This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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here are the solutions to question 1. Q1. a) Determine the surface temperature of the chip, in ^. Given: • Chip dimensions: 5 mm on a side, 1 mm thickness. • Electrical power input (P_in): 0.225 W. • Coolant temperature (T_f): 20^ C. • Heat transfer coefficient (h): 150 W/m^2K. • Heat transfer formula: Q = -hA(T_s - T_f), where T_s is the surface temperature. • At steady state, the electrical power input equals the heat transferred out (Q = P_in). Step 1: Calculate the surface area (A) of the chip exposed to the coolant. The problem states "the top surface of the chip is exposed to a coolant". The chip measures 5 mm on a side. A = (5 mm) × (5 mm) = (0.005 m) × (0.005 m) = 0.000025 m^2 Step 2: Set the heat transfer rate equal to the electrical power input. At steady state, the heat generated by the electrical power input is transferred to the coolant. Q = P_in = 0.225 W Step 3: Substitute the known values into the heat transfer formula and solve for T_s. Q = -hA(T_s - T_f) 0.225 W = -(150 W/m^2K)(0.000025 m^2)(T_s - 20^ C) 0.225 = -0.00375 (T_s - 20) Divide both sides by -0.00375: (0.225)/(-0.00375) = T_s - 20 -60 = T_s - 20 T_s = -60 + 20 T_s = -40^ C Q1. b) The rate of heat transfer between a certain electric motor and its surrounds varies with time as Q = -0.2(1 - e^-0.05t). Where t is in seconds and Q is in kW. The shaft of the motor rotates at a constant speed of 955 rev/min, and applies a constant torque of 18 Nm to an external load. The motor draws a constant electric power input equal to 20 kW. For the motor, plot Q and W, each in kW and the change in energy E in kJ, as function of time from t=0 to t=120 s. To "plot" these quantities, we will define their functions of time and calculate their values at the start (t=0 s) and end (t=120 s) of the given time interval. Step 1: Define the heat transfer rate Q(t). The heat transfer rate from the motor to its surroundings is given directly: Q(t) = -0.2(1 - e^-0.05t) kW At t=0 s: Q(0) = -0.2(1 - e^0) = -0.2(1 - 1) = 0 kW At t=120 s: Q(120) = -0.2(1 - e^-0.05 × 120) = -0.2(1 - e^-6) Q(120) ≈ -0.2(1 - 0.00247875) ≈ -0.1995 kW Step 2: Define the work done by the motor, W(t), in kW. The work done by the motor refers to its power output. First, convert the rotational speed to radians per second: = 955 revmin × 2 rad1 rev × 1 min60 s ≈ 100.0 rad/s The power output (W) is the product of torque () and angular speed (): W = = (18 Nm)(100.0 rad/s) = 1800 W = 1.8 kW The work done by the motor (power output) is constant: W(t) = 1.8 kW Step 3: Define the change in energy E(t) in kJ. The First Law of Thermodynamics for the motor (as a system) states that the rate of change of its internal energy is the net power input minus the net power output: (dE)/(dt) = P_in - P_out - Q_loss Where: • P_in = 20 kW (electrical power input). • P_out = W = 1.8 kW (power output to the load). • Q_loss is the rate of heat transfer from the motor to the surroundings, which is -Q(t). Q_loss = -Q(t) = -(-0.2(1 - e^-0.05t)) = 0.2(1 - e^-0.05t) kW Now, substitute these into the energy balance equation: (dE)/(dt) = 20 kW - 1.8 kW - 0.2(1 - e^-0.05t) kW (dE)/(dt) = 18.2 - 0.2 + 0.2e^-0.05t (dE)/(dt) = (18.0 + 0.2e^-0.05t) kW To find the total change in energy E(t), integrate (dE)/(dt) from t=0 to t: E(t) = _0^t (18.0 + 0.2e^-0.05t') dt' E(t) = [18.0t' + (0.2)/(-0.05)e^-0.05t']_0^t E(t) = [18.0t' - 4e^-0.05t']_0^t E(t) = (18.0t - 4e^-0.05t) - (18.0(0) - 4e^0) E(t) = 18.0t - 4e^-0.05t - (-4) E(t) = (18.0t - 4e^-0.05t + 4) kJ At t=0 s: E(0) = 18.0(0) - 4e^0 + 4 = 0 - 4 + 4 = 0 kJ At t=120 s: E(120) = 18.0(120) - 4e^-0.05 × 120 + 4 = 2160 - 4e^-6 + 4 E(120) ≈ 2160 - 4(0.00247875) + 4 ≈ 2160 - 0.009915 + 4 ≈ 2163.99 kJ To plot these functions: • Q(t): This is an exponential function starting at 0 kW and approaching -0.2 kW as t increases. • W(t): This is a constant function at 1.8 kW for all t. • E(t): This is a linear function with an exponential correction, starting at 0 kJ and increasing over time. You can use these formulas to calculate values at various time points between 0 s and 120 s and then plot them on a graph. Send me the next one 📸