right\rceil=\left\lceil\frac{45{,}000}{2545}\right\rceil=\lceil 17.68\rceil$$
$$\boxed{n=18\ \text{welds}}$$
(b1) Electrical energy needed per weld. Model the nugget as a cylinder of the electrode's diameter through the combined thickness of both sheets ($l=2\times2\ \text{mm}=4\ \text{mm}$), heated from room temperature to the aluminum melting point (no superheat needed beyond melting), using the casting energy-balance formula from the formula sheet with the liquid-superheat term dropped ($T_{pour}=T_{melt}$):
$$V=A_{weld}\,l=28.3\ \text{mm}^2\times4\ \text{mm}=113\ \text{mm}^3\qquad m=\rho_{solid}V=2700\ \text{kg/m}^3\times1.13\times10^{-7}\ \text{m}^3=3.05\times10^{-4}\ \text{kg}$$
$$E_{useful}=m\big[C_s(T_{melt}-T_0)+\Delta H_f\big]=3.05\times10^{-4}\big[900(660-25)+396{,}000\big]\ \text{J}$$
$$E_{useful}\approx 295\ \text{J}\qquad E_{elec}=\frac{E_{useful}}{0.65}\approx 455\ \text{J per weld}$$
(b2) Weld resistance and time. Treating the given 4.0 $\mu\Omega\cdot$cm as a bulk resistivity acting over the same nugget path ($l=0.4$ cm, $A=0.283\ \text{cm}^2$):
$$R=\frac{\rho\,l}{A}=\frac{(4.0\times10^{-6}\ \Omega\cdot\text{cm})(0.4\ \text{cm})}{0.283\ \text{cm}^2}\approx 5.66\times10^{-6}\ \Omega$$
From $H=I^2Rt$, solving for the time to deliver $E_{elec}$ at the fixed weld current:
$$t=\frac{E_{elec}}{I^2R}=\frac{455\ \text{J}}{(5000\ \text{A})^2(5.66\times10^{-6}\ \Omega)}\approx 3.21\ \text{s per weld}$$
$$\boxed{t_{total}=n\times t=18\times3.21\ \text{s}\approx 57.8\ \text{s}\ (\approx 0.96\ \text{min})}$$
(c) Electricity cost. Total electrical energy across all 18 welds:
$$E_{total}=n\,E_{elec}=18\times455\ \text{J}\approx 8180\ \text{J}=\frac{8180}{3.6\times10^6}\ \text{kW-hr}\approx 2.27\times10^{-3}\ \text{kW-hr}$$
$$\text{Cost}=2.27\times10^{-3}\ \text{kW-hr}\times\$0.09/\text{kW-hr}$$
$$\boxed{\text{Cost}\approx \$0.0002\ (\approx 0.02\ \text{cents})}$$
The cost is small because resistance spot welding is inherently a brief, localized, low total-energy process — the total electrical draw for the whole assembly is under 10 kJ.
Part
Result
(a) Number of welds
18
(b) Total welding time
≈57.8 s
(c) Electricity cost
≈$0.0002 (≈0.02¢)
Check: two engineering assumptions were needed because the question leaves them implicit. (1) The nugget volume heated per weld is taken as a cylinder of electrode diameter spanning BOTH 2 mm sheets (4 mm total path) — a common simplifying idealization; a smaller true fused-nugget volume would reduce the useful energy, weld time and (very slightly) the cost. (2) The "resistivity... 4.0 $\mu\Omega$" is interpreted as 4.0 $\mu\Omega\cdot$cm (a physically reasonable value for hot aluminum, roughly 1.5–2× the room-temperature resistivity of pure Al, ≈2.65 $\mu\Omega\cdot$cm) combined with the nugget geometry via $R=\rho l/A$; the electricity-cost answer in part (c) is unaffected by this choice (it depends only on the required energy, not on $R$), but the weld-time answer in part (b) scales directly with whichever resistance value is assumed.
