Q1a. A silicon chip measuring 5mm on a side and 1mm in thickness is embedded in ceramic substrate. At steady state, the chip has an electrical power input of 0.22SW, the top surface of the chip is exposed to a coolant whose temperature is 20°C. The rate of energy transfer by

The problem states "the top surface of the chip is exposed to a coolant". The chip measures 5 mm on a side.

$$A = (5 \text{ mm}) \times (5 \text{ mm}) = (0.005 \text{ m}) \times (0.005 \text{ m}) = 0.000025 \text{ m}^2$$

At steady state, the heat generated by the electrical power input is transferred to the coolant.

$$Q = P_{in} = 0.225 \text{ W}$$

$$Q = -hA(T_s - T_f)$$ $$0.225 \text{ W} = -(150 \text{ W/m}^2\text{K})(0.000025 \text{ m}^2)(T_s - 20^\circ \text{C})$$ $$0.225 = -0.00375 (T_s - 20)$$

Divide both sides by $-0.00375$:

$$\frac{0.225}{-0.00375} = T_s - 20$$ $$-60 = T_s - 20$$ $$T_s = -60 + 20$$ $$T_s = \boxed{-40^\circ \text{C}}$$

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 $955 \text{ rev/min}$, and applies a constant torque of $18 \text{ Nm}$ to an external load. The motor draws a constant electric power input equal to $20 \text{ kW}$. For the motor, plot $Q$ and $W$, each in kW and the change in energy $\Delta E$ in kJ, as function of time from $t=0$ to $t=120 \text{ s}$.

To "plot" these quantities, we will define their functions of time and calculate their values at the start ($t=0 \text{ s}$) and end ($t=120 \text{ s}$) of the given time interval.

The heat transfer rate from the motor to its surroundings is given directly:

$$Q(t) = -0.2(1 - e^{-0.05t}) \text{ kW}$$

At $t=0 \text{ s}$:

$$Q(0) = -0.2(1 - e^0) = -0.2(1 - 1) = 0 \text{ kW}$$

At $t=120 \text{ s}$:

$$Q(120) = -0.2(1 - e^{-0.05 \times 120}) = -0.2(1 - e^{-6})$$ $$Q(120) \approx -0.2(1 - 0.00247875) \approx -0.1995 \text{ kW}$$

The work done by the motor refers to its power output.
First, convert the rotational speed to radians per second:

$$\omega = 955 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \approx 100.0 \text{ rad/s}$$

The power output ($W$) is the product of torque ($\tau$) and angular speed ($\omega$):

$$W = \tau \omega = (18 \text{ Nm})(100.0 \text{ rad/s}) = 1800 \text{ W} = 1.8 \text{ kW}$$

The work done by the motor (power output) is constant:

$$W(t) = \boxed{1.8 \text{ kW}}$$

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:

$$\frac{dE}{dt} = P_{in} - P_{out} - \dot{Q}_{loss}$$

Where:

• $P_{in} = 20 \text{ kW}$ (electrical power input).
• $P_{out} = W = 1.8 \text{ kW}$ (power output to the load).
• $\dot{Q}_{loss}$ is the rate of heat transfer from the motor to the surroundings, which is $-Q(t)$.

$$\dot{Q}_{loss} = -Q(t) = -(-0.2(1 - e^{-0.05t})) = 0.2(1 - e^{-0.05t}) \text{ kW}$$

Now, substitute these into the energy balance equation:

$$\frac{dE}{dt} = 20 \text{ kW} - 1.8 \text{ kW} - 0.2(1 - e^{-0.05t}) \text{ kW}$$ $$\frac{dE}{dt} = 18.2 - 0.2 + 0.2e^{-0.05t}$$ $$\frac{dE}{dt} = (18.0 + 0.2e^{-0.05t}) \text{ kW}$$

To find the total change in energy $\Delta E(t)$, integrate $\frac{dE}{dt}$ from $t=0$ to $t$:

$$\Delta E(t) = \int_0^t (18.0 + 0.2e^{-0.05t'}) dt'$$ $$\Delta E(t) = \left[18.0t' + \frac{0.2}{-0.05}e^{-0.05t'}\right]_0^t$$ $$\Delta E(t) = \left[18.0t' - 4e^{-0.05t'}\right]_0^t$$ $$\Delta E(t) = (18.0t - 4e^{-0.05t}) - (18.0(0) - 4e^0)$$ $$\Delta E(t) = 18.0t - 4e^{-0.05t} - (-4)$$ $$\Delta E(t) = \boxed{(18.0t - 4e^{-0.05t} + 4) \text{ kJ}}$$

At $t=0 \text{ s}$:

$$\Delta E(0) = 18.0(0) - 4e^0 + 4 = 0 - 4 + 4 = 0 \text{ kJ}$$

At $t=120 \text{ s}$:

$$\Delta E(120) = 18.0(120) - 4e^{-0.05 \times 120} + 4 = 2160 - 4e^{-6} + 4$$ $$\Delta E(120) \approx 2160 - 4(0.00247875) + 4 \approx 2160 - 0.009915 + 4 \approx 2163.99 \text{ kJ}$$

To plot these functions:

• $Q(t)$: This is an exponential function starting at $0 \text{ kW}$ and approaching $-0.2 \text{ kW}$ as $t$ increases.
• $W(t)$: This is a constant function at $1.8 \text{ kW}$ for all $t$.
• $\Delta E(t)$: This is a linear function with an exponential correction, starting at $0 \text{ kJ}$ and increasing over time.

You can use these formulas to calculate values at various time points between $0 \text{ s}$ and $120 \text{ s}$ and then plot them on a graph.