Answer
(a) 896 Btu/s
Work Step by Step
We know that [ dE/dt = dQ/dt - dW/dt + d($m_{in}$)/dt*( $h_{in}$ +$Vin^{2}$/2 +gZin) -d($m_{out}$)/dt*( $h_{out}$ +$Vout^{2}$/2 +gZout) ]
Here, we are asked to neglect kinetic and potential effects, and also steady state is stated, so dQ/dt=0, also d($m_{in}$)/dt=d($m_{out}$)/dt.
also velocity and height terms will be 0.
so final equation turns out as,
0=dQ/dt -dW/dt +d($m_{out}$)/dt*($h_{in}$ -$h_{out}$ )
or
0=dQ/dt -dW/dt +d($m_{in}$)/dt*($h_{in}$ -$h_{out}$ )
here heat flow rate is given as 40 btu/s and putting the other values of mass flow rate as 5 lb/sec and specific heats in and out we get
dW/dt = 896 btu/s
