Recent content by joejoe121

Is what I did in part C correct or should I say the average enthalpy of the withdrawn vapor just be the saturated vapor value at 100 F? And I'm really confused on finding quality at this point.

Is it correct to say that the initial internal energy can be used to find the final quality?

I found my internal energy for Tank B by finding my values at state 2. u = u_f + x(u_g - u_f) but I need x so I solve quality in state 2 by using saturated volume values at 20 F and solved for specific volume by using V2 = V_a + V_b = 2 meter cubed. I used the mass of the initial tank and since...

for your first formula you posted. What do I equate -(m_i-m_f)h_{avg} to? And did i solve for my average enthalpy correctly?

I'm confused about the $$v_{Lf}$$ and $$v_{Vf}$$

I'm sorry, I don't follow.

in D, i used the formula to find quality using internal energy values and in E I'm wanting to find mass, so I use the volume of the tank divided by the specific volume of the tank at 50 F which I used the same quality formula after I found the quality.

I'm so sorry i forgot to post the questions. I'll edit it.

I've attached all my work and data table I used to answer the questions but there isn't an answer key so I would like a second opinion. a) The initial specific internal energy is.......Btu/lbm b) The initial mass is....lbm c)The average enthalpy of the withdrawn vapor is.....Btu/lbm d)The final...

I've attached all my work and data table I used to answer the questions but there isn't an answer key so I would like a second opinion.

3200 kj/h is the heat removal rate when running. and is input power work? If so, then I believe its 1454.54 kw/h or 0.404 kw/s.

If im working only for 15 minutes at a time or 0.25 hours moving 80 boxes is 320 boxes per hour

is to four times the power while running to compensate for the time not running?

I'm confused about the wording "runs one-fourth of the time". My professor said we multiply the Q_L by four, but I don't get it conceptually.

I understand that when volume change is zero, there is no work done so dU = W. However, how do I know in this process in a piston cylinder, the volume is constant? I also have one question. If the first law of thermodynamics is dU = Q - W. Why is w positive in this case?