Question

A closed tank, V = 0.35 ft3, containing 10 lbm of water initially at 77 F is heated to 350 F by a heat pump that is receiving heat from the surroundings at 77 F. Assume that this process is reversible. Find the heat transfer to the water and the work input to the heat pump.

Answer 1

C.V.: Water from state 1 to state 2.

Process: constant volume (reversible isometric)
1: v1 = V/m = 0.35/10 = 0.035 ft3/lbm

⇒ x1 = 2.692 x 10-5

u1 = 45.11 Btu/lbm, s1 = 0.08779 Btu/lbm R Continuity eq. (same mass) and constant volume fixes v2

State 2: T2, v2 = v1 ⇒ x2 = (0.035 - 0.01799) / 3.3279 = 0.00511

u2 = 321.35 + 0.00511×788.45 = 325.38 Btu/lbm

s2 = 0.5033 + 0.00511×1.076 = 0.5088 Btu/lbm R

Energy eq. has zero work, thus provides heat transfer as

1Q2 = m(u2 - u1) = 10(325.38 - 45.11) = 2802.7 Btu

Entropy equation for the total control volume gives for a reversible process:

m(s2 - s1) = QL/T0

⇒ QL = mT0(s2 - s1)

= 10(77 + 459.67)(0.5088 - 0.08779)

= 2259.4 Btu

and the energy equation for the heat pump gives

WHP = 1Q2 - QL

= 2802.7 - 2259.4 = 543.3 B