Given: V = 4 m³ (rigid), P1 = 3 MPa, T1 = 400 K, T2 = 320 K, R = 0.287 kPa·m³/kg·K
Initial mass: m1 = P1*V/(R*T1) = 3e6*4/(287*400) = 104.5 kg
Final pressure: P2 = P1 * T2/T1 = 3 * 320/400 = 2.4 MPa
Final mass: m2 = P2*V/(R*T2) = 2.4e6*4/(287*320) = 104.5 kg - same as initial (rigid, no venting)
If vented to maintain rigid tank but at 320 K: mass lost = 0 (no vent) or if allowed to escape to maintain pressure? Assume rigid closed tank - mass constant
Answer: P2 = 2.4 MPa, mass lost = 0 kg

Given: m = 6 kg, k = 0.42 m, N = 1500 rpm, t = 10 s
Moment of inertia: I = m * k² = 6 * 0.42² = 1.058 kg·m²
Angular velocity: w = 2n *N/60 = 2 *1500/60 = 157.08 rad/s
Angular acceleration: = w / t = 157.08 / 10 = 15.71 rad/s²
Torque: T = I * = 1.058 * 15.71 = 16.6 N·m
Answer: T = 16.6 N·m

Given: L = 150 m, D = 0.08 m, Q = 0.04 m³/s, f = 0.02, = 1000 kg/m³, g = 9.81 m/s²
Velocity: V = Q / ( *D²/4) = 0.04 / ( *0.08²/4) = 7.96 m/s
Head loss: h_f = f * (L/D) * V²/(2*g) = 0.02*(150/0.08)*(7.96²/(2*9.81)) = 61.2 m
Pressure drop: P = *g*h_f = 1000*9.81*61.2 = 600 kPa
Answer: h_f = 61.2 m, P = 600 kPa

Given: L = 5 m, P = 25 kN, E = 210 GPa, I = 1.5e-5 m
Tip deflection: deflection = P*L³/(3*E*I) = 25000*5³/(3*210e9*1.5e-5) = 3.3 m
Tip slope: slope = P*L²/(2*E*I) = 25000*25/(2*210e9*1.5e-5) = 0.099 rad
Answer: deflection =3.3 m, slope = 0.099 rad

Given: r = 0.6 m, t = 0.012 m, P = 6 MPa, _allow = 280 MPa
Hoop stress: _h = P * r / (2 * t) = 6e6 * 0.6 / (2*0.012) = 150e6 Pa
Factor of safety: FS = _allow / _h = 280e6 / 150e6 = 1.87
Answer: _h = 150 MPa, FS = 1.87